ksaimouli
  • ksaimouli
A block rests on a horizontal surface and weighs 425 N. A force is applied to the block and has a magnitude of 142 N. The force is directed upward at an angle relative to the horizontal. The block begins to move horizontally when the angle is 60 degrees. Determine the static coefficient.
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1349552917635:dw|
ksaimouli
  • ksaimouli
Correct me if I am wrong but here goes: Yea I get the x component to be 71N as well. The point at where the block is about to move is where the applied force is equal to the friction force. Friction = static coefficient * normal force I get 71 = static coefficient * 425 static coefficient = 71/425 = 0.167
anonymous
  • anonymous
The normal force, N\[N=425\] The force along the horizontal, Fx, is\[F_x=142\cos(\theta)\]

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anonymous
  • anonymous
http://en.wikipedia.org/wiki/Friction#Dry_friction When it is just about to move, the force of friction equals the normal force times the coefficient of friction. So \[142 \cos(60)=71= \mu 425\]
anonymous
  • anonymous
So yes, you are correct.
ksaimouli
  • ksaimouli
no but the answer is .235
anonymous
  • anonymous
|dw:1349553164246:dw|
ksaimouli
  • ksaimouli
i know i am wrong somewhere
ksaimouli
  • ksaimouli
answer should be .235
anonymous
  • anonymous
|dw:1349553244013:dw|
ksaimouli
  • ksaimouli
ok how do u solve it
anonymous
  • anonymous
\[\sum_{F}^{?}=142\]
anonymous
  • anonymous
@Decart is correct- I did not take into account F_y, the fact that the force's y-component weakens the normal force. @ksaimouli , just subtract 142sin60 from the normal force.
ksaimouli
  • ksaimouli
i am confused
anonymous
  • anonymous
|dw:1349553460977:dw|
anonymous
  • anonymous
Okay- you push the block at an angle. This does 2 things- push the block ALONG (which is resisted by friction) and UP (opposite to gravity, so this reduces the block's force on the ground, as the force of the block DOWN is decreased, so it reduces the normal force the ground exerts on the block). It's as if you have a man pushing horizontally with force Fx and a cable with tension Fy pulling up on the block.
anonymous
  • anonymous
tan theta
ksaimouli
  • ksaimouli
what is force of friction
anonymous
  • anonymous
The normal force changes, so the friction changes. @Decart , I'm not sure what you're doing, but Fy is defintely using sine This gives the right answer by the way.
ksaimouli
  • ksaimouli
instead of that can ii say 142cos60?
ksaimouli
  • ksaimouli
fx=71
ksaimouli
  • ksaimouli
i got i got ohhhhhhhh it is simple guys
ksaimouli
  • ksaimouli
|dw:1349554073776:dw|
ksaimouli
  • ksaimouli
so friction is 71 by finding the Fn=302
ksaimouli
  • ksaimouli
|dw:1349554155311:dw|
ksaimouli
  • ksaimouli
thx u for helpinggggggggggggg me
ksaimouli
  • ksaimouli
i have only one question how did u get the equation Fn=mg-fy
ksaimouli
  • ksaimouli
@henpen
anonymous
  • anonymous
http://gyazo.com/cef28f38aa4b15271c0397dce1f96da1
ksaimouli
  • ksaimouli
i got u thx

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