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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.
 one year ago
 one year ago
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.
 one year ago
 one year ago

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ksaimouliBest ResponseYou've already chosen the best response.1
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
 one year ago

henpenBest ResponseYou've already chosen the best response.1
The normal force, N\[N=425\] The force along the horizontal, Fx, is\[F_x=142\cos(\theta)\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
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 \[142 \cos(60)=71= \mu 425\]
 one year ago

henpenBest ResponseYou've already chosen the best response.1
So yes, you are correct.
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
no but the answer is .235
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
i know i am wrong somewhere
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
answer should be .235
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
ok how do u solve it
 one year ago

henpenBest ResponseYou've already chosen the best response.1
@Decart is correct I did not take into account F_y, the fact that the force's ycomponent weakens the normal force. @ksaimouli , just subtract 142sin60 from the normal force.
 one year ago

henpenBest ResponseYou've already chosen the best response.1
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.
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
what is force of friction
 one year ago

henpenBest ResponseYou've already chosen the best response.1
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.
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
instead of that can ii say 142cos60?
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
i got i got ohhhhhhhh it is simple guys
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
dw:1349554073776:dw
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
so friction is 71 by finding the Fn=302
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
dw:1349554155311:dw
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
thx u for helpinggggggggggggg me
 one year ago

ksaimouliBest ResponseYou've already chosen the best response.1
i have only one question how did u get the equation Fn=mgfy
 one year ago

henpenBest ResponseYou've already chosen the best response.1
http://gyazo.com/cef28f38aa4b15271c0397dce1f96da1
 one year ago
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