How to find force that a seat might exert on a book.

[Joyci116]

Homework Statement

Your 3.80-kg physics book is next to you on the horizontal seat of your car. The coefficient of static friction between the book and the seat is 0.650, and the coefficient of kinetic friction is 0.550. Suppose you are traveling at 72 km/h=20.0m/s and brake to a stop over a distance of 45.0m. (a)Will the book start to slide over the seat? (b)What force does the seat exert on the book in the process?

Homework Equations

F=$\mu$N
F=ma_x
N=mg=ma_y

The Attempt at a Solution

I know:
m=3.80kg
$\mu$_s=0.650
$\mu$_k=0.550
Vi=72.0 km/h=20.0m/s
Vf=0
x=45.0m

Well, coefficient of static friction means stationary and coefficient of kinetic friction means moving. Since in this case $\mu$_s>$\mu$_k, that means that the book is stationary.
So part a is no, because the static friction is greater than the kinetic friction.
But I had trouble with part b. I drew a free-body diagram. I have the weight force(earth,book) pointing down, normal force(earth,book) pointing up, and friction force point to the left. I have the weight force and normal force the same length, but the friction force is shorter. I don't know what to do after that. It asks for the force that the seat exert on the book, so would that be a normal force that the car seat might exert on the book?

Thank you,

Joyce

 

[Doc Al]

Well, coefficient of static friction means stationary and coefficient of kinetic friction means moving. Since in this case $\mu$_s>$\mu$_k, that means that the book is stationary.
So part a is no, because the static friction is greater than the kinetic friction.

That reasoning is flawed as it doesn't depend on what the car does. Do you think that it's impossible for the book move no matter what?

Instead, ask yourself: What acceleration is required of the book? Is static friction sufficient to provide that acceleration?

But I had trouble with part b. I drew a free-body diagram. I have the weight force(earth,book) pointing down, normal force(earth,book) pointing up, and friction force point to the left. I have the weight force and normal force the same length, but the friction force is shorter. I don't know what to do after that. It asks for the force that the seat exert on the book, so would that be a normal force that the car seat might exert on the book?

The seat exerts both the normal force and the friction. So what's the total force that the seat exerts on the book?

 

[Joyci116]

Hello Doc Al,

According to Newton's first law, objects in motion stays in motion and objects at rest stays at rest (stays in motion unless acted upon by another force). I think that if the acceleration increases, the book will start to move, for $\mu$_k will start to increase.

For the total force, would that be the sum of the normal force plus the friction force (because those are the forces that the seat is exerting on the book)?

Thank you,

Joyce

 

[Doc Al]

According to Newton's first law, objects in motion stays in motion and objects at rest stays at rest (stays in motion unless acted upon by another force). I think that if the acceleration increases, the book will start to move, for $\mu$_k will start to increase.

$\mu$_k is just the coefficient of kinetic friction--it's not going to change.

Ask yourself: How much friction force is required to give the book the same acceleration as the car? (That way the book is able to move with the car and not slide.) Will static friction be enough? (What's the maximum possible value of static friction?)

For the total force, would that be the sum of the normal force plus the friction force (because those are the forces that the seat is exerting on the book)?

Yes, the vector sum of those forces. You need to calculate the friction force and the normal force.

First things first: What's the acceleration of the car?

 

[Aggression200]

Hey, Joyce.

First, you must find the acceleration of the car. So, how would you find that?

Well, you have V$_{i}$=20.0 m/s and V$_{f}$=0.00 m/s. You also know that $\Delta$x is 45.0 m.

So, in order to find the acceleration of the car, you would use the equation...

V$_{f}$$^{2}$=V$_{i}$$^{2}$+2a$\Delta$x

So your acceleration would be what?

Then, find the maximum friction force. Once you have that, find the acceleration required to overcome that force.

Is the car's acceleration greater or less than that acceleration?

 

[Joyci116]

Doc Al,

Ok, so I use the equation F=$\mu$N
f=-w+N
w=N
w=mg or N=mg
F=$\mu$*mg
F=0.650(3.80)(-9.8)=-24.2
So, the books would have to overcome 24.2 static friction before the books can slide over the seat.

Joyce

 

[Doc Al]

Ok, so I use the equation F=$\mu$N
f=-w+N
w=N
w=mg or N=mg
F=$\mu$*mg
F=0.650(3.80)(-9.8)=-24.2

What you've calculated is the maximum value of static friction. Good! But how much friction do you actually need to accelerate the book?

Calculate the acceleration of the car and figure out how much force is needed to give the book that same acceleration.

So, the books would have to overcome 24.2 static friction before the books can slide over the seat.

It's true that if more than 24.2 N of force is required to accelerate the book, then static friction will not be enough and the book will start to slide. But first you need to know how much force is needed.

 

[Joyci116]

I found that the acceleration is -4.44 m/s².
And to find the maximal friction force I would use the equation, F=ma.
F=(3.80)(-4.44)=-16.9?

Would that be correct so far?

 

[Doc Al]

I found that the acceleration is -4.44 m/s².

Good. Don't worry about the sign, all we need is the magnitude.

And to find the maximal friction force I would use the equation, F=ma.
F=(3.80)(-4.44)=-16.9?

That's the force needed, not the maximum friction force. (You already calculated the maximum force.) So, will static friction be enough?

Would that be correct so far?

You're getting there.

 

[Joyci116]

I'm so sorry, but I'm really confused now.
Ok, so the maximum value is -24.2N. That is the force that is required to accelerate the book, but how can I find the force that is needed?

 

[Doc Al]

Ok, so the maximum value is -24.2N. That is the force that is required to accelerate the book, but how can I find the force that is needed?

No, that's the maximum value of static friction. The force needed--which you calculated using F = ma--is only 16.9 N.

So, is there enough static friction to provide the needed force? (That should be an easy question now.)

 

[Joyci116]

Since 16.9N does not exceed the maximum value of static friction, the books will not slide over the seat. There is not enough force, right?

 

[Doc Al]

Since 16.9N does not exceed the maximum value of static friction, the books will not slide over the seat.

Right!

There is not enough force, right?

Look at it from the opposite point of view--there's plenty of friction to do the job.

So the actual static friction force exerted on the book is only 16.9 N.

(If, for example, the car had a greater acceleration the force needed may have been too much for static friction to provide. In that case, the book would start to slide.)

 

[Joyci116]

I see! Thank you very much.