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Calculating angle with coefficient of static friction value! |
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| Feb27-12, 08:51 AM | #1 |
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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!
![]() ![]() 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 |
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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!
![]() 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 |
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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 ?
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| Feb27-12, 12:28 PM | #6 |
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![]() That's why I say "the normal direction", so there's no confusion. |
| Feb27-12, 12:46 PM | #7 |
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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 |
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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 |
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would be...
Fnet=Fn-Ff-Fg 0=32.34cosθ-98 ? |
| Feb27-12, 01:50 PM | #12 |
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but i'm missing
sintheta right? |
| Feb27-12, 01:56 PM | #13 |
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yes.
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| Feb27-12, 01:59 PM | #14 |
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0=32.34costheta-sintheta-98
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| Feb27-12, 01:59 PM | #15 |
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im not sure how i would isolate theta...
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| Feb27-12, 02:11 PM | #16 |
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put = in the middle and divide
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| Feb27-12, 02:17 PM | #17 |
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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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