# Homework Help: Minimum horizontal force needed so book does not fall?

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1. Jun 8, 2016

### Kloud

1. The problem statement, all variables and given/known data
Part C of problem Please see attachment : What is the minimum horizontal force needed to keep the book from slipping. Note that P=515N, and P=normal force, Fs=131.32N, Us=.255, M=9.85kg

2. Relevant equations
Fn=(m*g)/Us

3. The attempt at a solution

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2. Jun 8, 2016

### TSny

Hello and Welcome to PF!

Your answer to b is incorrect. Keep in mind that the formula $f_s^{\rm {max}} = \mu_s N$ is a formula for the maximum possible static friction force for the two surfaces. Is the static friction force in part b necessarily equal to the maximum possible static friction force?

3. Jun 8, 2016

### Kloud

Yes it is. Thats why I used the max formula. Part C on the other hand I am not sure about? What do you think about my answer for it?

4. Jun 8, 2016

### Parixit

Something to make u think instead of giving the answer. Why should u compare ur answer in c with ur answer in b?

5. Jun 8, 2016

### TSny

Why do you say that the friction force must be at its maximum value in part (b)? When $f_s$ is at its maximum value, the object is on the verge of slipping. But there is nothing in the statement of problem that says the object is on the verge of slipping in part (b).

You should be able to deduce the value of the friction force in part (b) without using $f_s^{\rm {max}} = \mu_s N$.
Set up ΣFy = may.

6. Jun 9, 2016

### Kloud

So the for part b would the friction force be fs=m*g*us?

Last edited: Jun 9, 2016
7. Jun 9, 2016

### Parixit

U r getting confused. All your calculations are correct. U just have to understand them fundamentally. Think about whether the answer in c should be compared with b or with a?

8. Jun 9, 2016

### Kloud

Ye your right, I am confused. Im pretty sure part b is correct cause the only equation I have for static friction is the max formula, while for part C Im kinda sure its correct cause initially we apply 515N to the book, so if I am getting 378.54, thats less force so I assume it to be the mininmum Normal force. Also Im not sure how to derive
the Friction force from Fy = may?

9. Jun 9, 2016

### Parixit

The frictional force will balance the weight of the book F equals mg

10. Jun 9, 2016

### TSny

Kloud, it might help to go back to part (a). How did you determine the value of the normal force?

11. Jun 9, 2016

### Kloud

Fn=P cause both forces is horizontal.
@Parixit, so are you saying for part B, Fs=(9.8m/s^2)*(9.85kg), not Fs=Us*N? Cause the problem does not specify max static friction or not.

I am confused now. How did we go from Fy=may to Fy=mg?, I know g is the gravitational acceleration, but if you change a to g wouldn't it be Fy=-mg?

12. Jun 9, 2016

### TSny

Just because two forces are horizontal doesn't mean that they have to be equal. Try to give a logical argument based on principles of physics. I have attached a free body diagram below. See if you can use Newton's second law ΣFx = max to deduce that Fn = P.

#### Attached Files:

• ###### FBD.png
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13. Jun 9, 2016

### Parixit

No, I thought u were talking about frictional force in part c. For part b frictional force would be Us * N. And since the book is not moving horizontally, it means the horizontal forces are balanced. So N equals P.

Also there is no need of considering max static friction in this problem. Hope that clarifies.

Last edited: Jun 9, 2016
14. Jun 10, 2016

### CWatters

I agree with your answer for the normal force (515N). The book isn't accelerating horizontally so the horizontal forces must sum to zero.

For question b apply the same reasoning. First you check that the MAX friction force (131N) is greater than the force due to gravity (approx 98N). That shows the book isn't sliding. If it's not sliding then it's not accelerating so the vertical forces must sum to zero. The actual friction force is NOT 131N.

For question c...You know the minimum friction force required to stop the book sliding and the coefficient of friction so you can calculate the minimum normal and minimum applied force required.