Push down on book, frictional force

[SilentBlade91]

Homework Statement

Place a book flat on a table and press down on it with your hand. Now suppose the hand-to-book and table-to-book coefficients of kinetic friction are 0.48 and 0.40 respectively. The book's mass is 0.80 kg and your downward push on it is 10.90 N. How much horizontal force is needed to keep the book moving at a constant speed if the hand is stationary with respect to the book?

Homework Equations

F=ma I am guessing
Possibly the frictional force equation

The Attempt at a Solution

I tried using the frictional force equation and doing (.48)(10.90N)=5.232N thinking that would be the force you would need but I was wrong

 

[pat666]

what about the books weight?

 

[SilentBlade91]

Ok so here is my attempt.

.48(10.90N)=5.232N
.40(.80kg*9.8m/s^2)=3.136N

So I just added 5.232N+3.136N=8.368N for an answer. Does this look right?

 

[collinsmark]

Ok so here is my attempt.

.48(10.90N)=5.232N
.40(.80kg*9.8m/s^2)=3.136N

So I just added 5.232N+3.136N=8.368N for an answer. Does this look right?

No not quite. The book is not sliding relative to the hand. From the original problem statement, "the hand is stationary with respect to the book." So for this part of the problem, you don't have to worry about any forces that cause the hand and book to move relative to each other.

What is the normal force exerted by the table onto the book?

 

[SilentBlade91]

So the .48 is just there to trick you? You don't really need it?

And the normal Fg of the book is .80kg*9.8m/s^2=7.84N so the Fn or the force the table exerts back to the book would be the same correct? I am not sure where to go from there though.

 

[SilentBlade91]

And actually the professor just changed the problem a bit, I don't know if this affects anything but...

You place a book flat on a table and press down on it with your right hand. The hand-to-book and table-to-book coefficients of kinetic friction are 0.48 and 0.40 respectively. The book's mass is 0.80 kg and your downward push on it is 10.90 N. Now, you use your left hand to push the book along the table at constant speed. Assuming that your right hand is stationary with respect to the table, what is the horizontal force exerted on the book by your left hand?

 

[collinsmark]

So the .48 is just there to trick you? You don't really need it?

Well, the fact that the coefficient of friction between your hand and the book is higher than the coefficient for the book and the table, means that it is a lot easier to keep your hand steady relative to the book, if a horizontal force is applied. If it was the other way around, it would be tougher to keep your hand from sliding around on top of the book.

But essentially, yes, you don't need to use the number.

And the normal Fg of the book is .80kg*9.8m/s^2=7.84N so the Fn or the force the table exerts back to the book would be the same correct? I am not sure where to go from there though.

Not quite, not. What you have calculated is the weight of the book; which is the normal force exerted by the table if the book was just sitting on the table all by itself.

Try this: take a book and put it on a bathroom scale. The scale's reading is the normal force exerted by the scale on the book; in this case, the weight of the book. Now, without moving anything, carefully put your hand on top of the book (which is still on the scale) and press down really hard. What happens to the scale's reading?

 

[collinsmark]

And actually the professor just changed the problem a bit, I don't know if this affects anything but...

You place a book flat on a table and press down on it with your right hand. The hand-to-book and table-to-book coefficients of kinetic friction are 0.48 and 0.40 respectively. The book's mass is 0.80 kg and your downward push on it is 10.90 N. Now, you use your left hand to push the book along the table at constant speed. Assuming that your right hand is stationary with respect to the table, what is the horizontal force exerted on the book by your left hand?

Okay, this changes things. Now there are two forces of friction since the book is sliding relative to the table and sliding relative to your hand!

What is the normal force that the table exerts on the book?

What is the normal force that the book exerts on your hand?

(Hint: Draw a free body diagram. Since nothing is accelerating, the sum of all the forces in any given direction must be zero. )

 

[SilentBlade91]

Ugggh this is so confusing. So I used the weight of the book (7.84)and the force by the hand downward (10.90) and added them and multiplied by .40 to get 7.496N but I'm not even sure that is right and what to do with the .48 now. I've been trying to figure this stupid answer out for like 7 hours

 

[collinsmark]

Ugggh this is so confusing. So I used the weight of the book (7.84)and the force by the hand downward (10.90) and added them and multiplied by .40 to get 7.496N

That's great! Just make sure you know which force this is. (I'll give you this one: it is the force of friction between the table and the book.)

but I'm not even sure that is right and what to do with the .48 now. I've been trying to figure this stupid answer out for like 7 hours

It sounds to me like there are two problems here. One problem where the hand is stationary with respect to the book, and another problem where the hand is stationary with respect to the table.

What do your free body diagrams look like? Everything will make more sense after putting the forces on the free body diagrams. To help, I'll list out the forces, but you need to put them on your diagram.

(I) Case where the hand is stationary relative to book:
o Force of hand on book
o Weight of book
o Normal force of table on book
o Force of friction between table and book.
o External horizontal force applied on book.

(II) Case where the hand is stationary relative to table:
o Force of hand on book
o Weight of book
o Normal force of table on book
o Force of friction between table and book.
o Force of friction between hand and book.
o Force of left hand applied on book.