Block on Block Dynamic Question

[wby]

I was wondering if anyone knows the answer to this question:

Homework Statement

consider 2 blocks of masses m1 and m2. the block of mass m1 is resting on top of the block m2 which in turn is resting upon a horizontal surface. The coefficient of static friction between the 2 blocks is u1 and the coefficient of the kinetic friction between the lower block and the horizontal surface is u2. A force, F, is being applied to the bottom block at angle0 to the horizontal, and the 2 blocks are moving together in the horizontal direction.

Homework Equations

a. Draw a free body diagram (Best if drawn, but I think I can still manage without it)

b. write equations representing Newton's 2nd law for each of the above diagrams. For each diagram add equation(s) representing the relationship between frictional and normal forces

c. wut is the acceleration of the bottom block?

d. What is the maximum force, F max, that can be applied before the top block slips?

The Attempt at a Solution

No idea...

Thanks a lot for the help!

 

[tiny-tim]

Welcome to PF!

Hi wby! Welcome to PF!

Show us what you've tried, and where you're stuck, and then we'll know how to help.

 

[baba89]

I would be happy to assist you with this question. First, let's start by drawing a free body diagram for the system. This will help us visualize the forces acting on the blocks and understand the problem better.

Next, let's apply Newton's second law to each block. For the top block, we have F - u1N = m1a, where F is the applied force, u1 is the coefficient of static friction, N is the normal force, m1 is the mass of the top block, and a is the acceleration of the system.

For the bottom block, we have u1N - u2N = m2a, where u2 is the coefficient of kinetic friction and all other variables are the same as above.

To find the acceleration of the bottom block, we can solve these equations simultaneously. Since we know that the blocks are moving together, we can set their accelerations to be the same (a = a1 = a2). This will give us a value for the acceleration.

To find the maximum force that can be applied before the top block slips, we need to use the concept of limiting friction. This means that the applied force (F) must equal the maximum frictional force that can be exerted (u1N). So, we can rearrange the equation for the top block to solve for F max, which will give us the maximum force that can be applied before the top block slips.

I hope this helps answer your questions and provides a solution to the problem. If you have any further questions or need clarification, please don't hesitate to ask. Good luck!