# Applied Force ,Kinetic Friction, Static Friction involving two stacked masses

by abpandanguyen
Tags: applied, force, friction, involving, kinetic, masses, stacked, static
 P: 33 1. The problem statement, all variables and given/known data A block of mass m1 = 2.0 kg is stacked on top of a block of mass m2 = 3.0 kg which is resting on top of a table Both blocks and the table are made of the same material; so the coefficients of friction between all surfaces are the same They are (attempt at a mew) ms = .60 and mk = .40. A constant force F is applied horizontally to m2. When F is small, the two blocks move together with the same acceleration. At some point as F is increased, the top block will no longer move with m2. a. What is the minimum force Fmin that is needed to start the two blocks moving? b. What is the maximum force Fmax that allows the two blocks to move together? 2. Relevant equations F=ma Fs = msn Fk = mkn 3. The attempt at a solution For a, I just added the two masses together and multiplied by g to get a total N then multiplied that by the coefficient of static friction to get the force required to get the two blocks moving. I hope that is right. For b, I am having a difficult time imagining what m1 (mass on top) is doing when the horizontal force applied to m2 (mass on bottom) gets larger. I'm assuming this problem requires the use of F=ma. I'm having trouble picturing all the forces in motion here as the horizontal force on m2 increases, particularly what is happening to the mass on top. Any help would be much appreciated! Please and thank you! Sorry I didn't make a picture for this. I'm on a computer where paint isn't even an option. >__<;;
 PF Patron HW Helper P: 3,440 Part (a) is correct. If you want to picture part (b), stack two books on top of another and put them on a table. Now push gently on the bottom book with your finger. What does the top book do? It starts from rest and moves together with the bottom book as if it were glued on it. In other words, the acceleration of the top book is the same as the acceleration of the bottom book. Now push on the bottom book a bit harder. The same thing happens. Whats going on? The force of static friction between the books adjusts itself to provide to the top book the acceleration that your hands have given to the bottom book. Now grab the bottom book with both hands and yank as hard as you can. The top book will fall off the bottom book. What's going on now? The static friction friction between books is not enough to provide the same acceleration to the top book that the bottom book is given by your hands. So the top book will slide when the acceleration that the external force (your hand) provides to the two book system is such that the net force on the top book is larger than the maximum value of static friction between books.
P: 33
 Quote by kuruman So the top book will slide when the acceleration that the external force (your hand) provides to the two book system is such that the net force on the top book is larger than the maximum value of static friction between books.
the maximum value of static friction between the books? I get the concept but I don't quite get what I would be doing mathematically. What is the equation for what I am solving for going to look like?

PF Patron
HW Helper
P: 3,440

## Applied Force ,Kinetic Friction, Static Friction involving two stacked masses

Here are the steps that you must complete to get to that equation.

1. Find an expression for the common acceleration of the two masses.
2. Use the common acceleration to find the force of static friction on the top mass which is the net force acting on the top mass. Note that your expression predicts that as the pushing force increases, the force of static friction must increase.
3. Read in your textbook about static friction. It has a maximum value beyond which it can grow no more. That is the maximum value of static friction. Set that value equal to the net force on the top mass.
4. Solve for the pushing force.
P: 33
 Quote by kuruman 1. Find an expression for the common acceleration of the two masses. 2. Use the common acceleration to find the force of static friction on the top mass which is the net force acting on the top mass. Note that your expression predicts that as the pushing force increases, the force of static friction must increase.
Okay, so I have the common acceleration for the two masses. How am I going to be finding the force of static friction for the top mass with it? Is my expression going to look like some form of F=ma with a slight change in the mass or am I thinking of this as if the masses were adjacent to each other horizontally versus vertically as in this problem.
 P: 367 you know that MuS = net force onto the top block, which is the bottom block pushing/rubbing onto the top block. So if you know the total F=ma for the bottom block, and you also know MuS, solve for the only other force
PF Patron
HW Helper
P: 3,440
 Quote by abpandanguyen Okay, so I have the common acceleration for the two masses. How am I going to be finding the force of static friction for the top mass with it? Is my expression going to look like some form of F=ma with a slight change in the mass or am I thinking of this as if the masses were adjacent to each other horizontally versus vertically as in this problem.
How many and what forces are acting on the top mass in the horizontal direction?
 P: 33 Okay, I think what I have been confused no is that the only force acting on the top book is a force of friction and also that it is acting in the same direction as the acceleration until that main force pulling on the bottom force gets larger. It is hard for me to think of the force of friction acting in the same force as the pushing/pulling force on the bottom block, but I guess it makes sense since SOMETHING had to be moving it. Well, knowing that the answer seems to come out a lot easier. You would have the force of static friction on the top block, and then with that you could find the acceleration. Once you have that acceleration, you would plug that in to the F=ma equation for the bottom block factoring in kinetic friction and you would be solving for F and that should give the maximum force for them to move together.
PF Patron
HW Helper
P: 3,440
 Quote by abpandanguyen Okay, I think what I have been confused no is that the only force acting on the top book is a force of friction and also that it is acting in the same direction as the acceleration until that main force pulling on the bottom force gets larger. It is hard for me to think of the force of friction acting in the same force as the pushing/pulling force on the bottom block, but I guess it makes sense since SOMETHING had to be moving it. Well, knowing that the answer seems to come out a lot easier. You would have the force of static friction on the top block, and then with that you could find the acceleration. Once you have that acceleration, you would plug that in to the F=ma equation for the bottom block factoring in kinetic friction and you would be solving for F and that should give the maximum force for them to move together.
You got it backwards. First you find the common acceleration of the masses. Then you say that the force of static friction between the two masses is the net horizontal force on the top mass. Therefore

fs=mTopa

Now you see that, as you increase the pushing force, the acceleration increases which in turn means that the force of static friction must increase if the masses are to be moving as one. Of course, the static friction cannot increase forever. When it reaches its maximum value, the top mass will start sliding.
 P: 33 Yeah I think I did what you said, not what I said. Thanks much!

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