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You need to move a 500 N box across the floor. You pull on the box

  1. Jun 13, 2010 #1
    You need to move a 500 N box across the floor. You pull on the box with a rope which is also horizontal. The floor is not frictionless. You move the box so that it moves with a constant v of 3m/s by applying a 200 N force.

    Coefficient of Friction?

    Also,

    same question except pull the box at an angle of 30 degrees from the horizontal, use 0.4 coefficient.

    what is the tension in the rope?

    I just really cant even get started on this one, i have no idea, my professor really didnt cover to much of this but gave us 4 similar questions on it, any help would help me out....
     
    Last edited: Jun 13, 2010
  2. jcsd
  3. Jun 13, 2010 #2

    cepheid

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    Re: Friction

    This should have been posted in the homework help sub-forum using the associated template for homework threads (which would have provided a systematic framework for you to articulate and approach the problem).

    The basic principle you need to understand in order to solve this problem is Newton's Second Law of Motion. Since the box's velocity is constant (in the horizontal direction) and zero (in the vertical direction), it is not accelerating, which implies that the forces on it are balanced (in both the horizontal and vertical directions). Therefore the NET horizontal force on the box must be zero. Since there are only two horizontal forces, the applied force due to the tension in the rope (forwards), and friction (backwards), these two must be equal to each other (in magnitude)

    Therefore, you KNOW what the magnitude of the frictional force must be in this situation. You also know what the normal force is (because you are given the weight and you know that the vertical forces are balanced). So, you have enough information to determine the coefficient of friction for sliding motion between the box material and floor material.
     
  4. Jun 13, 2010 #3

    jack action

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    Re: Friction

    Friction force = Coeff. friction X Normal force

    The normal force is perpendicular to the friction force (the one pressing on the ground)

    When there is a slope, the weight of the box is still pointing down and can be divided in two components: A horizontal one and a vertical one (with respect to the slope). Determine the friction force and carefully add the force horizontally (wrt the slope) to find the tension.
     
  5. Jun 13, 2010 #4

    cepheid

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    Re: Friction

    For the second part of the problem, the applied force (or tension) is now NOT entirely horizontal. It has both horizontal and vertical components. You can figure out what these components are in terms of the unknown quantity (the tension), because you know the angle at which this force is applied.

    Then it's just the same thing again. Newton's second law requires that if the acceleration is zero (horizontally and vertically) then:

    - the sum of the forces in the horizontal direction is zero

    - the sum of the forces in the vertical direction is zero

    The balance of the forces in the vertical direction is a little different from in the first part of the problem, because instead of just two vertical forces (normal force and weight), there are now three (the third being the upward component of the tension). This means that the normal force will change -- it will not be the same as it was in the first part of the problem.
     
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