# Homework Help: Calculating angle with coefficient of static friction value!

1. Feb 27, 2012

### Fireant

There's a box sitting on a plank. The coefficient of static friction is 0.33. I want to figure out at which angle the box will begin to slide if the plank is tilted.

I know the factors that come into play are FN, Fg or Fgx, and Fs.

So far I have

Fnet= FN-Fs-Fg
=FN- μs-sin∅
=cos-0.33-sin∅

This doesn't look right... can anyone help me out?

2. Feb 27, 2012

### tiny-tim

Hi Fireant!
I don't understand that at all.

First, use components in the normal direction to find N.

Then you know the friction is 0.33*N, and you can use components in the slope direction to find θ.

Show us what you get.

3. Feb 27, 2012

### Fireant

Hi thank you for responding!

The mass of the box is 10kg. So the normal force here would have to be mg...?

normal force=98N

Ff=μs(FN)
=0.33(98)
=32.34 N

and then
tanθ= 97.5/98
θ=44°

I got 97.5 (for the x component) from pythagorean theory...

I'm pretty lost!

4. Feb 27, 2012

### tiny-tim

Hi Fireant!
No, the weight, W, is mg (98 N).

The normal force, N, is the normal component of the reaction force.

You find N by taking components in the normal direction (of N and W), and equating the sum to zero …

what do you get?

5. Feb 27, 2012

### Fireant

I think i'm getting confused when you say 'the components in the normal direction'... do you mean on the y axis? Isn't that just FN? Which would be -98N ?

6. Feb 27, 2012

### tiny-tim

Some people call that the y axis, some people call the vertical the y axis

That's why I say "the normal direction", so there's no confusion.
What is the normal component of W ?

7. Feb 27, 2012

### Fireant

Sorry I'm at work, that's why it's taking so long to reply.

the normal component of W would be mgcostheta?

8. Feb 27, 2012

### tiny-tim

Yes, so N = mgcosθ.

ok, now find the friction force, and then write the equation for components along the plank.

9. Feb 27, 2012

### Fireant

so Ff=μsFN
=0.33(mgcosθ)
=0.33(98)(cosθ)
=32.34cosθ

10. Feb 27, 2012

### tiny-tim

ok

now the equation along the plank

11. Feb 27, 2012

### Fireant

would be...

Fnet=Fn-Ff-Fg
0=32.34cosθ-98

?

12. Feb 27, 2012

### Fireant

but i'm missing

sintheta

right?

13. Feb 27, 2012

### tiny-tim

yes.

14. Feb 27, 2012

### Fireant

0=32.34costheta-sintheta-98

15. Feb 27, 2012

### Fireant

im not sure how i would isolate theta...

16. Feb 27, 2012

### tiny-tim

put = in the middle and divide

17. Feb 27, 2012

### Fireant

lol i'm hopeless...! thank you so much for bearing with me

ok gonna try this out...

32.34costheta=sintheta-98

sintheta=32.34costheta+98
sintheta/32.34=costheta+98

Sorry, what am i dividing by and how?

18. Feb 27, 2012

### Fireant

or would it be easier if:

costheta=(sintheta-98)/32.34

19. Feb 27, 2012

### tiny-tim

sin/cos = tan ?

20. Feb 27, 2012