Challenging Problem - System of Masses - Try It

[coolchewy]

Homework Statement

A 0.5 kg wooden block is placed on top of a 1.0 kg wooden block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping?

Homework Equations

Fnet = Ff + Fg + Fapplied
Ff= Mew(Fn)

The Attempt at a Solution

So I tried doing this question many different ways and couldn't figure out the answer. The answer is supposed to be 8.1N. I find that there are too many variables to solve for. Thanks for the help I really appreciate it!

 

[berkeman]

Homework Statement

A 0.5 kg wooden block is placed on top of a 1.0 kg wooden block. The coefficient of static friction between the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20. What is the maximum horizontal force that can be applied to the lower block without the upper block slipping?

Homework Equations

Fnet = Ff + Fg + Fapplied
Ff= Mew(Fn)

The Attempt at a Solution

So I tried doing this question many different ways and couldn't figure out the answer. The answer is supposed to be 8.1N. I find that there are too many variables to solve for. Thanks for the help I really appreciate it!

We don't do your homework for you here on the PF. We can offer tutorial assistance if you show us your work so far.

Show us the equations that you are setting up, and how you are approaching the problem. What do your free body diagrams look like for each of the blocks?

 

[coolchewy]

Okay so for the first block 0.5 kg block I only have a force of friction and force applied because the force of gravity and the force of normal cancel out. Then for the second block 1.0 kg I have a force of normal, a force of friction, force of gravity and a force applied. The thing is that I have a force of normal coming out of the top but I am not sure if its right because the 0.5 block is acting on it. The first equation I used was that for the first block the Fnet = Ff + Fa, because the question doesn't give acceleration I was stuck. For the second block the equation I used was Fnet = Ff+ Fg+Fn+Fa and again I got stuck because there was no acceleration. I found out that the force of friction for the first 0.5 block is 1.72 N and the second 1kg block to be 2.94 N. This is what I have so far.