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Calculating angle with coefficient of static friction value!

 
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Feb27-12, 08:51 AM   #1
 

Calculating angle with coefficient of static friction value!


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?
 
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Feb27-12, 11:07 AM   #2
 
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Hi Fireant!
Quote by Fireant View Post
Fnet= FN-Fs-Fg
=FN- μs-sin∅
=cos-0.33-sin∅
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.
 
Feb27-12, 11:25 AM   #3
 
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!
 
Feb27-12, 12:07 PM   #4
 
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Calculating angle with coefficient of static friction value!


Hi Fireant!
Quote by Fireant View Post
Hi thank you for responding!

The mass of the box is 10kg. So the normal force here would have to be mg...?
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?
 
Feb27-12, 12:22 PM   #5
 
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 ?
 
Feb27-12, 12:28 PM   #6
 
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Quote by Fireant View Post
I think i'm getting confused when you say 'the components in the normal direction'... do you mean on the y axis?
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.
Isn't that just FN? Which would be -98N ?
What is the normal component of W ?
 
Feb27-12, 12:46 PM   #7
 
Sorry I'm at work, that's why it's taking so long to reply.

the normal component of W would be mgcostheta?
 
Feb27-12, 01:30 PM   #8
 
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Yes, so N = mgcosθ.

ok, now find the friction force, and then write the equation for components along the plank.
 
Feb27-12, 01:40 PM   #9
 
so Ff=μsFN
=0.33(mgcosθ)
=0.33(98)(cosθ)
=32.34cosθ
 
Feb27-12, 01:42 PM   #10
 
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ok

now the equation along the plank
 
Feb27-12, 01:47 PM   #11
 
would be...

Fnet=Fn-Ff-Fg
0=32.34cosθ-98

?
 
Feb27-12, 01:50 PM   #12
 
but i'm missing

sintheta

right?
 
Feb27-12, 01:56 PM   #13
 
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yes.
 
Feb27-12, 01:59 PM   #14
 
0=32.34costheta-sintheta-98
 
Feb27-12, 01:59 PM   #15
 
im not sure how i would isolate theta...
 
Feb27-12, 02:11 PM   #16
 
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put = in the middle and divide
 
Feb27-12, 02:17 PM   #17
 
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?
 
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