free body diagram

2013-09-23 2:23 pm
The graph is http://blog.yimg.com/3/KcutVhp7s58pQ5Vx_PowSzy3i98NvOevXdlckjayvtQThTvtxzASgQ--/8/l/QA2CG.lvFOljqIX8Hrv3JQ.jpg

Two crates, A and B are in an elevator as shown. The mass of crate A is greater than the mass of crate B. The elevator is initially moving downward at a constant speed but shows as it approaches its destination.

a) Draw a free-body diagram for crate B while the elevator is descending at constant speed. Use the length of the vector to indicate the relative magnitude of the forces(ie. if one force is greater in magnitude than another, its vector should be longer)

b) How would the forces change, if at all, when the elevator is slowing?(make a list of the forces acting and indicate whether they have increase, decreased, or stayed the same.)

For question a, is it only two forces acting on the crate B, one is gravitational force, another one is Normal force which is exerted by the floor of the elevator?

I am not sure for question b, hope someone can help!

Thanks

回答 (1)

2013-09-23 7:28 pm
✔ 最佳答案
(a) There are 3 forces acting on crate B
The weight (downward), normal reaction (upward) given by the elevator floor, and force (downward) given by crate A.

(b) Let M and m be the masses of crates A and B respectively
Ra be the normal reaction on A. Hence, the force acting on crate B by crate A is Ra (but in opposite, i.e. downward, direction)
Rb be the normal reaction on crate B given by the elevator floor
a be the deceleration of the elevator

Consider crate A:
Ra - Mg = Ma where g is the acceleration due to gravity
i.e. Ra = M(g+a) ----------- (1)

Consider crate B,
Rb - (Ra + mg) = ma
i.e. Rb - (Mg + Ma) - mg = ma
Rb = ma + mg + Mg + Ma
Rb = (M+m)g + (M+m)a -------------- (2)

When the elevator is moving at constant speed, a = 0
thus, Ra = Mg [from (1)]
Rb = (M+m)g [from (2)]

When the elevator is decelerating, a > 0
thus, Ra increases
Rb increase

Therefore, the answers are:
weight no change
normal reaction Rb increases
force given by crae A increases


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