Friction, Mass and Acceleration: Analyzing Block Motion

[as2528]

Homework Statement

Two blocks of mass m2 and m1 are put on a frictionless level surface as shown in the figure below. The static coefficient of friction between the two blocks is µ. A force F acts on the top block m2.

(c) Find the magnitude of the force F above which the block m1 starts to slide relative to the block m1.

Relevant Equations

F=ma
F=u*N

The block starts to slide if friction can no longer hold the block.

F=u*n and F=(m1+m2)a
so: (m1+m2)a=uN=>am1+am2=uN=>am2=(uN)/(am1)

So:am2=(uN)/(am1) is the force.

The answer is F=(u*m1g(m1+m2))/m2
I do not see how the acceleration terms are canceled. Is my answer equivalent to this?

 

[kuruman]

There is no figure below. Please post it.

 

[as2528]

There is no figure below. Please post it.

Added it!

 

[kuruman]

Please use the "Attach files" link on the lower left to put the figure in your post. It is inconvenient for people to download pdf files just to look at a figure. You will have to do that because you need to post your free body diagram that resulted in your equation.
Can you identify what each "F" in your equation stands for and explain your reasoning?
F=u*n and F=(m1+m2)a

(Fill in the blanks)
The first F stands for _________________________

The second F stands for _________________________

I set the two equal because _____________________________

 

[as2528]

Please use the "Attach files" link on the lower left to put the figure in your post. It is inconvenient for people to download pdf files just to look at a figure. You will have to do that because you need to post your free body diagram that resulted in your equation.
Can you identify what each "F" in your equation stands for and explain your reasoning?
F=u*n and F=(m1+m2)a

(Fill in the blanks)
The first F stands for _________________________

The second F stands for _________________________

I set the two equal because _____________________________

I have now attached the pictures. The first F is for Newton's Second Law. The second F is for the force of friction. When the force of friction breaks, and I set it equal to Newton's Second Law, I can find the magnitude of the force acting on object 2 at that instant.

 

[kuruman]

Do you understand how the first equation in part (c) was put together? The acceleration does not "cancel out". It is replaced by its value found in part (b). Just before the two masses start sliding relative to each other, they have the common acceleration from part (b). In your expression you left the acceleration as $a$ which doesn't do much for you. The two masses have common acceleration $a=\frac{F}{m_1+m_2}$ until the top mass starts sliding on the bottom mass.

 

[as2528]

Do you understand how the first equation in part (c) was put together? The acceleration does not "cancel out". It is replaced by its value found in part (b). Just before the two masses start sliding relative to each other, they have the common acceleration from part (b). In your expression you left the acceleration as $a$ which doesn't do much for you. The two masses have common acceleration $a=\frac{F}{m_1+m_2}$ until the top mass starts sliding on the bottom mass.

I see now. I did not realize that, I understand now. Thanks!