- 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.

- chestercat

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- anonymous

|dw:1349552917635:dw|

- 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

The normal force, N\[N=425\]
The force along the horizontal, Fx, is\[F_x=142\cos(\theta)\]

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## More answers

- 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

So yes, you are correct.

- ksaimouli

no but the answer is .235

- anonymous

|dw:1349553164246:dw|

- ksaimouli

i know i am wrong somewhere

- ksaimouli

answer should be .235

- anonymous

|dw:1349553244013:dw|

- ksaimouli

ok how do u solve it

- anonymous

\[\sum_{F}^{?}=142\]

- 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

i am confused

- anonymous

|dw:1349553460977:dw|

- 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

tan theta

- ksaimouli

what is force of friction

- 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

instead of that can ii say 142cos60?

- ksaimouli

fx=71

- ksaimouli

i got i got ohhhhhhhh it is simple guys

- ksaimouli

|dw:1349554073776:dw|

- ksaimouli

so friction is 71 by finding the Fn=302

- ksaimouli

|dw:1349554155311:dw|

- ksaimouli

thx u for helpinggggggggggggg me

- ksaimouli

i have only one question how did u get the equation Fn=mg-fy

- ksaimouli

- anonymous

http://gyazo.com/cef28f38aa4b15271c0397dce1f96da1

- ksaimouli

i got u thx

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