# Problem involving kinetic friction and relative accelerations

1. Jun 1, 2008

### tfmfyn

1. The problem statement, all variables and given/known data
I am having trouble with a problem from Physics for Scientists and Engineers by Tipler and Mosca, 6th edition, 5.56

A 100-kg mass is pulled along a frictionless surface by a horizontal force F such that its acceleration is a1=6.00m/s^2. A 20.0-kg mass slides along the top of the 100-kg mass and has an cceleration of a2=4.00m/s^2. (It thus slides backward relative to the 100-kg mass. (a)What is the frictional force exerted by the 100-kg mass on the 20.0 kg mass? (B)What is the net force acting on the 100-kg mass? What is the force F? (C) After the 20.0-kg mass falls off the 100-kg mass, what is the acceleration of the 100-kg mass? (Assume that the force F does not change).

2. Relevant equations
For my free body diagram, postive x and y are to the right and up, respectively.
I attempted to solve this problem using relative acceleration and setting a=0.
a=a1-ar
a=a2+ar
a1=6.00, a2=4.00. Therefore, ar=1.00m/s^2 (?)

3. The attempt at a solution
(a) Fnetm2y=Fn- m2 g=0
So Fn=m2g
Fnetm2x=m2a2
-fk=-$$\mu$$k m2 g=m2a2
-$$\mu$$kg=a2=a-ar
Therefore, $$\mu_{}$$k=.1019
And fk=(.1019)(20kg)(9.81m/s^2)=20.0N (?)

(b)
Force F=(m1+m2)ar=(120kg)(1.00m/s^2=120N (?)

(c) F=m ar when m2 falls off
so 120N=(100kg)ar
ar=1.20m/s^2
ar=a1-a
a=4.80m/s2

I am not sure about my answers because I don't know whether or not I used the relative acceleration concept correctly. Are my answers correct? Is this the only way to do the problem? Is there a way that it can be solved without using relative accelerations?

I googled this problem and found a site that solved it very differently and got different answers. I am not allowed to insert the link here. It did not use relative accelerations.

Any comments and/or answers would be very much appreciated. Thanks!!

Last edited: Jun 1, 2008
2. Jun 1, 2008

### Raze2dust

I haven't gone through your solution yet..but i think you dont need relative accelaration..

100 kg block moves with a= +6 m/s^2 .. So you can easily calculate NET force on 100 kg block(say this is Fnet)..Block on the top moves with 4 m/s^2...and its mass is given so you can fin the net force on it.
This force(say f) must be equal to the frictional force, since thats the only force acting on the block..right? The top box also exerts the same force f in the backward direction on the 100 kg block..So

Fnet=F-f, where F is the force we apply on the bottom block

try it by this method..

3. Jun 2, 2008

### tfmfyn

OK, I've looked over the problem again and realized my first attempt was wrong. Here is the revision:

m1=100kg
m2=20kg
a1=6.00m/s^2
a2=4.00m/s^2
Also, I used F=Fapplied and f=Ffriction for my variables.

(a)
$$\Sigma$$Fnetm2y=Fn-(m2)(g)=0
So Fn=(m2)(g)
$$\Sigma$$Fnetm2x=m2a2
-fk=-$$\mu$$m2g=-m2a2
-$$\mu$$g=-6.0+2.0
$$\mu$$-.408
so f=$$\mu$$m2g=80N

(b)
$$\Sigma$$netx=Fapplied-f=m1a1
$$\Sigma$$netx=m1a1=(100kg)(6.00m/s^2)=600N

(C)
$$\Sigma$$Fnet=Fapplied-f=m1a1
Fapplied=600N+80N=680N

What is really confusing me is why the force of friction that m2 resting on m1 experiences in the negative x direction (to the left) plays a factor in the net force acting on m1. Why is Fnetm1=Fapplied-f and not just Fnetm1=Fapplied? It makes sense because Newton's 3rd law says that every force has an equal and opposite force, but it seems kind of counterintuitive. Does it have to do with the concept of center of mass and how there is one particle on which all forces act?

4. Jun 3, 2008

### Raze2dust

Think about it this way...suppose both the blocks were stuck with glue..then your applied force will act on M=(m1+m2) right?
And their accelaration will be (F/M)..and they'll move together. Now in this case consider only the top block.. its accelaration is F/M . So net force on it has to be (F/M)*m1 and not F. So you say that friction exerts backward force=F-(F/M)*m1