Finding Constant Force to Pull Board from Mass 1 Box

[Toranc3]

Homework Statement

A small box of mass 1 is sitting on a board of mass 2 and 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, as usual, less than Us.

Find 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).

Homework Equations

F=ma

The Attempt at a Solution

I have a couple of questions:
1. The acceleration does not go to 0 in the x direction because the board is pulled with a constant force and therefore a constant acceleration because force and acceleration are both constant. Is this right? Or do they go to zero because they were at rest?(Inertia, Newton's first law)

[url=http://www.freeimagehosting.net/aavu4][PLAIN]http://www.freeimagehosting.net/t/aavu4.jpg[/url][/PLAIN]

 

[SammyS]

Homework Statement

A small box of mass 1 is sitting on a board of mass 2 and 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, as usual, less than Us.

Find 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).

Homework Equations

F=ma

The Attempt at a Solution

I have a couple of questions:
1. The acceleration does not go to 0 in the x direction because the board is pulled with a constant force and therefore a constant acceleration because force and acceleration are both constant. Is this right? Or do they go to zero because they were at rest?(Inertia, Newton's first law)

[url=http://www.freeimagehosting.net/aavu4][PLAIN]http://www.freeimagehosting.net/t/aavu4.jpg[/url][/PLAIN]

There are two objects involved, each with potentially different acceleration. Which object's acceleration are you asking about?

 

[Toranc3]

There are two objects involved, each with potentially different acceleration. Which object's acceleration are you asking about?

I meant that the board is pulled with a constant force and therefore a constant acceleration because force and acceleration are both proportional. Well for the box its zero right? But for the board, is the acceleration constant too? it is being pulled. or is the acceleration for the box and board zero because they were at rest?

also are my body diagrams correct?

 

[SammyS]

I meant that the board is pulled with a constant force and therefore a constant acceleration because force and acceleration are both proportional.

Yes.

Well for the box its zero right?

Isn't there a net force (on the box) to the right?

But for the board, is the acceleration constant too? it is being pulled. or is the acceleration for the box and board zero because they were at rest?

also are my body diagrams correct?

Your diagrams look good enough to answer this question. I'm not sure what all of your symbols mean.

 

[Toranc3]

Sorry, I labeled the box A with mass m1 and board is B with mass m2. N is my normal force, w is my weight, fs is my static friction force in the opposite direction of motion, and F is my force towards motion. When I do the body diagram for the board do i have to account the weight of the box? or is that only if i do a system body diagram? I trying to solve this by doing individual body diagrams.

 

[SammyS]

Sorry, I labeled the box A with mass m1 and board is B with mass m2. N is my normal force, w is my weight, fs is my static friction force in the opposite direction of motion, and F is my force towards motion. When I do the body diagram for the board do i have to account the weight of the box? or is that only if i do a system body diagram? I trying to solve this by doing individual body diagrams.

What you have in your diagrams looks correct to me.

Added in Edit:
What is the maximum force which can be exerted on the Box?

 

[Toranc3]

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What you have in your diagrams looks correct to me.

Added in Edit:
What is the maximum force which can be exerted on the Box?

Well I get the correct answer when I do not include my normal force of a on b for the board. But I can't figure it out when I am including it. Why?

Here is my work:

the box is just m1=1
board is m2=2

For the box
ƩFy: N(2on1) -W(1)=0 nothing going on in the verical
ƩFx: fs(2on1) =m(1)*ax

This turns out to be us*g=ax I substituted my n from my summation of forces in my y to my n from my summation of forces of x on fs(2on1)...fs=us*n


For the board
ƩFy: N(2)-W(2)-N(1on2)=0 Nothing going on in the vertical
ƩFx: F-fs(1on2)=m2*ax

What am I doing wrong here?

 

[SammyS]

Ʃ

Well I get the correct answer when I do not include my normal force of a on b for the board. But I can't figure it out when I am including it. Why?

Here is my work:

the box is just m1=1
board is m2=2

For the box
ƩFy: N(2on1) -W(1)=0 nothing going on in the verical
ƩFx: fs(2on1) =m(1)*ax

This turns out to be us*g=ax I substituted my n from my summation of forces in my y to my n from my summation of forces of x on fs(2on1)...fs=us*n

For the board
ƩFy: N(2)-W(2)-N(1on2)=0 Nothing going on in the vertical
ƩFx: F-fs(1on2)=m2*ax

What am I doing wrong here?

You're doing nothing wrong if you're analyzing the situation of the maximum force, F, which can be applied without slipping.

For a force, F⁺, just very slightly greater than this F, the frictional force (kinetic friction) will be μ_km₁g, and a₁ ≠ a₂ . μ_k < μ_s .

 

[Toranc3]

You're doing nothing wrong if you're analyzing the situation of the maximum force, F, which can be applied without slipping.

For a force, F⁺, just very slightly greater than this F, the frictional force (kinetic friction) will be μ_km₁g, and a₁ ≠ a₂ . μ_k < μ_s .

So i am doing is finding the max force but I need the minimum force. How would I change things differentl?. I know fs is less than or equal to us*n but how would I apply that in this case?

 

[Toranc3]

I am only stuck in the part where I need to substite my n in my fs=us*n for the board.. I want to know why is it that when I take n1on2 out of my body diagram i can get the answer. But it should be there because of Newtons third law.

 

[SammyS]

So i am doing is finding the max force but I need the minimum force. How would I change things differently?. I know fs is less than or equal to us*n but how would I apply that in this case?

Well, you found the greatest force, for which the book doesn't slide on the board. the minimum force for which it will slide on the board is infinitesimally larger than the max force with no slipping --- so they are essentially the same value.

The big difference is that the frictional force produced by kinetic friction will be slightly less than the maximum that can be produced by static friction. So the box will have a little bit smaller acceleration.

What about the acceleration of the board?

 

[Toranc3]

Well, you found the greatest force, for which the book doesn't slide on the board. the minimum force for which it will slide on the board is infinitesimally larger than the max force with no slipping --- so they are essentially the same value.

The big difference is that the frictional force produced by kinetic friction will be slightly less than the maximum that can be produced by static friction. So the box will have a little bit smaller acceleration.

What about the acceleration of the board?

That is where I get stuck. So i am not sure if I have my body diagram correct. I have this for the board: F-fs1on2=m*ax
F-us*n=m*ax. What would go in my n according to my body diagram? I have n2-n1on2-w2=0 along my y forces.

as far as what you said about kinetic friction, do I include that in there? Since the board will be sliding or no?

 

[Toranc3]

i got most of it now, thanks. But why is it that you don't need kinetic friction when it slides?

 

[SammyS]

That is where I get stuck. So i am not sure if I have my body diagram correct. I have this for the board: F-fs1on2=m*ax
F-us*n=m*ax. What would go in my n according to my body diagram? I have n2-n1on2-w2=0 along my y forces.

as far as what you said about kinetic friction, do I include that in there? Since the board will be sliding or no?

i got most of it now, thanks. But why is it that you don't need kinetic friction when it slides?

Well, yes the box slides on the board, and you should be using kinetic friction.

But acceleration, a₁, for the box, is not the same as acceleration, a₂, for the board.