1. hobbies and interests
2. games
3. board games and puzzles

Oregon State University
PH 211
Fall Term 2014
HW 6-7
Recommended finish date: Monday, November 17
The formats (type, length, scope) of these HW problems have been purposely created to
closely parallel those of a typical exam (indeed, the HW problems have been taken from
past exams). To get an idea of how best to approach various problem types (there are
three basic types), refer to these example HW problems.
1. Evaluate the following statements (T/F/N). As always, explain your reasoning.
a.
b.
A 30-kg child standing on the earth exerts a gravitational force on the earth that is twice as much as that of
a 15-kg child standing next to her.
True. The FG acting on each—therefore by each (this is Newton’s Third Law)—is simply mchildg. So twice the
mchild would give twice the force magnitude.
If one object is accelerating, there must be another object that’s accelerating.
True. If a net force has been exerted on mass 1 (since it’s accelerating) it must have been applied by another
object (mass 2): Fapp.21 But by Newton’s Third Law, a force was applied on mass 2 —equally but oppositely:
Fapp.12 = –Fapp.21. So mass 2 must be accelerating, also.
c. Units of acceleration could be correctly written also as N/kg.
True. The units of force, F, must be equivalent to m·a, so a is equivalent to F/m, which are N/kg in the SI system.
d. A 6-kg block on earth would be a 1-kg block on the moon.
False. The mass of an object is not affected by which planet it’s near. Mass in intrinsic to the amount of matter
contained in an object. If that hasn’t changed, the mass hasn’t changed.
e. A body that has zero acceleration may have forces acting on it.
True. Zero acceleration indicates only that the (vector) sum of all forces acting on the object is zero. That sum
may be (and is often) comprised of two or more force vectors.
f. A car is waiting at rest at a stop light on a westbound road. When the light turns green and the car
accelerates forward (to the west), its rear bumper exerts a force on the rest of the vehicle, and that force is
directed to the east.
True. Analyze the bumper and the rest of the car as two connected masses accelerating
together. If the bumper is accelerating to the west, the rest of the car must be pulling
on the bumper in that direction. So, by Newton’s 3rd Law, the bumper is pulling
oppositely (east) on the rest of the car.
<--- a
(car)
(rear
bumper)
g. An object (m = 5.00 kg), initially at rest, experiences a net force of [1.5t2, –2t] N (where t is in seconds)
throughout the interval 0 ≤ t ≤ 4.00 s. Then all force ceases. At t = 7.50 s, the object’s speed is
approximately 5.54 m/s.
False. asdflj
Pf = Pi + DP, but Pi = 0 (object starts at rest), so Pf = DP.
J
4
4
Thus: mvf = J
= ∫ F(t) dt = [∫ 1.5t2 dt , ∫ –2t dt] = [(1/2)t3|0 , –t2|0 ] = [32 , –16] kg·m/s
Therefore: |J| = √[322 + (–16)2] √(1280)
So: |vf| = |J|/m
So: |vf| = |J|/m = √(1280)/5.00 = 7.16 m/s
2
2. For each item, be sure to show your work and/or explain your reasoning.
a. A 150,000 kg jet airliner is moving with constant velocity at an angle of 12.7° above the horizontal, at an
altitude of 10 km and a speed of 200 m/s. What is magnitude of the total force acting on the airplane?
Zero. By Newton’s First Law, if the object’s acceleration is zero (true here), so is the net force acting on it.
b. A 2-kg object has the acceleration vector [–1, 6] m/s2. What additional force must be applied to the
object to put it into equilibrium? Express this as a magnitude and direction (using the +x-axis as 0°).
Right now the net force on the object is SF = ma = 2[–1, 6] = [–2, 12] N
Equilibrium is when SF = 0. To make this true, we must add the opposite of the current net force.
Thus, we must add Fadd = [2, –12] N = √(22 + (–12)2)∠[tan-1(–12/2)] = 12.2 N∠–80.5°
c. A 2-kg object is constrained to moving along a straight line.
The velocity as a function of time is shown here.
(i) What is the net force acting on the object at t = 5s?
Explain your reasoning.
Fnet = SF = ma. a(5) is the slope of the v-t graph at t = 5:
So a(5) = [v(5) – v(1)]/(5 – 1) = – 5 m/s2.
Thus Fnet(5) = 2(–5) = –10 N
(ii) What is the average net force acting on the object between t = 0s and t = 7s? Explain your reasoning.
(Fnet)avg = maavg = m(vf – vi)/Dt = 2(–10 – 0)/7 = –2.9 N
d.
Boxes A and B are sliding to the right across a frictionless table.
The hand, H, is slowing them down. The mass of A is larger than
the mass of B. Rank, from largest to smallest, these horizontal force
magnitudes: FAB, FAH, FBH, FBA, FHA, FHB
First, note: The hand isn’t touching A, so FAH = FHA = 0.
Then note: The hand must exert enough force to accelerate the combined mass,
but B must exert on A only enough force to accelerate the mass of A: FBA < FHB = 0.
Final ranking (using Third Law equivalencies): FBH = FHB > FBA = FAB > FAH = FHA = 0.
3
3. For each item, be sure to show your work and/or explain your reasoning.
a. The figure shows two masses at rest. The string is massless, and the pulleys are
frictionless. The spring scale reads in kg. What is the reading of the scale?
Be sure to fully explain your reasoning.
The reading is 5 kg. A scale reads only the force it is exerting in one direction. If it were to add or subtract the
forces it is exerting at both ends (yes, a FBD will show this), it would read either double or zero, and that would
make it useless.
b. In the situation shown, the horizontal surface is frictionless, and the pulley is massless
and frictionless. Is the tension in the cable less than, greater than, or equal to 88.2 N?
Do a full analysis with a FBD to support your argument.
The hanging mass (m1):
FT.21
The sliding mass (m2):
x
y
FG.P2
The hanging mass (m1): The sliding mass (m2):
SF1.y = m1a1.y
SF2.x = m2a2.x
FG.P1 – FT.21 = m1a1.y
FT.12 = m2a2.x
m1g – FT.21 = m1a1.y
FT.12 = m2a2.x Thus: m1g – FT = m1a
Solve one equation for a:
a = FT/m2
Substitute and solve for FT:
m1g – FT = m1FT/m2
FT = m1g/(1 + m1/m2)
c.
9 kg
FN.2
FT.12
FG.P1
5 kg
But note:
FT.12 = FT.21 (These are magnitudes and so are equal.)
a2.x = a1.y
(Their magnitudes and signs are equal.)
So call the above simply FT and a.
and: FT = m2a
= (9)(9.80)/(1 + 9/5) = 31.5 N (less than 88.2 N)
Rank the tension in the string, from greatest to
least, in the cases shown here. Assume in each
case that the hanging mass is the same, that
friction is negligible, and that the pulley and
rope are both massless. Use FBDs!
The solution for FT here is the same as in 2B, above: FT = m1g/(1 + m1/m2)
Here m1 is the same in all three cases, but not m2. And when m2 is larger the denominator here is smaller.
That is, when m2 is larger, FT is larger. Since m2A > m2B > m2C, this means that FTA > FTB > FTC.
d. An object of mass M is held in place by an applied force F and a pulley system,
as shown in the figure. The pulleys are massless and frictionless, as are the cables.
a. Find the tension in rope section T1. b. Find the tension in rope section T2. c. Find the tension in rope section T3. d. Find the tension in rope section T4. e. Find the tension in rope section T5. f. Find the magnitude of F. And looking at the whole pulley contraption as one object, we have just one upward force
(FT.4) and two downward forces (F and Mg) acting on it. Thus: FT.4 = F + Mg.
FT.1 = F = Mg/2 (See below for reasoning.)
FT.2 = F = Mg/2 (See below for reasoning.)
FT.3 = F = Mg/2 (See below for reasoning.)
FT.4 = 3Mg/2 (See below for reasoning.)
FT.5 = Mg (FT.5 holds M at rest against FG.PM.)
F = Mg/2 (FT.2 + FT.3 = FT.5. So: F + F = Mg)
For any massless cable, the tension at any point in it is the same (which you can prove
by using a FBD to analyze any two parts of it). And if such cables are connected to ideal
(massless, frictionless) pulleys, their tension is redirected without alteration.
T4
T3
T2
T1
T5
M
F
4
3. e.
For the situation shown, where four ideal (massless) pulleys
support two masses hanging at rest, evaluate (T/F/N) the
following statement. As always, you must justify your answer
with a valid mix of words, drawings and calculations.
FTB/FTA = 8
FTB
FTA
m2
m1
False. Each ideal pulley transmits the tension force throughout any continuous
wire. (This is how we know that the tension in the wire connecting the
second pulley to the ceiling is also FTA.) Thus, each pulley experiences twin
upward forces; so the tension each supports downward is twice that.
Hence the labels shown here. Thus: FTB/FTA = 4
FTA
FTB
2FTA
FTA
m1
4FTA
m2
5
4. a. Evaluate the following statements (T/F/N). As always, explain your reasoning.
(i) If an elevator is moving vertically upward, then the weight of a person standing in that elevator is
always greater than her weight when she is at rest.
False. If she is slowing down or moving at constant speed, her weight is either less than or equal to her rest
weight.
(ii) The weight of an astronaut orbiting the earth is greater than his weight when orbiting the moon.
False. Not greater than—equal to. The astronaut is weightless in either case: 0 = 0.
(iii)The static friction force exerted by one surface on another depends only on the friction coefficient (mS)
and the normal force that exist between the two surfaces.
False. The static friction force varies (from 0 to FSmax) in response to the applied force(s) that would other wise be causing the surfaces to slide past each other. Only FSmax is calculated via mSFN.
(iv)A stationary object can exert a kinetic friction force.
(Any answer OK here). As some students have alertly pointed out, there’s no object in the universe that is
known to be stationary. So by that assumption, this statement has no meaning—it’s either false or not enough
information. However, if we designate something such as the earth as “stationary,” then certainly it can exert
a kinetic friction force: Slide a book across the floor and watch it slow down and stop.
(v) A static friction force can cause acceleration.
True. For example, look at the force FS.12 in exercise 9 of Lab 5-I (the FBD exercise set).
(vi)When a block slides at 4 m/s on a surface with friction, there is twice the friction force acting on it as
when it slides on that surface at 2 m/s.
False. No, for low speeds, mK (and therefore FK) is about equal for any non-zero speed (so long as FN has not
changed).
(vii) A friction force between two surfaces acts on each surface equally in magnitude but opposite in
direction.
True. This is just a re-statement of Newton’s Third Law: Any force acting between two objects acts on each
with the same magnitude but opposite direction.
(viii) The coefficient of kinetic friction is a force.
False. mK is a unitless ratio of two force magnitudes: mK = FK.S1/FN.S1
(ix)A normal force is defined as a force exerted vertically upward by a surface.
False. A normal force is defined as a force exerted by a surface in a direction perpendicular to that surface.
That direction is not necessarily vertical.
(x) A man standing on the level floor of an elevator that is moving vertically downward could have a weight
greater than the gravitational force acting on him.
True. The man (and elevator) could be accelerating upward if the elevator were slowing while moving
downward. And an upward acceleration would mean that the normal force upward by the elevator floor
(what the man feels to judge his weight) must be must be greater than the downward gravitational force.
(xi)The gravitational force exerted by the earth on the orbiting space shuttle is greater than the
gravitational force exerted by the earth on the pilot’s seat inside that shuttle.
True. |FG.P1| = m1g, where g is the local value of gravitational free-fall acceleration. And the same local g
value applies to the entire space shuttle and to all of its parts, so a larger mass experiences a larger
gravitational force acting on it.
6
4. b. Helen is pushing a box so that it slides across the floor. The speed of the box is increasing.
Evaluate the following statements (T/F/N). As always, explain your reasoning.
(i) The frictional force is negligible (or effectively zero.)
Not enough information. All we know from the given information is that the box is speeding up, so the
horizontal component of her pushing force exceeds the kinetic friction force opposing the sliding.
That is: FApp.HBcosq > FK, where q is the angle of her push, as measured from the horizontal.
In other words: FApp.HBcosq > FK. But we simply don’t know how much FK is.
(ii) The frictional force is balanced by Helen’s applied force.
False. The box is accelerating, so the friction force is not being balanced; it is being exceeded.
(iii)Helen is pushing with a force greater than the normal force on the box.
Not enough information. As argued in (i), above, FApp.HBcosq > mKFN. And FN = mg – FApp.HBsinq.
Thus, FApp.HBcosq > mK(mg – FApp.HBsinq). But we don’t know the value of either mK, m or q.
(iv)Helen’s pushing force on the box is greater than the frictional force on the box.
True. As argued in (i), above, FApp.HBcosq > FK, and so even if Helen’s entire force is directed horizontally
(q = 0), that would still mean FApp.HB > FK.
(v) The box’s weight is less than Helen’s applied pushing force.
Not enough information. The box’s weight is the vector sum of all contact forces (FApp.HB, FN and FK) being
exerted on it. And we don’t know the magnitudes of any of those forces (nor the direction of FApp.HB).
7
5. a. The block (m = 30 kg) is on a level surface with friction
coefficients ms = 0.750 and mk = 0.500. If FT1 = 100 N,
q1 = 40.0°, and q2 = 25.0°, find FT2 so that:
FBD of m—situation (a)
– the block is still at rest but “just about” to start
sliding to the right.
FT1
– the block slides to the right at constant velocity.
FN
q1
max
FT1
Fmaxs
FT2
F
becomes Fk
in situation (b).
FT2
q2
s
q2
q1
m
coord. axes
y
x
x-analysis—situation (a)
FG
y-analysis—situation (a)
SFx = max
SFy = may
FT2x – FT1x – Fmaxs = max
FN + FT1y + FT2y – FG = may
FT2cosq2 – FT1cosq1 – msFN = 0
FN + FT1sinq1 + FT2sinq2 – mg = 0
x-analysis—situation (b)
y-analysis—situation (b)
SFx = max
SFy = may
FT2x – FT1x – Fk = max
FN + FT1y + FT2y – FG = may
FT2cosq2 – FT1cosq1 – mkFN = 0
FN + FT1sinq1 + FT2sinq2 – mg = 0
These are two equations in two unknowns (FN and FT2)—solve simultaneously:
Solve for FN:
FN = mg – FT1sinq1 – FT2sinq2
Substitute:
FT2cosq2 – FT1cosq1 – ms(mg – FT1sinq1 – FT2sinq2) = 0
Collect terms:
FT2cosq2 + msFT2sinq2 = FT1cosq1 + msmg – msFT1sinq1
Solve:
FT2 = (FT1cosq1 + msmg – msFT1sinq1)/(cosq2 + mssinq2)
For (b):
= [(100)(cos40) + (.75)(30)(9.80) – (.75)(100)(sin40)]/[cos25 + (.75)(sin25)] = 203 N
FT2 = (FT1cosq1 + mkmg – mkFT1sinq1)/(cosq2 + mksinq2)
= [(100)(cos40) + (.50)(30)(9.80) – (.50)(100)(sin40)]/[cos25 + (.50)(sin25)] = 171 N
8
5. b. A 3-kg block is free to slide vertically up or down a
frictionless wall. A 20 N force is applied as shown. Find
the block’s acceleration (both magnitude and direction).
y
60°
20 N
SFx = max SFy = may FN – Fsin60° = 0
Fcos60° – mg = may
FN – Fx = max 2
Fy – FG = may (Fcos60°)/m – g = ay FG
ay = –6.47 m/s (that’s 6.47 m/s downward)
SFx = max FN
2
A 3-kg block is held at rest against a wall by a two identical
applied forces F, as shown. The wall has a coefficient of static
friction of 0.65 with the block. Find the magnitude of F at
which the block would be ready to slip down the wall.
F
c.
x
FN – F = max FN – F = 0
FN = F
F
y
F
x
SFy = may F
F + Fsmax – FG = may
F + msFN – mg = 0
FN
F
F + msF – mg = 0
F + msF – mg = 0
FSmax
FG
F(1+ ms) = mg
F = mg/(1+ ms) = 17.8 N
d.
A box of mass m is initially sliding at a speed of v on a horizontal surface. Wind is applying a constant force
magnitude F at an angle q down from the horizontal, in the opposite direction of the box’s motion. The box
comes to rest after traveling a distance d. Find the coefficient of kinetic friction, mK, between the box and
the surface, expressed in terms of m, v, F, q , d and g.
02 = v2 + 2axd
Thus: ax = –v2/(2d)
SFx = max First, do kinematics to find the x-acceleration value necessary to bring the box to rest in distance d:
–FK – Fx = max 2
SFy = may FN – Fy – FG = may
mKFN – Fcosq = –mv /(2d)
Solve y-equation for FN: FN = Fsinq + mg
Solve for mK:
mK = [Fcosq – mv2/(2d)]/(Fsinq + mg)
FN
Now do a FBD analysis (referring to diagrams here—using conventional x- and y- axes):
Substitute into x-equation:
FN – Fsinq – mg = 0
mK(Fsinq + mg) – Fcosq = –mv2/(2d)
F
q
FK
q
F
FG
9
5. e.
Two identical masses are connected by a massless cord, as shown.
They are being pulled to the right on a level, frictionless surface by
q
horizontal force F. The tension in the cord, the angle q of the cord,
and F all remain constant. Evaluate (T/F/N) the statement below.
As always, you must justify your answers with a valid mix of words, drawings and calculations.
The difference in magnitude between the two normal forces being exerted in this situation is F·tanq.
True.
m1
FN.1
m2
FN.2
FT.21
q
FG.1
F
q
FT.12
SF1.x = m1a1.x
SF1.y = m1a1.y
SF2.x = m2a2.x
FTcosq = ma
FN.1 + FTsinq – mg = 0
F – FTcosq = ma
FT.21.x = m2a1.x FN.1 + FT.21.y – FG.1 = m1a1.y
So:
Or:
FN.1 = mg – FTsinq
F
FG.2
F – FT.12.x = m2a2.x
SF2.y = m2a2.y
FN.2 – FT.12.y – FG.2 = m2a2.y
FN.2 – FTsinq – mg = 0
FN.2 = mg + FTsinq
FTcosq = F – FTcosq
FT = F/(2cosq)
Thus: FN.2 – FN.1 = (mg + FTsinq) – (mg – FTsinq)
= 2FTsinq
= F·tanq
= 2sinq[F/(2cosq)]
10
6. In each case below, the acceleration magnitude on the 5-kg box (M) is 4 m/s2, which is as large as possible without the box sliding. For each case, find the coefficient of static friction between the box and the surface beneath
it.
SFx = max SFy = may msFN = max
FN – mg = 0
Fs
max
= max Therefore:
Thus:
msmg = max
FN – FG = may
SFx = max msFN = m(acos30°)
Fs
max
= max Therefore:
Thus:
FSmax
ms = ax/g = 4/9.80 = 0.408
b. The box is in a ski gondola on its way to the top of the mountain.
(The gondola does not swing during the motion; its attachment
to its cable remains vertical at all times.)
FN
a. The box is on the bed of a truck that is accelerating on level ground.
SFy = may FG
FN
FN – FG = may
FN – mg = m(asin30°)
ms(masin30° + mg) = m(acos30°)
ms = (acos30°)/(asin30° + g)
= (4cos30°)/(4sin30° + 9.80) = 0.294
FSmax
FG
11
7. a. Block 2 is hanging freely (at rest) by a thread from block 1.
Block 1 is not attached to the ceiling, but it is being held at
rest against the ceiling, due to the effects of the force, F.
mS
Find the maximum force magnitude, F, which can be applied
so that block 1 won’t slip.
q = 25.0°;
m1 = 4.06 kg; m2 = 1.37 kg
For m2:
SFx = m2ax
For m1:
0 = 0
FT1.2
FG.2
SFy = m2ay
FT1.2 – FG.2 = m2ay
FT1.2 – m2g = 0
Thus: FT1.2 = m2g = 1.37(9.80) = 13.426 N
So:
FT2.1 = FT1.2 = 13.426 N
Solve for F:
q
SFx = m1ax
Fx – FSmax = m1ax
Fsinq – mSFN = 0
For m2:
q FT1.2
FG.2
FSmax
F
q
FT2.1
FN
FG.1
Also: FN = Fsinq/mS
So: Fcosq – FT2.1 – m1g – Fsinq/mS = 0
Fcosq – Fsinq/mS = FT2.1 + m1g
F(cosq – sinq/mS) = FT2.1 + m1g
F = (FT2.1 + m1g)/(cosq – sinq/mS)
= [13.426 + (4.06)(9.80)]/[cos25° – (sin25°)/0.890] = 123 N
F = 98.0 N;
q
SFx = m2ax
For m1:
FT1.2.x = m2ax
FN
FT1.2sinq = m2ax
FK
SFy = m2ay
FT1.2.y – FG.2 = m2ay
FT2.1
FT1.2cosq – m2g = 0q
F
m1
q
m1 = 4.65 kg; m2 = 1.73 kg
FT1.2 = m2g/cosq = 1.73(9.80)/cos25° = 18.707 N
FT2.1 = FT1.2 = 18.707 N
ax = FT1.2sinq/m2 = 18.707sin25°/1.73 = 4.5699 m/s2
Solve for mK:
m2
SFy = m1ay
Fy – FT2.1 – FG.1 – FN = m1ay
Fcosq – FT2.1 – m1g – FN = 0
Find the coefficient of kinetic friction, mK.
q = 25.0°;
F
mS.ceiling.block1 = 0.890
b. Block 2 is attached to the side of block 1 only by a thread.
The blocks are accelerating together as Block 1 slides along
a horizontal shelf.
m1
q
FG.1
mK
m2
SFx = m1ax
Fx – FK – FT2.1.x = m1ax
Fcosq – mKFN – FT2.1sinq = m1ax
SFy = m1ay
FN – FT2.1.y – FG.1 – Fy = m1ay
FN – FT2.1cosq – m1g – Fsinq = 0
FN = FT2.1cosq + m1g + Fsinq
= 18.707cos25° + 4.65(9.80) + 98sin25°
= 103.94 N
mK = (Fcosq – FT2.1sinq – m1ax)/FN
= [98cos25° – 18.707sin25° – 4.65(4.5699)]/103.94 = 0.574
12
8. a.
A block (m = 10.2 kg) sits at rest on a surface that has
µS and µK values (with the block of 0.75 and 0.36,
respectively. The surface is initially level but then is
gradually tilted until the block slides down the slope.
At that angle of tilt, how long does it take the block to
slide for 1.84 m along that slope—measured from the
moment it begins to move? (Ignore any sudden jerks or
irregular motion to begin with—treat the motion as a
smooth transition from rest.)
FBD
The first situation to analyze is the block while
it’s still at rest but about to slip:
FN
FSMax
m
q FG
q
coord. axes
y
x
x-analysis
y-analysis
SFx = max
SFy = may
FGx – FSMax = max
FN – FGy = may
mgsinq – mSFN = 0
FN – mgcosq = 0
The unknowns here are FN and q. And it’s really the q value we want to determine—the slope angle at which the
block will slide down the incline—so solve for FN first (using the y-equation) and substitute this into the x-equation:
FN = mgcosq (from the y-equation)
mgsinq – mS(mgcosq) = 0
(substituting into the x-equation)
mgsinq = mS(mgcosq)
(rearranging)
sinq = mScosq
(simplifying)
tanq = mS
(simplifying)
q = tan-1(mS) = tan-1(0.75) = 36.8699°
This is the slope at which the block will be sliding.
13
The next situation to analyze is the block while it’s sliding down the slope:
FBD
FN
FK
m
q
q FG
coord. axes
y
x
x-analysis
y-analysis
SFx = max
SFy = may
FGx – FK = max
FN – FGy = may
mgsinq – mKFN = max
FN – mgcosq = 0
The unknowns here are FN and ax. And it’s really the ax value we want to determine—the block’s acceleration—so
solve for FN first (using the y-equation) and substitute this into the x-equation:
FN = mgcosq (from the y-equation)
mg(sinq – mKcosq) = max
(simplifying)
mgsinq – mK(mgcosq) = max
(substituting into the x-equation)
g(sinq – mKcosq) = ax = (9.80)[sin(36.8699) – .36·cos(36.8699)] = 3.0576 m/s2
Now it’s just a kinematics problem. Knowing Dx = 1.84, v0 = 0, and a = 3.0576, we can solve for Dt:
Dx = v0(Dt) + (1/2)a(Dt)2
Dx = (1/2)a(Dt)2
(simplifying)
2Dx/a = (Dt)2
(rearranging)
[2Dx/a]1/2 = Dt = [2(1.84)/3.0576]1/2 = 1.10 s
14
8. b.
The block shown at right remains at rest while the
force F is present. Find the block’s speed 2.50 seconds
after F is suddenly removed. Be sure to show your
work and/or explain your reasoning.
F
g
ms =
0.50
20°
Does the block move when released? Is the slope sufficient
to cause it to slip? Ask that question first! If the gravity force down the slope
exceeds the maximum static friction force the surface can exert, it will slip.
SFx = ?
Fs
max
– FG.x = ?
ms FN – mgsinq = ?
FN
10 k
mk =
0.25
SFy = may
FN – FG.y = may
Fs
coord. axes
y
FN – mgcosq = 0
x
FN = mgcosq = 10(9.80)cos20° = 92.09
Since the gravitational pull down the slope is less than Fsmax, the block will not slip; its speed will remain zero.
c.
mgsinq = 10(9.80)sin20° = 33.52 N
So: Fs
max
q F
G
A 3-kg block is sliding at constant speed down
a sloped surface, as shown. The coefficient of
kinetic friction (mk) between the block and the
slope is 0.45. What is the angle q of the slope?
Show your work and/or explain your reasoning.
SFx = max FG.x – Fk = max mgsinq – mkFN = 0
= ms FN = .50(92.09) = 46.04 N
q°
mk = mgsinq/FN = mgsinq/(mgcosq)
= tanq d.
A 3-kg block is sitting at rest on a sloped surface, as shown.
The coefficient of static friction (ms) between the block and the
sloped surface is 0.55. What is the magnitude of the total force
exerted by the sloped surface on the block?
Be sure to show your work and/or explain your reasoning.
FG
q
Therefore: q = tan-1(mk) = 24.2°
SFx = max FG.x – FS = max mgsinq – FS = 0
FS = mgsinq = 10.055 N SFy = may FN – FG.y = may FN – mgcosq = 0
FN = mgcosq = 27.627 N
y
20°
x
FN
Fs
Fs
Fslope.block
The total force by slope on block is a vector sum: Fslope.block = FN + FS
But FN and FS are perpendicular to each other, so they form a
right triangle whose hypotenuse is the total force; its magnitude
can be calculated via the Pythagorean Theorem: 29.4 N
Fk
x
SFy = may FN – FG.y = may FN – mgcosq = 0
FN = mgcosq
FN
y
q
q
FN
FG
Also OK is clearly explaining: The only object supporting the block upward is
the sloped surface, so the sum of all forces exerted by the surface on the block must exactly
counter the downward gravitational force on the block (FG = mg = 3·9.80 = 29.4 N).
15
9. For each item, be sure to show your work and/or explain your reasoning.
a. A 200 g hockey puck is launched up a metal ramp that is inclined at a 30° angle. The coefficients of static
and kinetic friction between the hockey puck and the metal ramp are μS = 0.40 and μK = 0.30, respectively.
The puck’s initial speed is 3 m/s.
(i) Find the net force magnitude and direction on the puck while it is sliding up the ramp.
So:
Thus: Find a: SF1x = m1a1x
–FG.1.x – FK.1 = m1a1x
–m1gsinq – mK.1FN.1 = –m1a
30°
SF1y = m1a1y
FN.1 – FG.1.y = m1a1y
FN.1 – m1gcosq = 0
FN.1 = m1gcosq
–m1gsinq – mK.1(m1gcosq) = –m1a
a = g(sinq + mK.1cosq)
= (9.80)[sin30° + (0.30)cos30°] = 7.446 m/s
Just do kinematics, since the x-acceleration value is now known:
q
FG
(ii) How far up the ramp does the puck travel?
FK
2
This is the total acceleration magnitude of the puck (since ay = 0),
and it’s in the negative x-direction, so a = ax = –7.446 m/s2
SF1 = m1a = (0.200)(–7.446) = 1.49 N down the ramp
b.
x
FN
y
vf.x2 = vi.x2 + 2axDx
Thus: Dx = (vf.x2 – vi.x2)/(2ax) = (02 – 32)/[2(–7.446)] = 0.604 m
A 20 kg block on a table is connected by a string to a 12 kg mass, which is hanging
over the edge of the table (modeled here as hanging over an ideal pulley). The 20-kg
block is 3.1 m from the table edge. The coefficient of kinetic friction between the block
and the table is 0.1. The coefficient of static friction between the block and the table is 0.5.
20 kg
(i) Is the hanging block is heavy enough to cause the block on the table to slide?
Could FS.1max hold m1 against the tension of the string, FT (= FT.21 = FT.12),
if m2 were hanging from rest? Find out:
In that special, simple case: FN.1 = FG.P1 = m1g
And: FT.12 = FG.P2 = m2g
FN.1
FS.1max
Thus: FS.1max = mS(FN.1) = mS(m1g) = (0.5)(20)(9.80) = 98.0 N But: FT.21 = FT.12 = m2g = (12)(9.80) = 118 N
x
y
12 kg
FT.12
FT.21
FG.P1
FG.P2
The tension FT.21 would be greater than FS.1max could hold. The block will slip.
(ii) Regardless of your answer to part a, assume that the block on the table does slide and determine how
much time will pass before the block reaches the end of the table.
FN.1
The sliding mass (m1):
The hanging mass (m2):
SF1.x = m1a1.x
SF1.y = m1a1.y
FK.1
F
–
F
=
m
a
FG.P2 – FT.12 = m2a2.y
T.21
K.1
1 1.x
FT.21
FT – FK.1 = m1a
m2g – FT = m2a
FT = FK.1 + m1a
FT = m2g – m2a
FG.P1
So:
F
+
m
a
=
m
g
–
m
a
K.1
1
2
2
Or: a = (m2g – FK.1)/(m1 + m2) = (m2g – mKm1g)/(m1 + m2)
= [(12)(9.80) – (0.1)(20)(9.80)]/(20 + 12) = 3.063 m/s2
Now kinematics:
Thus:
Dx = vi.x(Dt) + (1/2)ax(Dt)2
FT.12
FG.P2
Since vi.x = 0, this simplifies to Dx = (1/2)ax(Dt)2
Dt = √[2(Dx)/ax] = √[(2)(3.1)/3.063] = 1.42 s
16
c.
The figure shows a block of mass m resting on a 20° slope. The block and
this surface have a coefficient of static friction of 0.86 and a coefficient of
kinetic friction of 0.76. It is connected via a massless string over a
massless, frictionless pulley to a hanging block of mass 2.0 kg.
(i) What is the minimum mass m that will stick and not slip?
FN.1
y
FT.12
y
FT.21
x
FS.1max
x
q
FG.P1
FG.P2
The sticking mass (m1):
The hanging mass (m2):
SF1.y = m1a1.y
FN.1 – FG.P1.y = m1a1.y
FN.1 – m1gcosq = 0
SF2.y = m2a2.y
FT.12 – FG.P2 = m2a2.y
FT – m2g = 0
SF1.x = m1a1.x
FT.21 – FG.P1.x – FS.1max = m1a1.x
FT – m1gsinq – mSFN.1 = 0
FN.1 = m1gcosq
So:
Or:
SF2.x = m2a2.x
0 = 0
FT = m2g
m2g – m1gsinq – mSm1gcosq = 0
m1 = m2/(sinq + mScosq) = 2/[sin20° + (0.86)(cos20°)] = 1.7389 = 1.74 kg
(ii) If a mass half of that value were placed there instead, find the acceleration of the blocks.
FN.1
y
x
FT.21
FK.1
x
q
FG.P2
FG.P1
FT.12
y
Sliding mass (m1 = 1.7389/2 = 0.8694):
Hanging mass (m2):
SF1.x = m1a1.x
FT.21 – FG.P1.x – FS.1max = m1a1.x
FT – m1gsinq – mKFN.1 = m1a
SF2.x = m2a2.x
0 = 0
FN.1 = m1gcosq
FT = m2g – m2a
SF1.y = m1a1.y
FN.1 – FG.P1.y = m1a1.y
FN.1 – m1gcosq = 0
SF2.y = m2a2.y
FT.12 – FG.P2 = m2a2.y
FT – m2g = –m2a
So: m2g – m2a – m1gsinq – mKm1gcosq = m1a
Or: m2g – m1gsinq – mKm1gcosq = a(m1 + m2)
Thus: a = [m2g – m1gsinq – mKm1gcosq]/(m1 + m2)
= [(2)(9.80) – (0.8694)(9.80)(sin20°) – (0.76)(0.8694)(9.80)(cos20°)]/(0.8694 + 2) = 3.69 m/s2
17
9. d. A box (m = 10.0 kg) on a level surface has an applied pushing force, F, being exerted on it at an angle, q, as
shown. The static and kinetic friction coefficients between box and surface are 0.700 and 0.350, respectively.
a. Find the friction force magnitude acting on the box bottom when F = 75.0 N and q = 5.00°.
b. Find the friction force magnitude acting on the box bottom when F = 75.0 N and q = 20.0°.
c. What minimum value of F would be required to move the box when q = 40.0°?
d. What is the maximum angle (0 ≤ q ≤ 90°) at which you could exert the force F
(of any magnitude) toward the box and still get the box to move?
FS.1max
x
q
SF1.x = m1a1.x SF1.y = m1a1.y mSFN.1 – Fslipcosqslip = 0
FN.1 – Fslipsinqslip – m1g = 0
FS.1
max
– Fx = m1a1.x
FN.1 – Fy – FG.1 = m1a1.y
Thus: FN.1 = m1g + Fslipsinqslip
FG.1
F
Thus: mS(m1g + Fslipsinqslip) – Fcosqslip = 0
Or:
Fslipcosqslip – mSFslipsinqslip = mSm1g
So:
Fslip = (0.7)(10)(9.80)/[cos(5°) – (0.7)sin(5°)] = 73.4 N
a. Find Fslip for qslip = 5.00°:
Thus: FK.1 = mK(FN.1) = mK(m1g + Fsinq)
= (0.350)[(10)(9.80) + (75.0)sin(5.00°)] = 36.6 N
b. Find Fslip for qslip = 20.0°:
Fslip = mSm1g/(cosqslip – mSsinqslip)
Since the actual F applied (75 N) is greater than Fslip for this q, the block slips; the friction force is kinetic.
But the above y-analysis doesn’t change for the sliding case, so: FN.1 = m1g + Fsinq
Fslip = (0.7)(10)(9.80)/[cos(20°) – (0.7)sin(20°)] = 98.0 N
Since the actual F applied (75.0 N) is less than Fslip for this angle, the block does not slip,
so the friction force is static but not maximum. So the analysis becomes:
FN.1
y
FS
x
q
F
FG.1
SF1.x = m1a1.x FS – Fx = m1a1.x
c. Find Fslip for qslip = 40.0°:
FS – Fcosq = 0
Thus: FS = Fcosq
q
Here’s a situation where there are two different values (F and q) that you can vary. For any given value of F,
if you want it to be the force Fslip that puts the block into “ready-to-slip” mode, there’s just one certain angle
(if any) that you could apply that force (a certain value of q)—call this qslip. So how do Fslip and qslip relate to
each other? Find out: Analyze the situation as if the box (call it m1) were about to slip:
FN.1
y
F
= (75.0)cos20° = 70.5 N
Fslip = (0.7)(10)(9.80)/[cos(40°) – (0.7)sin(40°)] = 217 N
d. Looking at Fslip = mSm1g/(cosqslip – mSsinqslip), notice that Fslip approaches infinity when the denominator
(cosqslip – mSsinqslip) approaches zero. In other words when (cosqslip – mSsinqslip) goes to zero, it doesn’t
matter how great the force you apply at that qslip, the block will still be only “about to slip”—still at rest.
So the maximum angle, qslip.max, is given by:
That is:
tanqslip.max = 1/mS Or:
cosqslip.max – mSsinqslip.max = 0
qslip.max = tan-1(1/mS)
= tan-1(1/0.7) = 55.0°
10. a. The figure shows a 100 kg block being released from rest from a height of 1.0 m.
The block takes 0.64 s to reach the floor. What is the mass of the block on the left?
First, do kinematics to find the y-acceleration value so that the box lands in 0.64 s:
Dy = vi.y(Dt) + (1/2)ay(Dt)2 Now do a FBD analysis (referring to diagrams here—using conventional x- and y- axes):
Or: ay = 2[Dy – vi.y(Dt)]/(Dt)2
= 2[–1 – 0(0.64)]/(0.64)2 = –4.883 m/s2
Let the unknown mass be m1 and the 100-kg mass be m2. And note that the magnitudes
of their accelerations will be the same (a = 4.883 m/s2); and that the tension forces
exerted will also be equal (because the pulley is ideal—massless). Thus:
m1:
m 2:
FT2.1
SF1y = m1a1y FT1.2
FT2.1 – FG.1 = m1a1y
FG.1
FT – m1g = m1a F
Then: m2g – m2a – m1g = m1a
So:
m1 = m2(g – a)/(a + g)
Or: m2(g – a) = m1(a + g)
G.2
SF2y = m2a2y
FT1.2 – FG.2 = m2a2y
FT – m2g = –m2a
So: FT = m2g – m2a
= 100(9.80 – 4.883)/(4.883 + 9.80) = 33.5 kg
b. The two blocks shown here are sliding down the incline. Find the tension in the (massless) string.
Now do a FBD analysis (referring to diagrams here—using axes as shown):
Let the 1-kg mass be m1 and the 2-kg mass be m2. Note that the magnitudes
of their accelerations will be the same (a). Thus:
m1:
F
N.1
FK
.1
FN
m2:
.2
FK
FT
.21
FT
.2
y
.12
q
FG.1
q FG.2
SF1x = m1a1x
FT2.1 + FG.1.x – FK.1 = m1a1x
FT + m1gsinq – mK.1FN.1 = m1a
SF2x = m2a2x
FG.2.x – FT1.2 – FK.2 = m2a2x
m2gsinq – FT – mK.2FN.2 = –m2a
And: FN.2 = m2gcosq
And: m2gsinq – FT – mK.2(m2gcosq) = –m2a
SF1y = m1a1y
FN.1 – FG.1.y = m1a1y
FN.1 – m1gcosq = 0
SF2y = m2a2y
FN.2 – FG.2.y = m2a2y
FN.2 – m2gcosq = 0
So:
Thus: FN.1 = m1gcosq
FT + m1gsinq – mK.1(m1gcosq) = m1a
So:
FT/m1 + gsinq – mK.1(gcosq) = FT/m2 – gsinq + mK.2(gcosq)
Find a: a = FT/m1 + gsinq – mK.1(gcosq)
x
And: a = FT/m2 – gsinq + mK.2(gcosq)
Find FT: FT(1/m1 – 1/m2) = (mK.1 + mK.2)(gcosq) – 2gsinq
FT = [(mK.1 + mK.2)(g)(cosq – 2sinq)]/(1/m1 – 1/m2)
= [(0.20 + 0.10)(9.80)(cos20° – 2sin20°)]/(1/1.0 – 1/2.0) = 1.50 N
19
10. c.
Two blocks (masses m1 = 26.5 kg; m2 = 38.9 kg) are moving together
horizontally to the right. Block 2 is touching, but not attached to,
block 1. The coefficient of static friction, ms, between the two blocks
is 1.47 (yes, they are quite sticky). Find the minimum coefficient of
kinetic friction, mk, between block 1 and the floor so that block 2
does not slip.
First, do a complete FBD and Newton’s Laws analysis of each mass,
as follows:
FBD of m1
FN.1
FN.21
m1
x-analysis
SFx = m1ax
FN.21 – Fk.1 = m1ax
FN.21 – mkFN.1 = m1ax
y-analysis
Fk.1
coord. axes
y
m2
Fs.21max
FG.1
x
SFy = m1ay
FN.1 – Fs.21max – FG.1 = m1ay
FN.1 – msFN.21 – m1g = 0
x-analysis
FBD of m2
Fs.12max
SFx = m2ax
–FN.12 = m2ax
–FN.12 = m2ax
y-analysis
FN.12
coord. axes
y
FG.2
x
SFy = m2ay
Fs.12max – FG.2 = m2ay
msFN.12 – m2g = 0
20
Now solve for mk, as follows:
I. Solve msFN.12 – m2g = 0 for FN.12:
FN.12 = m2g/ms
II. Solve –FN.12 = m2ax for ax (the x-acceleration of both masses): ax = –FN.12/m2 = –(m2g/ms)/m2 = –(g/ms)
III.Solve for FN.21:
By Newton’s Third Law, the magnitudes of FN.21 and FN.12 must be equal.
FN.21 = FN.12 = m2g/ms
IV. Solve FN.1 – msFN.21 – m1g = 0 for FN.1: FN.1 = msFN.21 + m1g = ms(m2g/ms) + m1g = (m1 + m2)g
V. Solve FN.21 – mkFN.1 = m1ax for mk:
mk = (FN.21 – m1ax)/FN.1 = [m2g/ms + m1(g/ms)]/[(m1 + m2)g] = 1/ms
21
10. d. As shown in this side (cutaway or “x-ray”) view, a sphere (msphere = 2.00 kg) rests in
one compartment of a two-compartment box (mbox = 5.00 kg). The box rests on a
level, frictionless surface. All surfaces of the box and sphere are also frictionless.
When the box and sphere are at rest (as shown), there are two normal forces acting
on the sphere, and one of those forces has a magnitude 20% greater than the other.
But when a tension force, FT (not shown here) is then exerted horizontally (to the right) on the right side of
the box, the two normal forces acting on the sphere become equal in magnitude. Find the magnitude of FT.
q
Let the two surfaces of the box that are exerting normal forces on the sphere be called “Wall” (W) and
“Divider” (D). Then here is the analysis of the sphere (m1) when everything is at rest, as shown above
(and note the definition of q now on that diagram):
y
FN.D1
SF1.x = m1a1.x SF1.y = m1a1.y x
FN.W1 – FN.D1cosq = 0
FN.D1sinq – m1g = 0
FN.W1
q
FN.W1 – FN.D1.x = m1a1.x
FG.1
FN.D1.y – FG.1 = m1a1.y
Thus:
But we also know:
Therefore:
Solve for q:
Now analyze the combined mass (box and sphere together—call this m1&2), when the force FT is applied:
FN.D1 = FN.W1/cosq
FN.D1 = 1.20(FN.W1)
1.20(FN.W1) = FN.W1/cosq
q = cos-1(1/1.20) = 33.557°
F
SF1&2.x = m1&2a1&2.x SF1&2.y = m1&2a1&2.y FT = m1&2ax
FN.1&2 – m1&2g = 0
y
N.1&2
FT
x
FG.1&2
FT = m1&2a1&2.x
Thus: ax = FT/m1&2
Now analyze just the sphere (m1) as it is accelerating to the right inside the box:
y
FN.D1
FN.W1
x
q
SF1.x = m1a1.x SF1.y = m1a1.y FN.W1 – FN.D1cosq = m1ax
FN.D1sinq – m1g = 0
FN.W1 – FN.D1.x = m1a1.x
FG.1
Thus:
Or:
Solve for FT:
FN.1&2 – FG.1&2 = m1&2a1&2.y
FN.D1.y – FG.1 = m1a1.y
Thus: FN.D1 = m1g/sinq
FN.D1 – FN.D1cosq = m1(FT/m1&2)
(m1g/sinq)(1 – cosq) = m1(FT/m1&2)
FT = (m1&2/m1)(m1g/sinq)(1 – cosq)
= (7.00/2.00)[(2.00)(9.80)/sin(33.557°)][1 – cos(33.557°)] = 20.7 N
11. a.
Two crates (m1 is known; m2 is unknown) are being dragged
together (one linked to the other via a horizontal cable, as
shown) across the floor of a large freight elevator. The crates’
horizontal velocity is constant. The elevator is accelerating
upward at a known rate, a. The pulling force F is known,
as is the angle q above the horizontal at which F pulls.
The coefficient of kinetic friction, mK, between each crate
and the elevator floor is also known. Find m2, expressed
in terms of m1, a, F, mK and g
First, do a complete FBD and Newton’s Laws analysis of each mass, as follows:
elevator
F
m2
FBD of m1
F
q
Fk.1
coord. axes
y
m1
q
x-analysis
FN.1
FT.21
(cable)
FG.1
x
SFx = m1ax
Fx – Fk.1 – FT.21 = m1ax
Fcosq – mkFN.1 – FT.21 = 0
y-analysis
SFy = m1ay
Fy + FN.1 – FG.1 = m1ay
Fsinq + FN.1 – m1g = m1ay
x-analysis
FBD of m2
SFx = m2ax
FN.2
FT.12 – Fk.2 = m2ax
FT.12 – mkFN.2 = 0
FT.12
Fk.2
coord. axes
y
FG.2
x
y-analysis
SFy = m2ay
FN.2 – FG.2 = m2ay
FN.2 – m2g = m2ay
23
Now solve for m2, as follows:
I. Solve the y-force equation (Fsinq + FN.1 – m1g = m1ay) for FN.1:
FN.1 = m1ay + m1g – Fsinq
II. Solve the x-force equation (Fcosq – mkFN.1 – FT.21 = 0) for FT.21:
FT.21
III.Solve for FT.12:
By Newton’s Third Law, the magnitudes of FT.21 and FT.12 must be equal.
FT.12 = FT.21
IV. Solve the x-force equation (FT.12 – mkFN.2 = 0) for FN.2:
FN.2 = FT.12/mk
= Fcosq – mkFN.1
= Fcosq – mk(m1ay + m1g – Fsinq)
= Fcosq – mk(m1ay + m1g – Fsinq)
= Fcosq/mk – (m1ay + m1g – Fsinq)
= F(cosq/mk + sinq) – m1(ay + g)
V. Solve the y-force equation (FN.2 – m2g = m2ay) for m2: m2 = FN.2/(g + ay) = F(cosq/mk + sinq)/(ay + g) – m1
24
11. b.
Walking through your neighborhood, you see Thor set down his hammer while working on the roof.
The hammer slides, starting from rest, a distance d down the roof, which is angled q from the horizontal.
The hammer leaves the roof from a height h, and lands a horizontal distance x from the point where it
becomes a projectile. What is the coefficient of kinetic friction between hammer and the roof?
You may consider these values as known: d, q, h, x, g.
This is an ODAVEST problem, but keep in mind that you’re not being asked to actually solve for the final expression. In fact, you’re not
being asked to do any math at all—not even any algebra. Rather, for the Solve step, you are to write a series of succinct instructions on
how to solve this problem. Pretend that your instructions will be given to someone who knows math but not physics. And for the Test step,
you should consider the situation and predict how the solution would change if the data were different—changing one data value at a time.
Objective:
Data:
An object starts from rest and slides a distance d down a roof surface that is inclined at an angle q
below the horizontal. The object leaves the roof at a height h above the ground and first impacts
the ground at a horizontal distance x from the point where it left the roof. We need to determine
mK (the coefficient of kinetic friction) between the object and the roof surface.
vi = 0
The object began its slide at rest.
d
The distance over which the object slid on the roof surface.
q
The angle of incline of the roof surface, with respect to the horizontal.
h
The vertical distance from the impact point to where the object left the roof.
x
The horizontal distance from the impact point to where the object left the roof.
g
The local free-fall acceleration magnitude.
Assumptions: Objects
We will treat the hammer (object) as a point mass—having no extent or rotation.
Surfaces
We will assume that the surfaces of both the roof and the hammer are uniform.
We will also assume that the roof is planar.
Slide
We assume that the slide was in a straight line and directly down the slope
(along the steepest possible path).
Projectile
We assume the g is constant for all relevant heights here.
We will disregard any effects of wind or air drag.
Visual Reps.:
FN
.S1
Free-body diagram of object sliding against kinetic friction down the roof slope:
FK
.S1
Kinematics diagram (and “visual” data inventory) of the slide:
vi = 0
Dx =
d
vf
q
q FG.P1
(x)
(y)
Dx = d
vi.x = 0
vf.x =
ax = ??
Dt =
25
Kinematics diagram (and “visual” data inventory) of the projectile motion:
Dy = –h
vi
qi
Dx = x
Equations:
I.
II.
III.
IV.
V.
qf
vf
m1gsinq – mkFN.S1 = m1ax
FN.S1 – m1gcosq = 0
vf.x2 = 02 + 2axd
x = vicosq(Dt)
–h = visinq(Dt) – (1/2)g(Dt)2
Solving:
Solve IV for Dt. Substitute that result into V.
Solve V for Dt. Substitute that result into IV.
Solve IV for vi. Substitute that result as vf.x into III.
Testing:
(x)
(y)
Dx = x
Dy = –h
vi.x = vicosq vi.y = visinq
vf.x =
vf.y =
ax = 0
ax = –g
Dt =
Solve III for ax. Substitute that result into I.
Solve II for FN.S1. Substitute that result into I.
Solve I for mk.
Dimensions:
The solution would need to be unitless (just a numerical value).
Dependencies:
In general, this number should be somewhere between 0 and 1 (or possibly a little
above 1 if the roof is very sticky or rough).
A greater vi that produces the same projectile motion would imply that the roof
slows down the slide a little more—a greater mk.
A greater slide distance d that produces the same projectile motion would imply that the acceleration is less (achieves same launch speed over a longer slide):
a greater mk.
A higher fall, h, with the same x as before, would imply a slower launch speed:
a greater mk.
Without explicitly solving the above equation set as prescribed, the effects of a
greater roof angle q are not clear.
A greater x travel for the same fall height would imply a faster launch speed:
a smaller mk.
Without explicitly solving the above equation set as prescribed, the effects of a
greater value of g are not clear. It would affect the friction, the slide and the fall.
(It’s possible, therefore, that it might even have no implications at all for mk.
26
12. a.
A 90-kg man boards an elevator and stands on an
accurate bathroom scale (which just happens to be
there). During his elevator ride, the scale reads a
steady 700 N. Also during this ride, he happens to
drop a coin from a height of 1 m above the elevator
floor. From the moment he releases the coin, how
long does it take to hit the floor of the elevator?
FBD of m
FN
elevator
m (person)
(scale)
y
coord. axes
FG
x
x-analysis
SFx = max
SFy = may
0 = 0
FN – FG = may
FN – mg = may
y-analysis
The x-direction has no forces and thus zero acceleration. In the y-direction, ay is easily solved for:
ay = (FN – mg)/m = [700 – (90)(9.80)]/90 = –2.022 m/s2
This is the acceleration of the elevator floor, the man, and the coin in his hand—they all move as one—until the
moment when he releases the coin. After that moment, the coin’s acceleration is –g. But the elevator floor’s
acceleration continues to be ay (–2.022). Essentially then, the question becomes this: How long does it take the
coin to “catch up” to the elevator floor as the both accelerate downward—at different rates?
Object A (coin):
DyA = vA.i(Dt) + (1/2)(–g)(Dt)2
Object B (elevator floor):
DyB = vB.i(Dt) + (1/2)(ay)(Dt)2
But DyA = DyB – 1 (It’s –1 because Dy downward is negative; A travels downward 1 meter more than B.)
And vA.i = vB.i
(When the coin is released, its velocity is the same as the elevator floor’s.)
Substituting: vA.i(Dt) + (1/2)(–g)(Dt)2 = vA.i(Dt) + (1/2)(ay)(Dt)2 – 1
Solve for Dt:
–(1/2)(ay)(Dt)2 – (1/2)(g)(Dt)2 = – 1
(Dt)2[(1/2)(ay) + (1/2)(g)] = 1
(Dt) = {1/[(1/2)(ay) + (1/2)(g)]}1/2
= {1/[(1/2)(–2.022) + (1/2)(9.80)]}1/2 = 0.507 s
27
12. b.
box (A)
dry glass (mS > mK > 0)
brick (C)
beam (B)
d
wet glass (≈ frictionless)
Careless construction workers have left for the holidays with their site as shown here.
A large, heavy plate of glass is lying flat on level ground, partly sheltered under a roof.
Sitting at rest on the glass plate are (A) a full box of bricks (mass = mA; height = hA)
and (B) a heavy iron beam (mass = mB). Sitting at rest on top of the box is (C) a single
extra brick (mass = mC).
It has rained all week. The wet section of the glass plate has become very slippery
(as in “hydroplaning”)—essentially frictionless. The dry glass sheltered by the roof has
its normal friction (mS and mK are known). The bucket has been collecting rainwater.…
cord
bucket
The workers have used the box as an “anchor” for a bucket of sand they’ve left hanging
—attached to the box by a (massless) cord suspended over a massless, frictionless pulley.
d
Finally, the bucket becomes just heavy enough that box A (together with brick C) slips.
It then slides along the dry glass and out onto the wet glass, where it collides with the
beam. As a result, B and A stick together and slide at speed vAB, but the loose brick (C)
flies off, landing on the glass a distance x ahead of the box.
As it happens, the bucket’s height above the ground is the same distance, d, as the box’s
travel distance from its original position to the edge of the wet section of glass. Find d.
Here is a summary of all known values: mA, hA, mB, mC, mS, mK, vAB, x, g
This is an ODAVEST problem, but keep in mind that you’re not being asked to actually solve for the final expression. In fact, you’re not
being asked to do any math at all—not even any algebra. Rather, for the Solve step, you are to write a series of succinct instructions on
how to solve this problem. Pretend that your instructions will be given to someone who knows math but not physics. And for the Test step,
you should consider the situation and predict how the solution would change if the data were different—changing one data value at a time.
Objective:
A flat plate of glass, lying on level ground, is partly dry and partly wet.
Initially at rest on the dry part of the glass is a box of bricks.
Initially at rest on the box is a single extra brick.
At rest on the wet part of the glass is an iron beam.
Attached to the box is a massless cord, whose other end is attached to
a bucket. The bucket is suspended over a frictionless, massless pulley
—so the box is “anchoring” the bucket.
The bucket gradually gets heavier as rain accumulates in it, until its pull
on the cord is sufficient to start the box sliding.
The box and beam stay together after the collision,
The box (together with the extra brick) slides and collides with the iron beam.
but the loose brick goes flying ahead (as a projectile) over the beam
(“leapfrog” style), landing back on the glass at a known distance from the box.
The distance fallen by the bucket is the distance the box slides on the dry glass.
We must find that distance.
Here is a summary of all known values: mA, hA, mB, mC, mS, mK, vAB, x, g
Data:
mA The mass of the box of bricks.
hA The height of the box of bricks (the distance from its top to the glass below).
mB The mass of the iron beam.
mC The mass of the single brick.
vA.i , vB.i, vC.i The initial velocities of the masses (all are zero).
mS The coefficient of static friction between the box and the dry glass.
mK The coefficient of kinetic friction between the box and the dry glass.
vAB The speed of the box and iron beam (which stay together) after their collision.
x The distance between the box and the brick at the moment of the brick’s impact on the glass.
g The local free-fall acceleration magnitude.
Assumptions: Surfaces
We assume the surfaces of the three masses are uniform and planar.
Alignment We will assume that the box “arrives at” the dry/wet boundary of the glass
when the center of the box crosses that boundary (i.e. that the box’s
width is essentially negligible).
Box Since the box’s height is relevant, we can’t model it completely as a particle.
But we can assume that it won’t tip or rotate at any time.
Beam We can model the beam as a particle—it doesn’t interfere with the
brick or the cord, nor does it rotate or tip.
Brick We model the brick as a particle for its projectile motion.
We assume it has no rotation at any moment as a projectile.
and that it leaves box A on a horizontal trajectory.
(If we choose not to treat the brick as a particle, then we should
probably assume that its hindmost point is the last to touch box A
and is still the hindmost point at impact—and how x is measured to A.)
Cord We assume the cord does not sag, wobble or produce anything other than a
steady, horizontal tension force.
We also assume the cord does not interfere with the collision of the box and
the beam—nor with any motion thereafter.
Bucket
We assume the bucket bottom stays level as it falls.
Air and g We assume the g is constant for all relevant heights here.
We disregard any effects of wind or air drag
Visual Rep(s):
Equations:
FBD of brick (C):
I.
FN.AC – mCg = 0
FBD of combined mass (AC), ready to slip:
II.
FT.Bucket.A – mSFN.SAC = 0
FBD of box (A), ready to slip:
III.
FN.SA – FN.CA – mAg = 0
Visual Rep(s), cont.: Equations, cont.:
FBD of bucket when box is ready to slip
IV.
FT.A.Bucket – FG.P.Bucket = 0
FBD of bucket, as combined mass (AC) slides on dry glass
V.
FT – FG.P.Bucket = mBucket aBucket
FBD of mass AC sliding on dry glass
VI.
FT – mKFN.SAC = (mA + mC) aAC
VII.
aBucket = –aAC
Kinematics diagram of box accelerating
from rest to wet glass
VIII. vAC2 = 02 + 2aACd
mCvC
x-momentum before/after analysis of
IX.
(mA + mC)(vAC) = (mA + mB)(vAB) +
projectile motion analysis of brick C
X.
hA = (1/2)g(DtC)2
XI.
DxC = vCDtC
XII.
DxAB = vABDtC
collision/explosion
kinematics analysis of AB slide
Solving:
Solve X for DtC. Substitute that result into XII and XI.
Solve XII for DxAB. Substitute that result into XIII.
Solve XI for vC. Substitute that result into IX.
Solve XIII for DxC. Substitute that result into XI.
Solve IX for vAC. Substitute that result into VIII.
Solve I for FN.AC. Substitute that result (as FN.CA) into III.
Solve III for FN.SA. Substitute that result into II and VI.
Solve II for FT.Bucket.A. Substitute that result (as FT.A.Bucket) into IV.
Solve IV for FG.P.Bucket. Substitute that result into V.
Solve V, VI and VII simultaneously for aBucket, FT and aA.
Substitute the result for aA into VIII.
XIII. x = DxC – DxAB
Solve VIII for d.
Testing:
Dimensions:
The solution for the distance should have dimensions of length.
Dependencies:
The effect on the calculation of d of increasing any one of the given
masses is not clear without explicitly solving the above equation set
as prescribed; the masses affect several parts of those calculations.
If mS were greater, this would imply a greater bucket weight and thus
a shorter d needed to achieve the same motion and collision.
If the height, hA of the box were greater, this would result in a lesser
value calculated for d because the same value for x achieved in a longer
fall time would imply a slower vC and thus a shorter slide to achieve the
extra momentum the brick C that carried away from the collision.
If mK were greater, this would imply a longer d needed to achieve the
same motion and collision.
If vAB were greater, this would imply a longer d needed to achieve that
motion and collision.
If x were greater, this would imply a longer d needed to achieve that
motion and collision.
Varying the value of local g would affect the bucket fall (therefore the
tension in the cord) but also the normal forces and brick C’s fall time,
so we would need an explicit solution to reveal the dependencies clearly.