# Finding the coefficients of friction

1. Apr 9, 2009

### wilson_chem90

1. The problem statement, all variables and given/known data
A box with a mass of 22 kg is at rest on a ramp inclined at 45 degrees to the horizontal. The coefficients of friction between the box and the ramp are uS = 0.78 and uK = 0.65.

a) Determine the magnitude of the largest force that can be applied upward, parallel to the ramp, if the box is to remain at rest.
b) Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.

2. Relevant equations

Fg = mg
Ff = ukFn
Ff = usFn

3. The attempt at a solution

I dont know where i should begin, i understand thati need to find the force of friction using static friction but I'm not sure how to start it.

2. Apr 9, 2009

### LowlyPion

You will need to draw a force diagram.
Friction is determined by the coefficient of friction times the weight that is normal (⊥ ) to the surface.

The force up the ramp then must offset both the component of weight (m*g) that points down the incline || to the surface and the force of friction determined by the coefficient of friction and the weight normal to the surface.

3. Apr 9, 2009

### wilson_chem90

alright well this is what i got for a)

Fg = mg sin45
= 22 kg (9.8 m/s^2) sin45
= 183.4 N

Ff = uSFn
= 0.78(183.4N)
= 143.1N

4. Apr 9, 2009

### LowlyPion

I think you will want to check your math again for Fg.

I will trust that you know that Fn is not based on Sin45 but Cos45, which in this case does happen to be the same.

5. Apr 9, 2009

### wilson_chem90

ohhhh sorry, it is cos eh. so it equals 113.3 N not 183.4 N.

so that means Ff = 0.78(113.3 N)
= 88.4 N

6. Apr 9, 2009

### LowlyPion

No because 183.4 in not correct.

22*9.8*(.707) is not 183.4.

7. Apr 9, 2009

### wilson_chem90

yeah sorry, i didn't have my calculator in degrees that was the problem

and i thought you mean't that it was cos, and i was pretty sure it was sin anyways

but yeah so Fg = 152 N
and Ff = 118.6 N

8. Apr 9, 2009

### LowlyPion

Well you still have to construct the equations necessary to determine the answers to a) and b). I will presume that you have done that correctly.

9. Apr 10, 2009

### wilson_chem90

yeah so what i would do for a) is add the static frictional force and the gravitational force and then for b) i would do the same but use kinetic friction and subtract the 2 forces

10. Apr 10, 2009

### LowlyPion

Careful. For a) and b) It says "to remain at rest". That suggests to me that kinetic frictional considerations are not relevant. So whatever excess of normal force that is needed (times the coefficient of static friction) to offset the force of gravity that is || along the incline.

11. May 18, 2009

### wilson_chem90

sorry its been awhile since i've replied, but i understand what you mean, and understand how to get normal force but do i have to sin or cos inorder to determine it? since its perpendicular

12. May 18, 2009

### LowlyPion

I suspect this link covers it in more detail than you want:
http://www.glenbrook.k12.il.us/gbssci/Phys/Class/vectors/u3l3e.html [Broken]

Last edited by a moderator: May 4, 2017
13. May 18, 2009

### wilson_chem90

thanks, i figured it was that..
anyways this is what i got

Fn = mgcos45
= 22kg(9.8m/s^2)cos45
= 152.4 N

Ffs = FnuS
= 152.4 N(0.78)
= 118.9 N

14. May 18, 2009

### wilson_chem90

sorry i would also then subtract Fn - Ff

so it would be 152.4 N - 118.9 N
which would be 33.5 N

15. May 18, 2009

### LowlyPion

You have the force of friction, but you are apparently subtracting it from its weight.

What is the force of gravity down the incline?

Isn't that mg*sinθ ?

So for a) you have F up the incline = 118.9 + 154*sinθ won't you, because you have to overcome both gravity and friction in the up incline direction.

16. May 18, 2009

### LowlyPion

For b) consider how much additional force needs to act through the coefficient of static friction to keep it in position.

Remember that while the normal force acts perpendicular to the incline, its effect results in resistance along the incline.