## ksaimouli Group Title 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

1. henpen Group Title

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2. ksaimouli Group Title

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

3. henpen Group Title

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

4. henpen Group Title

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 $142 \cos(60)=71= \mu 425$

5. henpen Group Title

So yes, you are correct.

6. ksaimouli Group Title

no but the answer is .235

7. henpen Group Title

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8. ksaimouli Group Title

i know i am wrong somewhere

9. ksaimouli Group Title

10. Decart Group Title

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11. ksaimouli Group Title

ok how do u solve it

12. Decart Group Title

$\sum_{F}^{?}=142$

13. henpen Group Title

@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.

14. ksaimouli Group Title

i am confused

15. henpen Group Title

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16. henpen Group Title

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.

17. Decart Group Title

tan theta

18. ksaimouli Group Title

what is force of friction

19. henpen Group Title

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.

20. ksaimouli Group Title

instead of that can ii say 142cos60?

21. ksaimouli Group Title

fx=71

22. ksaimouli Group Title

i got i got ohhhhhhhh it is simple guys

23. ksaimouli Group Title

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24. ksaimouli Group Title

so friction is 71 by finding the Fn=302

25. ksaimouli Group Title

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26. ksaimouli Group Title

thx u for helpinggggggggggggg me

27. ksaimouli Group Title

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

28. ksaimouli Group Title

@henpen

29. henpen Group Title
30. ksaimouli Group Title

i got u thx