Problem involving kinetic friction and relative accelerations

In summary, the 100 kg block is pulled along a frictionless surface by a horizontal force F and a 20 kg mass slides along the top of the 100 kg block. The 100 kg block exerts a frictional force on the 20 kg mass.
  • #1
tfmfyn
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Homework Statement


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).

Homework 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 (?)

The Attempt at a Solution


(a) Fnetm2y=Fn- m2 g=0
So Fn=m2g
Fnetm2x=m2a2
-fk=-[tex]\mu[/tex]k m2 g=m2a2
-[tex]\mu[/tex]kg=a2=a-ar
Therefore, [tex]\mu_{}[/tex]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!
 
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  • #2
I haven't gone through your solution yet..but i think you don't 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 that's 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
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)
[tex]\Sigma[/tex]Fnetm2y=Fn-(m2)(g)=0
So Fn=(m2)(g)
[tex]\Sigma[/tex]Fnetm2x=m2a2
-fk=-[tex]\mu[/tex]m2g=-m2a2
-[tex]\mu[/tex]g=-6.0+2.0
[tex]\mu[/tex]-.408
so f=[tex]\mu[/tex]m2g=80N

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

(C)
[tex]\Sigma[/tex]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
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
 

1. What is kinetic friction and how does it affect relative accelerations?

Kinetic friction is a type of friction that occurs between two surfaces in motion relative to each other. It is caused by the microscopic irregularities on the surfaces and acts in the direction opposite to the motion. This frictional force can affect the relative accelerations between the two surfaces by slowing down or changing the direction of their motion.

2. How is the coefficient of kinetic friction related to relative accelerations?

The coefficient of kinetic friction is a measure of the roughness or smoothness of the two surfaces in contact. It is directly proportional to the frictional force and therefore, affects the relative accelerations between the two surfaces. A higher coefficient of kinetic friction means a stronger frictional force, resulting in slower or more significant changes in relative accelerations.

3. Can kinetic friction be reduced or eliminated to improve relative accelerations?

Kinetic friction can be reduced by using smoother surfaces or adding a lubricant between the two surfaces. However, it cannot be completely eliminated. In fact, some amount of kinetic friction is necessary for objects to stay in motion and maintain a constant relative acceleration.

4. How does the mass of an object affect the relative accelerations in the presence of kinetic friction?

The mass of an object does not directly affect the relative acceleration in the presence of kinetic friction. However, a heavier object will experience a stronger frictional force, resulting in a slower or more significant change in its relative acceleration compared to a lighter object.

5. What other factors can influence the relative accelerations in a problem involving kinetic friction?

In addition to the coefficient of kinetic friction and the mass of the object, other factors that can influence relative accelerations in a problem involving kinetic friction include the angle of the surfaces, the applied forces, and the presence of other external forces such as air resistance or gravity.

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