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Total forces acting downward on the plane

  1. Nov 9, 2015 #1
    • Member warned about proper use of the template and showing an attempt at solution.
    1. The problem statement, all variables and given/known data
    the total force acting downwards the plane = 100(9.81)sin20 + 150= 485.5 N , should I take the tension in B into consideration ? how to get the tension on the rope ?

    2. Relevant equations


    3. The attempt at a solution
     

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  3. Nov 9, 2015 #2

    SteamKing

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    Do you feel that the tension has no effect on the block?

    The block is in equilibrium as shown. How does the block stay in equilibrium?
     
  4. Nov 9, 2015 #3

    Doc Al

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    It's a force acting on the block, so you'd better.

    You are told that point B is in equilibrium.
     
  5. Nov 9, 2015 #4

    PhanthomJay

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    tension and friction
    try a free body diagram of joint B. Please show your work. Is the block in equilibrium?

    2. Relevant equations


    3. The attempt at a solution[/QUOTE]
     
  6. Nov 9, 2015 #5
    [/QUOTE]

    so , the ttotal forces downwards should be
    100(9.81)sin20 +150+196.2 = 681.7N ?
     

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  7. Nov 10, 2015 #6

    PhanthomJay

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    No. But there seems to be a problem in the problem statement, because it states that joint B is in equilibrium which implies that the block must be in equilibrium also. But the solution gives a kinetic (moving) friction value, so something went wrong here, the joint cannot be in equilibrium with the given values.
     
    Last edited by a moderator: Nov 10, 2015
  8. Nov 10, 2015 #7
    so , the correct one should be 100(9.81)sin20 +150 only ?
     
  9. Nov 11, 2015 #8

    PhanthomJay

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    No, a free body diagram of the block will show that in addition to the two down-the-plane forces you have identified, there is a tension force acting up the plane and possibly a friction force acting parallel to the plane. Now the problem states that the joint B is in equilibrium (not accelerating) and thus the block also must not be accelerating. But when you draw a free body diagram of the joint at B to determine the tension force, and apply that force to the block, there is not enough available static friction force down the plane to keep the block from moving, which is in contradiction with the problem statement. As I see it, the problem is not correctly worded.
     
  10. Nov 11, 2015 #9

    Doc Al

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    I agree. A very poorly worded problem. (It reads as if translated from another language.)

    You can certainly get the given answer, but you have to disregard (or charitably reinterpret) some of the problem statement. You first assume equilibrium, then ask if that is possible. It's not--as PhanthomJay explains--so the friction must be kinetic.
     
  11. Nov 11, 2015 #10
    isn't it = 100(9.81)sin20 ?
     
  12. Nov 11, 2015 #11

    PhanthomJay

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    No, see Doc Al response above for clarification. If the block is moving, the friction force is just [itex]\mu_k(mgcos\theta) [/itex]. You have to charitably reinterpret the problem.
     
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