# Newton's 3rd law and kinetic friction

• sunbunny
In summary, the conversation is about solving a problem involving a small box and a board on a frictionless surface. The goal is to find the minimum constant force needed to pull the board out from under the box so that the box falls off the opposite end of the board. The conversation includes discussions about the equations and variables involved, as well as clarifications on how to properly draw a free body diagram for the board and the box. The final solution involves rearranging equations and substituting values to solve for the applied force.
sunbunny
Hi, I'm having problems solving this problem:

A small box of mass m1 is sitting on a board of mass m2 with length L . The board rests on a frictionless horizontal surface. The coefficient of static friction between the board and the box is us. The coefficient of kinetic friction between the board and the box is us.

Throughout the problem, use for the magnitude of the acceleration due to gravity.

Find Fmin., the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).

Express your answer in terms of some or all of the variables us ,m1 ,m2 ,g ,L and . Do not include Force of friction in your answer.

This is what I've done:

I know that acceleration of board must be larger than the acceleration of the box.

I've found the acceleration of the box to be a=Ff/m1 which then equals

a= usg

now i need to find the acceleration of the board.

I've found that Fx=F-Ff

now i need to find the acceleration of the board.

I changed the Ff into usm2g

F-usm2g=max

then i changed m into m1 and m2 because the board has both masses

f-usm2g=m1m2a

then for a I got:

a=f-usm2g/m1m2

however this is wrong. if anyone can help me with this that would be great thanks

sunbunny said:
a= usg

now i need to find the acceleration of the board.

I've found that Fx=F-Ff

now i need to find the acceleration of the board.

I changed the Ff into usm2g , incorrect, think Newton 3 for the friction force between box and board

F-usm2g=max

then i changed m into m1 and m2 because the board has both masses

f-usm2g=m1m2a

then for a I got:

a=f-usm2g/m1m2

however this is wrong. if anyone can help me with this that would be great thanks
You are mixing up your free body diagrams. If you are going to isolate the board, do not include the mass of the block. If you are going to isolate the system, do not include the friction force betwen the block and board.

So for the board, would i just include a normal force, weight force, and the force applied

sunbunny said:
So for the board, would i just include a normal force, weight force, and the force applied
No, if you're just looking at the board in a FBD, you have the applied force acting right, and the friction force between box and board acting left. You have the weight and normal force also, but you don't need them here to solve. The masss is just m2. Alternatively, look at the block and board system as a FBD. then you just have the applied force to consider, but the mass is (m1 + m2.)

okay so would this make the horizontal component of the FBD for the board:

max=Fapplied,x - Ffriction,x

(m1+m2)a=Fapplied,x-usm2g,x

a=Fapplied,x-usm2g,x/(m1+m2)

sunbunny said:
okay so would this make the horizontal component of the FBD for the board:

max=Fapplied,x - Ffriction,x

(m1+m2)a=Fapplied,x-usm2g,x

a=Fapplied,x-usm2g,x/(m1+m2)
Ah, there you go again mixing up your FBD's. For the board, m2a= Fapplied - usm1g
For the box, m1a = usm1g
and check your work with a FBD of the system, (m1 +m2)a = Fapplied

Now you've got more than enough equations to solve for F. Note that for applied forces greater than F, the block will start to slide. This is sometimes not easily understood.

okay, I'm starting to see how it's done and how I've mixed up my FBD. I think I just need to practice some more problems.

I rearanged the m1a=usm1g to a=usm1g/m1

i then substituted this value for a into:

Fapplied=m2a+usm1g

Thanks so much for your help, I really appreciated it! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!

okay, I'm starting to see how it's done and how I've mixed up my FBD. I think I just need to practice some more problems.

I rearanged the m1a=usm1g to a=usm1g/m1

i then substituted this value for a into:

Fapplied=m2a+usm1g

Thanks so much for your help, I really appreciated it! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!

okay, I'm starting to see how it's done and how I've mixed up my FBD. I think I just need to practice some more problems.

I rearanged the m1a=usm1g to a=usm1g/m1

i then substituted this value for a into:

Fapplied=m2a+usm1g

Thanks so much for your help, I really appreciated it! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!

okay, I'm starting to see how it's done and how I've mixed up my FBD. I think I just need to practice some more problems.

I rearanged the m1a=usm1g to a=usm1g/m1

i then substituted this value for a into:

Fapplied=m2a+usm1g

Thanks so much for your help, I really appreciated it! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!

okay, I'm starting to see how it's done and how I've mixed up my FBD. I think I just need to practice some more problems.

I rearanged the m1a=usm1g to a=usm1g/m1

i then substituted this value for a into:

Fapplied=m2a+usm1g

Thanks so much for your help, I really appreciated it! Now I'm going to try some more practice problems while it's fresh in my mind. thanks again!

oh wow I'm sorry i didn't mean to put that many up, i don't know what happened

sunbunny said:
oh wow I'm sorry i didn't mean to put that many up, i don't know what happened
$$you're welcome^5$$!

## What is Newton's 3rd law?

Newton's 3rd law states that for every action, there is an equal and opposite reaction. This means that when an object exerts a force on another object, the second object will exert an equal and opposite force back on the first object.

## How does Newton's 3rd law apply to kinetic friction?

When an object is moving across a surface, there is a force called kinetic friction that acts in the opposite direction of the object's motion. According to Newton's 3rd law, the object will exert an equal and opposite force back on the surface due to this kinetic friction.

## What is the relationship between Newton's 3rd law and the coefficient of kinetic friction?

The coefficient of kinetic friction is a measure of how much force is needed to keep an object moving across a surface. According to Newton's 3rd law, the force exerted by the object on the surface due to kinetic friction is equal to the force exerted by the surface on the object. Therefore, an increase in the coefficient of kinetic friction will result in an increase in the force needed to maintain the object's motion.

## Can Newton's 3rd law be violated?

No, Newton's 3rd law is a fundamental law of physics and has been observed to hold true in all physical interactions. It cannot be violated.

## How does Newton's 3rd law affect the motion of objects?

Newton's 3rd law plays a crucial role in determining the motion of objects. The equal and opposite forces exerted on each object involved in an interaction will result in a net force of zero, causing the objects to either remain at rest or continue moving at a constant velocity.

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