How Do You Calculate the Maximum Force on Stacked Blocks Without Slippage?

[Speedking96]

Homework Statement

Two masses M1 = 4.00 kg and M2 = 7.10 kg are stacked on top of each other as shown in the figure. The static coefficient of friction between M1 and M2 is μs = 0.430. There is no friction between M2 and the surface below it.

What is the maximum horizontal force that can be applied to M2 without M1 sliding relative to M2?

Homework Equations

F=ma

Ffriction = μFnormal

The Attempt at a Solution

For the 4 kg box:

Fnet = ma
Ff=ma
μmg = ma

μg=a
(0.43)(9.81) = a = 4.23 m/s/s

I understand that this must be the acceleration for the whole system, but when I calculate the max force, do I do it for the whole system's mass or for the 7.10 kg box only? Why or why not?

 

[physicsboy121]

Draw a free body diagram for the other box as well. Also, write the net force equations in the x and y directions.

 

[Speedking96]

Draw a free body diagram for the other box as well. Also, write the net force equations in the x and y directions.

For the y-direction:

∑forces = 0 ; since the object is not accelerating in that direction

For the x-direction:

Fnet= Fapp

Fapp=ma= (7.10)(4.23) = 16.88 Newtons

But, when we are applying the force to the lower box, aren't we essentially pushing the whole system? Therefore, shouldn't the mass of the whole system be considered?

 

[physicsboy121]

You are missing one force on the FBD you just drew. And instead of F normal of the system, I suggest that you think it as F12 (i.e. force of mass 1 onto mass 2) Also the friction force on the FBD of your initial post should be pointing in the other direction.

 

[Speedking96]

You are missing one force on the FBD you just drew. And instead of F normal of the system, I suggest that you think it as F12 (i.e. force of mass 1 onto mass 2) Also the friction force on the FBD of your initial post should be pointing in the other direction.

Is it the static friction of the top box?

 

[physicsboy121]

correct

 

[Speedking96]

correct

Fnet=ma
Fapp-Ff = ma

Fapp - μmg = ma

Fapp = ma + μmg

= (7.1)(4.23) + (0.43)(4)(9.81) = 46.91 Newtons ?

 

[Speedking96]

I don't know how to approach problems that deal with multiple masses. Frustrating.