Determining Impending Motion with Coefficients of Friction

[jaredmt]

Homework Statement

The coefficients of friction are Us = .4 and Uk = .3 between all surfaces of contact. Determine the force P for which motion of the 30kg block is impendinging if cable AB is attatched as shown

The Attempt at a Solution

wups i forgot to lable AB but there's only 1 cable so the picture is clear

im having trouble setting this one up. i don't know whether to start on the bottom block or the top. and I am not sure if I am setting up my forces correctly. here's what i have for the top block:

sum Fy = N - (20)(9.8) = 0 ---> N = 196N
sum Fx = Tab - Us(N) = 0 ---> Tab = Us(N) = .4(196) = 78.4N

is that much correct? if it is then I am not sure if I am setting up the next part right. for the bottom block i have:

sum Fy = N - (20)(9.8) - (30)(9.8) = 0 ----> N = 490N
sum Fx = -P + Tab + (Us)(N) = 0 ---> P = 78.4 + (.4)(490) = 274.4N


if i screwed up anything, its probably the horizontal forces. is Tab right or should it be (tab)(Us) ?

 

[LowlyPion]

If you look at it as being equivalent to the 30 kg and the 20 kg taken twice contributing to friction that's 70*g*.4 which is 274.4 as you've found.

I'd say you are ok.

 

[Doc Al]

for the bottom block i have:

sum Fy = N - (20)(9.8) - (30)(9.8) = 0 ----> N = 490N
sum Fx = -P + Tab + (Us)(N) = 0 ---> P = 78.4 + (.4)(490) = 274.4N

A couple of problems: For one, the cable only pulls on the top block, not the bottom block. So cable tension shouldn't appear explicitly in the analysis of the bottom block. Also, there are two friction forces acting on the bottom block.

Nonetheless, your answer is good. (Since the cable tension is equal to the other friction force, you get the right answer.)

 

[jaredmt]

hm. ok so what you're saying is i didnt really even need a freebody diagram for the top block, i could have just gone straight to the bottom and put (20kg)(9.8)(.4). ok thanks

 

[Doc Al]

That's right. All you need is a freebody diagram for the bottom block, which would show two friction forces. (Of course, you'd need a FBD for the top block to calculate the normal force between the two blocks, if it wasn't obvious.)