Newton's 3rd law and kinetic friction

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Homework Help Overview

The discussion revolves around a physics problem involving Newton's third law and kinetic friction. The scenario includes a box resting on a board, with the board on a frictionless surface, and participants are tasked with determining the minimum force required to pull the board from under the box without the box sliding off.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between the accelerations of the box and the board, questioning how to correctly apply Newton's laws and free body diagrams (FBDs). There are attempts to express the forces acting on both the box and the board, with some participants noting the need to separate the masses involved in their calculations.

Discussion Status

There is an ongoing exploration of the correct application of forces and accelerations in the context of the problem. Some participants have offered guidance on how to set up the FBDs correctly, while others are working through their misunderstandings and attempting to clarify their reasoning. The discussion is productive, with participants actively engaging in refining their approaches.

Contextual Notes

Participants are grappling with the implications of static and kinetic friction, as well as the necessity of distinguishing between the forces acting on the box and the board. There is a recognition of the need for practice with similar problems to solidify understanding.

sunbunny
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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.
 
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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[/color]

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 between 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!
 
  • #10
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!
 
  • #11
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!
 
  • #12
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!
 
  • #13
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!
 
  • #14
oh wow I'm sorry i didn't mean to put that many up, i don't know what happened
 
  • #15
sunbunny said:
oh wow I'm sorry i didn't mean to put that many up, i don't know what happened
[tex]you're welcome^5[/tex]!
 

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