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Another MCQ

  1. Sep 17, 2007 #1
    1. The problem statement, all variables and given/known data

    Consider a smooth vertical wall against which a block is placed.Another block is also placed in contact with the earlier inner block.An external force F is acting horizontally on the outer block towards the wall.There is friction between the two blocks,but no friction between the wall and the inner block.The friction on the outer block by the inner block in equilibrium is

    (i) upward
    (ii) downward
    (iii)zero
    (iv)the system cannot be in equilibrium

    2. Relevant equations
    3. The attempt at a solution

    I think it is upward,but the answer is (iv)

    I cannot accept it from the free body diagram consideration...
     
  2. jcsd
  3. Sep 17, 2007 #2

    learningphysics

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    Homework Helper

    Are the blocks supposed to be up in the air against the wall?
     
  4. Sep 17, 2007 #3
    draw a vertical line(wall).then draw a rectangle to its right so that its one side touches the line.then,draw another rectangle to its right whose one side toches the outer side of the inner rectangle.
    the blocks are in contact with air in their exposed surfaces.

    I am attaching a pdf image:
     

    Attached Files:

  5. Sep 17, 2007 #4

    Doc Al

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    Staff: Mentor

    Think of the two blocks as a single system. What vertical forces act on it? Is it in equilibrium?
     
  6. Sep 17, 2007 #5
    sorry,I thought it was downward,but last time,I wrote it upward.

    It is tempting to say,no force apart from weight is acting in the vertical direction.But there is,I think,a possibility of a frictional force acting in the upward direction.

    I am referring to the friction that will arise when you push the outer block towards the wall---this friction will be between the outer block and the agent applying the force(possibly,your hand).

    Is not that is going to make an equilibrium?

    By the way,everytime I am uploading a pdf image,it is showing that "pending".Yet,I do not think there is any problem in the procedure...Can anyone help?
     
  7. Sep 17, 2007 #6
    Let us consider the forces acting on the inner block: weight mg down wards and frictional force between the two blocks upwards.

    Forces on the outer block : weight mg down wards and frictional force downwards. Hence, outer block has no upwards force to support it. Hence. the system is unstable.
     
  8. Sep 17, 2007 #7

    Doc Al

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    Internal forces cancel and cannot influence equilibrium.

    Attachments must be approved by a mentor before they can be seen by others.
     
  9. Sep 17, 2007 #8
    How the friction I referred to is going to be an internal force?
    The same question to Vijay Bhatnagar.

    I told that the friction is due to the external force acting on the outer block.It must be a force external to the system...

    For instance,if the wall were not smooth,a friction would exist.And we must consider that friction to be an external force.Isn't it?
     
  10. Sep 17, 2007 #9

    Doc Al

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    Staff: Mentor

    Look at both blocks together as a single system. The only vertical force on them is their weight. No equilibrium.
    Sure, the friction from the wall would be an external force.
     
  11. Sep 17, 2007 #10
    I am referring to my point 3rd time:

    I am referring to the friction that will arise when you push the outer block towards the wall---this friction will be between the outer block and the agent applying the force(possibly,your hand).

    This is a outer force and I do not see why you people are repeatedly overlooking this.
     
  12. Sep 17, 2007 #11

    Doc Al

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    Staff: Mentor

    You seem to be overlooking this:
    I interpret that to mean that the outer force has no vertical component. (Imagine the outside of the outer block covered in slippery grease.) The only potential friction force is between the two blocks. But that won't hold up the blocks against gravity.
     
  13. Sep 17, 2007 #12
    Say the outer force if F,then,we should have a friction like (mu)F
     
  14. Sep 17, 2007 #13

    Doc Al

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    The "outer force" is the force exerted by some unknown entity on the outer block. If there were friction on that outer surface, the "outer force" would have a vertical component and wouldn't just act horizontally. The only force mentioned as acting on that outer block acts horizontally.

    Imagine a single block against a frictionless wall. I push on it. If there's friction between my hand and the block, then the force I exert on the block is not strictly horizontal--it must act at an upward angle. If there's no friction, and thus the only force I can exert is horizontal, the block slides down the wall no matter how hard I push.
     
  15. Sep 17, 2007 #14
    So,you say,if there is a friction,we must have some angle involved?

    Is it necessarily the case everywhere?

    Please don't mind,it might be trivial to you.But,somehow I could not apply the concept while doing this.

    Actually when we do problems,we always make the friction as (mu)N and think that friction is due to N only and never bother about the actual contact force-which certainly act in some angle.

    So,friction is due to the tilted contact force,not due to the normal force,OK?As I find,this should be the case as we regard friction as a component of the contact force.
     
  16. Sep 17, 2007 #15

    Doc Al

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    Sure. If there's friction, there must be both a normal force and a parallel force, then the total contact force between the two surfaces cannot just be normal to the surface.


    Depending on the problem, that's fine. You often have no need to calculate the direction of the contact force.

    I'd state it: If there's friction, then the contact force must be tilted.

    I don't know what you mean by friction not being due to the normal force. Assuming the usual model for friction, if there's no normal force then there can be no friction force.
     
  17. Sep 17, 2007 #16
    ok it is clear now...
    thank you very much...
     
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