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Horizontal component of force without angle given?

  1. Mar 12, 2013 #1
    1. The problem statement, all variables and given/known data
    A person is dragging a 60kg block at a constant velocity over flat ground. The force applied by the person to the block has components that are up and to the right. The coefficient of friction for sliding the block over the ground is μk=0.60.

    Calculate the rightward component of the person's pull force assuming the upward component of the pull force is 453N


    2. Relevant equations
    F vertical = Fsinθ
    F horizontal = Fcosθ


    3. The attempt at a solution
    I'm confused about how to approach this. Shouldn't there be an angle of force given in order to solve this?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 12, 2013 #2

    collinsmark

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    Homework Helper
    Gold Member

    Hello Hanyuu,

    Welcome to Physics Forums!
    You can solve this without being given an angle. Drawing a free body diagram would be a good way to start.

    There is one critical piece of information that you don't want to neglect: The block is being dragged with a constant velocity. In other words, the block is not accelerating. (And it's being dragged over flat ground too.)
     
    Last edited: Mar 12, 2013
  4. Mar 12, 2013 #3

    gneill

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    Staff: Mentor

    What are your thoughts about the net horizontal force on the block?
     
  5. Mar 12, 2013 #4
    Ah, Fnet should be 0, right?

    So Fnet would be adding up Fg, Ff, Fn and the two components?
     
  6. Mar 12, 2013 #5

    gneill

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    Staff: Mentor

    Yes; What is your argument for that?

    If you draw a Free Body Diagram for the block, what comprises the contributions to the net horizontal force? Which do you have enough information to compute?
     
  7. Mar 12, 2013 #6
    Fnet=ma, since acceleration is 0, then Fnet is also zero. Friction force also affects horizontal force.

    Solved the problem, thanks!
     
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