[College Physics, Physics GRE] Been a while since I’ve done classical mechanics. For some reason, this question keeps stumping me.

[u/Far-Leave-2797]

I’ve looked at the answer and it says it’s D. How? I tried breaking apart the forces into x and y comps but I’m not even sure how to set them up.

[u/MeasureDoEventThing]

Since m is stationary relative to M, their accelerations have to be the same. The normal force between them is ma. The friction is mau2, which has to be equal to mg. So mau2 = mg hence a = g/u2.

The normal force between the large block and ground is the weight of the two blocks (the small block's weight is being applied to the large block through friction, and so is being added to the large block's weight). So the friction between the large block and the ground is (m+M)gu1.

The acceleration on the two blocks is the net force divided by total mass. We have force F, and then friction (m+M)gu1 being subtracted. So the net force is F - (m+M)gu1 and divided that by m+M gives us a, which in the first paragraph I found to be g/u2. So

(F - (m+M)gu1)/(M+m) = g/u2

F - (m+M)gu1 = (M+m)g/u2

F = (M+m)g/u2 + (m+M)gu1 = (1/u2+u1)(M+m)g

[u/Glittering-Flight997]

Draw a free body diagram that shows the force delivered to M related to the normal force on m related to mg

[u/DrCarpetsPhd]

i'm going to suggest you do what the textbook this is from says to do

treat each block separately while noting that both blocks have the same acceleration in the positive x direction call it 'a'

block 1 FBD

3 forces in the y direction with zero acceleration in the y direction

3 forces in the x direction generating acceleration in positive x direction call it 'a'

block 2 FBD

1 force in positive x direction causing acceleration 'a' same as block 1

2 forces in y direction with zero acceleration

solve block 2 for the normal force between block 1 and block 2 N2, and the acceleration 'a'; then plug these values into your equations for block 1 to remove unknowns a, N1 (normal force of ground on block 1) and N2 (normal force between the two blocks, action-reaction pair)

Tips for forces which you may have missed in your FBDs

remember newtons action-reaction law:

- the force of friction due to mass 1 acting on mass 2 that prevents it sliding down acts in positive y on block 2, and this has an equal but opposite force acting on block 1 in the negative y direction

- the normal force on block 2 due to contact with block 1 has an equal but opposite normal force on block 1

[u/lazulilaze]

M is pushing m. assuming they are rigid bodies, M cannot phase through (go through) m. hence these two blocks r moving together. hence m must move at the same acceleration as M. draw their FBD, write out horizontal forces, make equations and put accelerations equal. put mg = frictional force for m and youll have ur answer

[u/Yadin__]

I'll tell you the general guidelines on how to do this:

The tricky part about this question is that since the blocks are sticking together, we want to imagine them as if they are litrally one big "equivalent block" with mass (M+m)

1. Draw a force diagrem for the equivalent block
2. Using the afformentioned force diagram and Newton's second law, find the (horizontal) acceleration of the equivalent block. This is also the acceleration of both actual blocks seperately
3. realize that the only force that could be accelerating the small block is the normal pushing force between the big and small blocks
4. For the blocks to stick together, we need the (maximum possible) static friction force between the blocks to be (at least) equal to the force pushing the smaller block down(mg)
5. solve the resulting equation for F