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Homework Help: Simple force down an inclined plane question.

  1. Dec 11, 2005 #1
    I'm using my teacher's study guide to study for a test, and he has the answers on it. I think he's just wrong with one.

    Here's the question: A book weighing 20. N slides at a constant velocity down a ramp inclined at 30 degrees to the horizontal. What is the force of friction between the book and the ramp.

    I know the force of friction has to be equal to the force of the book sliding down the ramp, since it's sliding with a constant velocity.

    As far as I know, all I should have to do is sin30 = 20N/x

    That gives me 40N.

    The answers say 10N is the FF between the book and ramp.

    Did my teacher just mess up?
  2. jcsd
  3. Dec 11, 2005 #2


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    It is the component of the WEIGHT along the inclined plane that the force of friction must balance, which is not what you have done.
  4. Dec 11, 2005 #3


    Staff: Mentor

    Just as a "reality check" you need to remember that the component of a vector in a particular direction will never be greater than the magnitude of the vector itself. In other words, if the weight is 20N, then the component of the weight must be less than 20N. So here it would let you know that 40N simply cannot be correct.

    That is a helpful hint that has saved my grade more than once!

  5. Dec 11, 2005 #4
    The answer is indeed 10 N.

    The clue is this : project all forces along the x and y axis.

    x-axis = along incline (downward is positive direction)
    y-axis = perpendicular to the incline

    The forces are :
    1) gravity : has both x and y components (can you calculate them)
    2) Normal force N : along the positive y-axis
    3) friction force F : directed upward the incline (opposite direction of the x-axis)

    Applying Newton's second Law in the x-direction gives you:

    0 = F - mgsin(30)


    F = 20sin(30)


    ps : be sure you know why i used 0 in the left hand side of the equation and be sure that you know how to calculate the components of each force along each direction. In this case, one does not need the equation of motion in the y-direction but it is instructive to calculate it anyway because all this problems with inclines are solved the same way.

    Good Luck
    Last edited: Dec 11, 2005
  6. Dec 11, 2005 #5
    Your teacher is right. The component, tangential the ramp, is 20N x sin30 = 10 N. Remember that the weight is always perpendicular the earth and
    its components in this case are one tangential and the other vertical the ramp.
  7. Dec 11, 2005 #6
    This may look a little misleading..

    The best way to do it will be by drawing out the ramp and the box.

    Draw the force and its components acting on the box. You will see that the force gravity itself has a vertical and horizontal component. The horizontal component will be the one pointing in the direction parallel to the line of action.. The vertical component will be the one that is 90 degrees perpendicular from its horizontal component. You will see that both force components add up to give gravity.. Not the other way around.
  8. Dec 12, 2005 #7


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    You should use the terms "tangential&normal directions/components" rather than "horizontal&vertical directions/components".
  9. Dec 12, 2005 #8
    perphaps its a little easier to understand.
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