Sample Examination #1 Closed book and notes. Calculators permitted For full marks you must a) insert relevant diagrams, b) explain in a few words what you are doing, c) present solutions neatly, d) indicate units where relevant, e) carry 3 significant figures. 1. The force F has a magnitude of 80 lb and acts within the volume shown. Determine the magnitudes of the x, y, and z components of F. (15 points) Write down the direction cosines first, and use them to find the magnitudes. We have α, β, but no γ yet, so start by finding γ. cos2α + cos2β +cos2γ = 1 ∴ (0.25 + 0.5 + cos2γ) =1 ∴ γ = 60° also Now magnitudes are Fx Fy Fz = 80 cos 60° = 40 lb = 80 cos 45° = 56.6 lb = 80 cos 60° = 40 lb 2. Determine the magnitude of the projected component of force F along the pole. (15 points) This is one of the uses of the dot product! we need a unit vector along AO? We need 2 vectors and only have 1: so AO = [-3i + 2j -6k] and its magnitude is 32 + 22 + 62 = 7 so a unit vector along AO is = 1/7[-3i + 2j -6k] and the dot product is F. AÔ = -20x-3/7 + 50x2/7 - 10x-6/7 = 60/7 + 100/7 + 60/7 =31.4 lb 3. Determine the stretch in each of the two springs required to hold the 20 kg crate in the equilibrium position shown. Each spring has an unstretched length of 2m and a stiffness of k = 300N/m. (20 points) FOC = FOA = FOB = F = 6i + 4j + 12k 62 + 42 + 122 - FOA j - FOB i - 196.2k FOC = 3/7 FOCi + 2/7 FOCj + 6/7 FOCk and ΣF = 0 Equating i, j, and k components 3/7 FOC - FOB = 0 2/7 FOC - FOA = 0 6/7 FOC - 196.2 = 0 and solving, we have FOC = 228.9 N and FOB = 98.1 N and Then using the spring formula, s = F/k: sOB = 98.1/300 = 0.327m sOA = 65.4/300 = 0.218m FOA = 65.4 N 4. Two men exert forces of F = 80 lb and P = 50 lb on the ropes. Determine the moment of each force about A. Will the pole rotate clockwise or counterclockwise? (20 points) P and F both can be resolved into vertical and horizontal components. The vertical components pass through A and contribute zero moment. Therefore, we only consider the horizontal components. ΣMP = P . cos 45° . 18 = 636 ft.lb (acw) ΣMF = F . 4/5 .12 = -768 ft.lb (cw) ΣMTOTAL = 636 - 768 = -132 ft.lb = clockwise (or, using r ^ F, for force F, r = 0i + 12j + 0k F = 80 4/5 i - 80 3/5 j + 0k so r^F = i j k 0 12 0 64 -48 0 = = 0i + 0j - 768k 768 lb.ft c.w. etc. .. same approach for the other force, P, then find resultant to see if it is cw (ve) or acw (+ve) Other questions that might have been asked A. Determine the coordinate direction angles of F1 and FR. (15 points) Unit vector of F1 and FR?? uF1 = 4/5 i + 3/5 k = 0.8i + 0.6k uFR = = cos 45° sin 30° i + cos 45° cos 30° j + cos 45° j 0.3536i + 0.6124j + 0.707k So the coordinate direction angles are: F1 cos α = 0.8, ∴ α = cos β = 0 ∴ β = cos γ = 0.6 ∴ γ = FR cos α = 0.3536 cos β = 0.6124 cos γ = 0.707 ∴α = ∴β = ∴γ = 36.9° 90° 53.1° 69.3° 52.2° 45° B. Determine the moment created by the force F = [50i + 100j - 50k] N, acting at D, about each of the joints at B and C. (20 points) To find moments, use M = r ^ F where r = each r in turn. rBD = (0.75i + 0j - 0.3k) m rCD = (0.75i + 1.25j - 0.3k) MB = i j 0.75 0 50 100 MC = ................... k -0.3 -50 = {30i + 22.5j + 75k} N.m = {-32.5i + 22.5j + 12.5k} N.m