Homework #2 Solutions Essay

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Physics 221 Summer 2012 HOMEWORK #2 Due Friday June 22, 2012 1 A 70.0-kg person stands on a scale placed on the floor of an elevator. Find: - the weight of the person (magnitude and direction), - the normal force by the scale on the person (magnitude and direction), - and what the scale reads (in kilograms) in the following cases: (a) The elevator moves up with a constant speed of 2.0 m/s2 . (b) The elevator has a constant upward acceleration of 2.0 m/s2 . (c) The elevator has a constant downward acceleration of 2.0 m/s2 . (d) The cable snaps and the elevator falls freely (ignore friction and the bloody end!). SOLUTION: The weight of the person depends on Mass and g, therefore: W = M g = (70.0 kg) 9.80 m/s2 = 686 N downward in all…show more content…
The rope and pulleys are very light and ideal. Determine the weight of the object B that maintains the equilibrium given that the angle between the rope and the horizontal is 35◦ on both sides of the movable pulley. TR TL 35◦ 35◦ TR B WA WB SOLUTION: Let is draw free-body diagrams for the point of suspension above Object A and free-body diagram for object B (a) Since the system is in equilibrium, then a = 0 in all newton’s second law equations. For the point just above object A x − direction : ⇒ 0 = TR cos (35◦ ) − TL cos (35◦ ) TR = TL ≡ T Use results above for equation of motion in y direction: y − direction : ⇒ For object B: 0 = TR − WB = 0 ⇒ WB = TR = T = 17.4 N 0 = 2T sin (35◦ ) − WA WA T = = 17.4 N 2 sin (35◦ ) 5 Physics 221 Summer 2012 HOMEWORK #2 Due Friday June 22, 2012 6 A plane which just took off flies at an angle θ = 15◦ with the horizontal and it is gaining speed at 1.5 m/s2 . The 75-kg pilot sits on a seat whose bottom and back are perfectly perpendicular to each other, and the bottom is parallel to the direction of the motion. Find the magnitude of the forces exerted on the pilot by: (a) The back of the seat. (b) The bottom of the seat. SOLUTION: It is most convenient to chose the coordinate system to line up with
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