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baddinlol

  • 4 years ago

A block of mass 10kg is pulled along a horizontal plane by a force of 10N inclined at 30* to the plane. The coefficient of friction between the block and the plane is 0.05, Find the acceleration of the block.

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  1. baddinlol
    • 4 years ago
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    |dw:1327647830040:dw|

  2. IsTim
    • 4 years ago
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    Wrong group. There is a physics group though. Post your question there.

  3. baddinlol
    • 4 years ago
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    It is in my mathematics books though.... sorry

  4. dumbcow
    • 4 years ago
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    well, F = m*a Force in the horizontal direction is 10*cos(30) = 5sqrt3 not sure where the coefficient of friction goes though...been awhile since physics

  5. Xishem
    • 4 years ago
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    This question is acceptable here -- don't worry. But you have more like in the physics group.

  6. Xishem
    • 4 years ago
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    Luck, rather.

  7. baddinlol
    • 4 years ago
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    Do I subtract the friction coeffecient from the opposing force? I am confused...

  8. baddinlol
    • 4 years ago
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    I will try the physics one if it doesn't work out =)

  9. Xishem
    • 4 years ago
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    I believe you need to multiply the normal force by the friction coefficient.

  10. Xishem
    • 4 years ago
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    And doing that will tell you the frictional force. You add that frictional force as a force vector, if I remember correctly.

  11. dumbcow
    • 4 years ago
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    hmm i think i got it now the vertical force = 10*sin(30) = 5 the downward force = 10g = 98 therefore R = 98 - 5 = 93 R* friction = 93*.05 = 4.65 but thats going in opposite direction net forward force = 5sqrt3 - 4.65 = 4.01 now apply F=ma 4.01 = 10a a = 0.401

  12. baddinlol
    • 4 years ago
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    Thanks, that looks right.

  13. baddinlol
    • 4 years ago
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    Can I do this in i-j notation?

  14. dumbcow
    • 4 years ago
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    i guess, you could write the vectors in i-j notation

  15. baddinlol
    • 4 years ago
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    Thanks a lot, really helped

  16. Xishem
    • 4 years ago
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    \[F_n=mg \cos \theta =(10kg)(9.81\frac{N}{kg})(\cos(30 \deg))=85.N\]\[F_f = μN\]\[F_f=(0.05)(85.N)=4.2N\]\[F=85.N-4.2N=80.8N\]\[F=ma\]\[80.8N=(10kg)a \rightarrow a=8.08\frac{N}{kg}=8.08\frac{m}{s^2}\]This is what I get. I don't think you found the force normal correctly, crow.

  17. dumbcow
    • 4 years ago
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    oh oops, i thought that the Net force had to be 0 downward force, mg = 98N so upward force must be 98N, no?

  18. Xishem
    • 4 years ago
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    If the net force were 0, the block wouldn't be accelerating would it?

  19. dumbcow
    • 4 years ago
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    no i was speaking of vertical forces, the block is staying on the ground. isn't the only Force pushing it forward is the 10N applied at the 30* angle ?

  20. Xishem
    • 4 years ago
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    Yes, and so it wouldn't make sense for the normal force to be more than 10. Is the 10N even enough to overcome gravity? \[F_n=mgcos \theta=(10kg)(-9.81\frac{N}{kg})(\cos(30\deg))=-85.0N\]

  21. dumbcow
    • 4 years ago
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    i dunno, that doesn't seem right the normal force is pointing straight up because the block is moving on a horizontal plane so you don't need to multiply mg by cos* the normal force will just be enough to cancel gravity -> Fn = mg - 10sin(3)

  22. Xishem
    • 4 years ago
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    Oh! I misread this question, and the picture only confused me more. I was thinking of it as a 30degree incline. Geez. Ok.

  23. dumbcow
    • 4 years ago
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    ahh, yeah so you're right if the block is moving up the incline

  24. Xishem
    • 4 years ago
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    Yep, and your old solution is correct :P

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