# Block sliding on another block (static friction problem)

1. A block of mass m2 = 4.00 kg block is placed on top of another block of mass m1 = 12.0 kg block that rests on a frictionless, horizontal table. The coefficient of static friction between the two blocks is 0.600.
http://www.flickr.com/photos/43001762@N07/3963579065/

(a) What is the maximum horizontal force that can be applied before the 4.00 kg block begins to slip relative to the 12.0 kg block, if the force is applied to the more massive block?

(b) What is the maximum horizontal force that can be applied before the 4.00 kg block begins to slip relative to the 12.0 kg block, if the force is applied to the less massive block?

2. Homework Equations : F=ma; Force from friction= meu(Normal force);

3. Part a): I got the right answer for this part. I set up a free body diagram for the small block and figured out what acceleration would overcome the force of friction. Then multiplied that acceleration times the weight of both blocks. So something like this:

.600(Normal Force) = ma
Normal Force = mg
.600mg=ma
a=5.88 m/s^2
5.88(12+4)=94.1 N (which is correct)

Part b)" I tried to set up a free body diagram for the small block again (this time including the force P).
i have -force from Friction +P (force applied to block) = m(both blocks)a(acceleration from part a)
here is my equation:
-(.600)(m2g)+p=m(total)a
Solve for p and get 118 N which is wrong.

kuruman
Homework Helper
Gold Member
here is my equation:
-(.600)(m2g)+p=m(total)a
Hi axlr0se, welcome to PF.

There is a problem with the equation quoted above.

You say this comes from the free body diagram of the top mass. Ok, then your system is the top mass m2 not mtotal.

Last edited:
Hi axlr0se, welcome to PF.

There is a problem with the equation quoted above.

You say this comes from the free body diagram of the top mass. Ok, then your system is the top mass m2 not mtotal.

First of all thanks for responding, I had given up on this problem. So If I plug in 4 instead of 16, like you suggest P comes out to 47.04 N, which is also wrong.

This leads me to suggest that you do NOT use 5.88m/s^2 for the acceleration. (the acceleration from part a). I have two more guesses on my homework, and I keep coming to the same answers I've already tried. Seriously if anyone has ANY thoughts whatsoever. They would be greatly appreciated. I have utterly given up on this problem, physics has broken my brain.

I FINALLY GOT THE RIGHT ANSWER!!! unfortunately I have no idea why it's right. If somebody could please explain it to me. So what I did to get the right answer was set up a free body diagram for the top box first, with the force pushing to the right (P) and friction pushing to the left (-Ff) and set them equal to zero (again I have no idea why this works) so P=uM2G=23.52N.
Then I drew another Free body diagram for the larger box, and put the mystery force from above (23.52N) pointing to the right, and since there's nothing to the left, I set it equal to M1a and solved for acceleration (1.96)
Then i multiplied that by 16 (m2+m1) and got 31.36 N, which is correct.
So my math went something like this:
FBD top box
Fx: P-uM2G=0
P=uM2G
P=23.52N
FBD bottom box
Fx:23.52N=M1a
a=1.96m/s^2
and finally

1.96(M2+M1)=31.4 N

kuruman
Homework Helper
Gold Member
You can't use the same acceleration in part (b) because you are calculating a new pulling force which means that there should be a new acceleration. The "mystery" force is the force of static friction. It acts to the left on the top mass, but on the bottom mass it acts to the right and has the same magnitude. That's because according to Newton's 3rd Law forces come in pairs.

Perhaps an easier way to understand the solution is this way. First you draw a free body diagram for both masses. Force P (whatever it is) is the net force acting on the two masses, so the acceleration is

a = P/(m1+m2)

If the force is pulling on the bottom mass, then static friction is the net force on the top mass. The max value of this is μsm2g. So for part (a)

μsm2g=m2a=m2P/(m1+m2)

If the force is pulling on the top mass you follow the same procedure noting that the acceleration is given by the same expression

a = P/(m1+m2)

But now static friction is the net force on the bottom mass so

μsm2g=m1a=m1P/(m1+m2)

In either case you solve for P.

You can't use the same acceleration in part (b) because you are calculating a new pulling force which means that there should be a new acceleration. The "mystery" force is the force of static friction. It acts to the left on the top mass, but on the bottom mass it acts to the right and has the same magnitude. That's because according to Newton's 3rd Law forces come in pairs.

Perhaps an easier way to understand the solution is this way. First you draw a free body diagram for both masses. Force P (whatever it is) is the net force acting on the two masses, so the acceleration is

a = P/(m1+m2)

If the force is pulling on the bottom mass, then static friction is the net force on the top mass. The max value of this is μsm2g. So for part (a)

μsm2g=m2a=m2P/(m1+m2)

If the force is pulling on the top mass you follow the same procedure noting that the acceleration is given by the same expression

a = P/(m1+m2)

But now static friction is the net force on the bottom mass so

μsm2g=m1a=m1P/(m1+m2)

In either case you solve for P.

Thanks, it makes sense now that you spell it out for me. Man I'm bad at this stuff!

kuruman