Statics – Sample Test 3 Solution 1. The bent gate of width 1 m (perpendicular to the plane of the paper) has water (density = 1000 kg/m3) on one side. Neglecting the weight of the gate, determine the reactions at pin A and the smooth surface at B. 2. For the truss shown, the applied loads are: F1 = 600 lb, and F2 = 300 lb. Using the method of sections, and only two equations of equilibrium, find the force in member BD. Answer: FBD = 500 lb (C) Plan of Solution: a. Section a-a: Summation of moments about G Æ FAE b. Section b-b: Summation of moments about C Æ FBD 3. FBD Section ABFE: It is not apparent which counter is active, so we guess that FEH = 0; ΣM F = 0: ⎛ 24 ⎞ FEG + 40 kN ⎟ − ( 4.8 m )( 20 kN ) = 0 ⎝ 745 ⎠ ( 3.2 m ) ⎜ FEG = −11.3728 kN FEG = 11.37 kN C W ΣM G = 0: ⎛ 24 ⎞ FFH ⎟ − ( 7.2 m )( 20 kN ) − (1.3 m )( 40 kN ) = 0 ⎝ 745 ⎠ ( 5.8 m ) ⎜ FFH = 38.432 kN, ΣFx = 0: 20 kN − FFH = 38.4 kN C W 13 15 ( 38.432 kN − 11.3728 kN ) − FFG = 0 17 745 FFG = 8.0604 kN, FFG = 8.06 kN T W Since FEG is in tension, our guess was correct. A negative answer would be impossible, indicating an incorrect guess. 3. The device shown raises the load W by extending the hydraulic actuator DE. The bars AD and BC are 4 ft long, and the distances b = 2.5 ft and h = 1.5 ft. If W = 300 lb, what force must the actuator exert to hold the load in equilibrium? Answer: 742 lb 6. Determine the internal forces and moment at point B. Answers: 9 kN, -4 kN, -2 kN-m