What Is the Maximum Car Acceleration Without the Book Slipping?

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The discussion centers on calculating the maximum acceleration of a car without a book slipping off its top, given a static coefficient of friction of 0.45. Participants clarify that the only horizontal force acting on the book is static friction, which determines the maximum force before slipping occurs. By applying Newton's second law, they derive that the maximum acceleration is 4.41 m/s², confirming it as one of the provided answer choices. The conversation emphasizes understanding the relationship between friction, weight, and acceleration in this context. Overall, the participants successfully solve the problem while seeking deeper insight into the coefficient of static friction.
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Hi I'm new to this forum, I've seen it before but since I'm now taking physics courses at my university, I've decided to register.

I'm working on an assignment and I'm stuck on two or three questions. Some of them I've already solved, but there's one in particular that I'm not certain there's enough data given to solve it. Here it goes:

A book sits on a horizontal top of a car as the car accelerates horizontally from rest. If the static coefficient of friction between the car top and the book is 0.45, what is the maximum acceleration the car can have if the book is not to slip?
(a) 3.21 m/s2 (b) 3.76 m/s2 (c) 4.02 m/s2 (d) 4.41 m/s2
Now I can do certain friction problems, but I'm clueless on this one. I've tried to acquire the mass of the book form the static coefficient, but something tells me that I'm going in the wrong direction. Can anyone tell me what I should do to solve this problem? Thanx
 
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There's enough data.

What is the friction between the book and the car's top ?
What is the accelerating force on the book ?
Apply Newton's 2nd law.
 
Fermat said:
There's enough data.
Ok, that confirms it.

Fermat said:
What is the friction between the book and the car's top ?
The friction cofficient constant is 0.45, so I gather this is a ratio based on the equation. Does it mean that a force required to push ths object is 9N for the book to weigh 20N?


Fermat said:
What is the accelerating force on the book?
Since the book is in an initial state of equlibrium, I gather it's 0 or do you mean the 9N described above?


Fermat said:
Apply Newton's 2nd law.
a = 9/2.04
= 4.41 m/s^2

It's one of the choices. I guess I solved it. But can you explain me about the coefficient more in detail? I feel like I'm doing this blindfolded. :rolleyes:
 
*bump*
Anyone can give me insight on this?
 
The book is not in equilibrium.

Hint: The only horizontal force on the book is the static friction. So what's the maximum force on the book? (Any more and it will start to slip.) Then apply Newton's 2nd law to find the acceleration.
 
I got it, so the coefficient of static friction is just a ratio of the maximum friction force over the book's weight.
 
That's right.

You've got one eqn involving friction and the weight of the object.

Newton's 2nd law will give you the other eqn you need.
 
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