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Inclined Plane Help

  1. Jan 29, 2009 #1

    Using that picture above and this data.... A block of mass M slides down a plane at an angle of 37o to the horizontal. The coefficient of kinetic friction = 0.20. The horizontal lines in the picture represent magnitude of forces, in units of the force of gravity on the block. The arrows show directions for forces. DATA: sin(37o)=0.60 cos(37o)=0.80, tan(37o)=0.75. I need to find

    A)The 2-letter combination for the total force on the block.
    B)The direction of the force on the ramp due to the sliding block. If it is along one of the arrows, select that arrow. If not, select the two adjacent arrows between which that force is directed.

    I have no clue what to do or what equation to use....
  2. jcsd
  3. Jan 29, 2009 #2


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    Your first job is to draw a nice neat picture. The purpose of this is to understand what's happening. Draw an x-axis parallel to the plane and a y-axis perpendicular to it. Put the origin in the center of the block. Now draw some arrows representing the forces. One vertical representing gravity down. Notice that gravity acts to accelerate the block down the plane and also acts to hold the block onto the plane. Draw in arrows to represent these two components of the force of gravity. Draw in an arrow to represent the force of the plane on the block. Notice that this is equal and opposite to the force of gravity holding the block on the plane. Is friction a part of your problem? If it is it will act up the plane.
    The force of gravity on your block is mg. You will have to use some trigonometry to figure out the component of gravity down the plane and the component of gravity perpendicular to the plane.

    That should get you started. Remember: the first step is not to find an equation. The first step is to understand what's happening.
  4. Jan 29, 2009 #3
    I already did all that I dont understand how I figure out the total force do you add everything up or what??
  5. Jan 30, 2009 #4
    Often people get confused by problems with just a variable for the mass. If "m" throws you off for mass, go ahead and plug in 10 kg, 50 kg, or whatever number turns you on. Then when you've solved it, just divide by that number to get the value in terms of "m".

    If the style of the answers to the problem is giving you troubles: The two letters refer to one to match the angle, and another to match the length of the vector.

    Hope this helps. This problem is a fairly straight-forward one; use the tips AEM offered. "Total force" is just that, the vector addition of the forces. Remember that some of them will cancel each other out.
  6. Jan 30, 2009 #5


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    Well, I agree that the request for the "total force" is confusing. If I were writing the problem, I wouldn't have worded it that way. So let's take the problem apart. The force of gravity holding the block on to the plane is balanced by the force of the plane on the block, so there's no motion in the y direction (the direction perpendicular to the plane). Along the plane you have the component of gravity acting down the plane and the force of friction up the plane. The force of friction is given by the coefficient of friction time the normal force (that's the force holding the block onto the plane). If the component of the force of gravity along the plane is larger than the force of friction the block moves. So you need to compute the difference between this frictional force and the component of gravity acting down the plane. I'll bet this difference is what the person writing the problem had in mind.
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