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Forces - A block slides upward on a rough, vertical wall

  1. Mar 31, 2014 #1
    forces -- A block slides upward on a rough, vertical wall

    1. The problem statement, all variables and given/known data

    A block slides upward on a rough, vertical wall at constant velocity when a force F of 60 N acts on it at an angle Ɵ = 60ᵒ to the horizontal. The mass of the block is 3.0 kg. See picture below.

    dDOgI4J.png

    a) Using Newton’s laws, find the normal force on the block.

    b) Determine the force of kinetic friction on the block.

    2. The attempt at a solution

    a) N = mg - Fy = (3.0)(9.8) - 60sin60° = -23

    b) Fcos60° - μN = 0
    60cos60° + 23μ = 0
    μ = -1.3

    a negative coefficient?
     
  2. jcsd
  3. Mar 31, 2014 #2

    TSny

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    Hello, santoki.

    Be sure to draw a good free body diagram. Is the normal force on the block vertical or horizontal?

    Note that you are not asked to find the coefficient of friction. You just need to find the force of friction.

    Which way does the force of friction act?
     
  4. Mar 31, 2014 #3
    For some reason you assumed the wrong direction for the normal force. Did you draw a free body diagram?
     
  5. Mar 31, 2014 #4
    Oh sorry. I drew by FBD like this:

    fenr5Jh.jpg
     
  6. Mar 31, 2014 #5

    TSny

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    In the original diagram, the applied force, F, points up and to the right. But in the above drawing, you have it pointing up and to the left.

    What surface produces the normal force on the block? Why is it called a "normal" force?
     
  7. Mar 31, 2014 #6
    Isn't N the normal force? That's why I drew it pointing to the right because that's the direction the object is exerting force on the wall.
     
  8. Mar 31, 2014 #7

    TSny

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    On the free body diagram of the block, you should only be drawing forces on the block (not on the wall). You are going to apply Newton's laws to the block, so you should be considering the forces on the block.

    The normal force that you draw should be the normal force that the wall exerts on the block. What would be the direction of that force? Again, think about the meaning of the word "normal" here.
     
  9. Mar 31, 2014 #8
    So the normal force would be on the right because that's the direction the wall is exerting it's force on the block and the 60N should be on the left because that's the direction it's being exerted on the block? Or would N be pointing to the left horizontally because that's the direction of the force it's exerting on the block?
     
    Last edited: Apr 1, 2014
  10. Apr 1, 2014 #9

    TSny

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    A normal force is a force that is exerted on an object by a surface in a direction that is perpendicular (normal) to the surface.

    When you stand on the floor, the floor exerts an upward force on you to support you. In this case the surface of the floor is horizontal while the normal force exerted on you by the floor is vertical (i.e., perpendicular to the floor).

    As another example, look at the attached figure which shows a block on an inclined surface. Note how the normal force on the block from the surface is perpendicular to the surface.

    In your problem you have a block against a vertical surface (the wall). In what direction does the wall exert a force on the block: upward, downward, to the right, to the left, or none of these?
     

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  11. Apr 1, 2014 #10
    to the right?
     
  12. Apr 1, 2014 #11

    TSny

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    No. The block pushes on the wall to the right. What does Newton's third law tell you about the force that the wall exerts on the block?
     
  13. Apr 1, 2014 #12
    oh okay, I get it. The wall pushes on the block to the left whilst the block pushes on the wall to the right. The third law states they would be exerting the same force in opposite directions.
     
  14. Apr 1, 2014 #13

    TSny

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    Yes, that's right. The normal force on the block from the wall will be horizontal and toward the left.

    In addition, the wall exerts a friction force on the block. Can you see what direction the friction force will be on the block? (Remember, the block is sliding upward).
     
  15. Apr 1, 2014 #14
    Frictional force would be downward in this case. How can I incorporate all of this to finding the normal force and kinetic friction?
     
  16. Apr 1, 2014 #15

    TSny

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    Good. So, now you know that there are 4 forces acting on the block:

    (1) Gravity (mg) acting downward
    (2) Friction (f) acting downward
    (3) Normal (N) acting horizontally to the left
    (4) The applied force (F) acing upward and to the right at an angle of 60o above the horizontal.

    Draw all of these carefully on a free body diagram. You are given that the block slides with constant velocity. So, think about the acceleration of the block.

    What does Newton's 2nd law then tell you about [itex]\sum[/itex]Fx and [itex]\sum[/itex]Fy?
     
  17. Apr 1, 2014 #16
    If it slides with constant velocity, then ay would be 0 so ∑Fy would be 0. And Fx = max, but wouldn't it also equal to 0?
     
  18. Apr 1, 2014 #17

    TSny

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    Yes, good. ∑Fx = 0 and ∑Fy = 0.

    Use your free-body diagram to assist with getting the x and y components of the forces.

    You should find that ∑Fx = 0 and ∑Fy = 0 will give you two equations you can use to solve for N and f.
    (Remember, you do not need to find the coefficient of friction μ, just the force of friction, f.)

    Well, I'm off to bed. :zzz: Will check back tomorrow. Good luck.
     
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