# Inclined plane problem. Help please?

1. Oct 30, 2006

### cak

Here is the problem I'm having trouble with...

The coefficient of static friction between the 32 kg crate and the 35.0° incline of Figure 4-34 is 0.2. What is the minimum force, F, that must be applied to the crate perpendicular to the incline to prevent the crate from sliding down the incline?

I did a force diagram and figured out the variables:

1. I got normal Force as 257.152
2. force of friction Fk=µk*N = .2 * 257.152 = 51.43
3. perpendicular force = 32 * 9.8 * cos(35) = 256.89
4. parallel force = 32 * 9.8 * sin(35) = 179.87 = 179.87
5. Net force=parallel force - force friction = 179.87 - 51 = 128.49

But that isn't right. Can someone help me?

2. Oct 30, 2006

### drpizza

?? How is your normal Force different from your perpendicular component of force?

You want the force of friction to be equal to the parallel force in order for it not to start sliding, right? How do you increase the force of friction?

3. Oct 30, 2006

There is another force F acting (the unknown one) in the direction of the weight. So, what does the normal force equal?

4. Oct 30, 2006

### cak

umm by increasing the normal force or µk??

I think this is right since the force of friction=µk*N
Considering the fact that the force of friction is directly related to µk and/or the normal force.

am i right???

thanx for helping me!!

5. Oct 30, 2006

### cak

do you mean like the µk (.2) * normal force has to equal 179.87?
so like (the unknown force + the normal force of 257) * .2 = 179.87?
is that right?

6. Oct 30, 2006

Think about what's wrong in this step. (Hint: if the net force does not equal zero, what does Newton's 2nd law tell you?)

7. Oct 30, 2006

### cak

F=ma ....that if the force goes up then the acceleration goes up?
cause mass stays the same here.

8. Oct 30, 2006

### cak

newtons second law tells us that acceleration of an object is dependent upon two variables - the net force acting upon the object and the mass of the object. acceleration is directly related to the net force and inversely related to the mass...

9. Oct 30, 2006

What is important is that you had a net force calculated, which would mean that you have an acceleration. But you want the crate to rest. So, what does that mean? In what relation do the forces paralell to the incline have to be?

10. Oct 30, 2006

### cak

Since the crate is at rest... doesn't that mean that the acceleration has to be zero?
so would that mean that the forces paralell to the incline would have to be zero...possibly??

11. Oct 30, 2006

Acceleration is zero. The sum of the two forces equals zero (i.e the net force). Now, what does that mean? (Hint: what does that tell you about their directions and their magnitudes?)

12. Oct 30, 2006

### cak

Wouldnt that mean that the force of friction is equal to the parallel force? Doesnt that also mean that they're going in the opposite direction and their magnitudes are the same...

13. Oct 30, 2006

Yes, exactly. You're on the right track now.

14. Oct 30, 2006

### cak

ok well...
correct me if im wrong but...
if acceletation is zero, then wouldnt that mean
ma=Fparallel-Ffriction
0=32(-9.8)(sin35)-.2(-9.8)(32)(cos35)

but how do i apply that to get the force applied to prevent the crate from sliding down?

15. Oct 30, 2006

Ffriction = mu*N, right? Now, as I already asked before, what does N equal? It's not just 9.8*32*cos35. There is another force involved. Look at the text of your problem.

P.S. You don't have to write the minus sign in front of 9.8

16. Oct 30, 2006

### cak

Does that mean Im adding the unknown force to N?

17. Oct 30, 2006