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Push up ramp

  1. Oct 10, 2008 #1
    Can someone help with this problem I dont even know where to start I have my freebody diagram drew up but I dont know where to go from there

    A UPS worker is loading a truck by pushing boxes up a ramp that makes an angle of 32 degrees above the floor. The coefficients of friction between a box and the ramp are 0.62 and 0.43.

    A) What is the minimum push up the ramp that the worker must exert on the 100 box to keep the box at rest? Hint: First draw a freebody diagram for the box without the worker's push and then add the push.

    B) What push must the worker exert parallel to the floor (not the plane) to cause the box to move up the plane at a constant speed 0. 2
  2. jcsd
  3. Oct 10, 2008 #2
    I am going to assume that 0.62 is static friction and 0.43 is kinetic friction.

    The '100 box' is 100kg?
    The forces acting on the box are gravity, normal reaction, friction and the worker pushing.

    Force due to gravity = mg.

    The component parallel to the inclined plane that is going to make it harder for the worker is therefore going to be mgsin(32).

    The normal reaction force is going to be perpendicular to the inclined plane which will be equal to mgcos(32).

    Multiply this by the co-efficient of static friction to get .62mgcos(32).

    To remain stationary he must push enough force to get .62mgcos(32) and mgsin(32) to be equal.
  4. Oct 12, 2008 #3
    yes you assumed right. but I am still unsure as to how to solve this
  5. Oct 19, 2011 #4
    Any follow up on this? I have the same problem but 70kg instead of 100kg. I set up my equation:

    Fcos(32) = .62[Fsin(32)+wcos(32)]+wsin(32)
    w = mg = weight which for me would be 70(9.8) = 686N

    So I get... .848F = .329F + 360.7 + 363.5

    Then I solved for F and got F=1395N, which is definitely NOT the correct answer....

    Any help would be much appreciated.
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