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Nedd help finding weight over length

  1. Mar 12, 2009 #1
    what i mean is.... i need to know how much a table weighs when picked up only at one end

    the table is 1100lbs but we are only picking it up (one side only )at about a 35 or 40 degree angle. and also the table is 8 ft long

    theres gotta be a formula out there but ive been googling for about 30 min now and i havnt found anything

    thanks for the help
  2. jcsd
  3. Mar 12, 2009 #2
    i was just thinking, would the weight after lifting be half of the original
  4. Mar 12, 2009 #3


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    Staff: Mentor

    If the weight is distributed evenly, it is half when you first lift one end. As the angle increases, the amount of weight on the part on the ground increases, though how much depends on how the two sides are supported. The minimum weight the up side needs to support is, irrc, half the weight times the cosine of the angle. You might want to draw a diagram and check that - that's just off the top of my head. The minimum is with the person pushing perpendicular to the table surface (so at an angle when the table is angled to the floor).
  5. Mar 12, 2009 #4
    If we convert 1100 lbs we get approximately 498.9567 kg and then times the Fg which is 9.8 N which equals 4,889.776 N. The you will have to split the problem into a y and x component to get the force the the table exerts when you lift it 35 or 40 degrees up.
  6. Mar 12, 2009 #5
    yes u would haft to use force vectors and x and y components like the guy above me said.
  7. Mar 13, 2009 #6
    thanks alot guys
  8. Mar 17, 2009 #7
    im sorry about reviving this thread but im not going to lie to you....i have no idea how to do that. ive googled and tried on my own but i cant teach my self physics

    i hate to ask but could someone do this for me:confused:

    thank you all so much
  9. Mar 17, 2009 #8
    If we assume the table is perfectly balanced and even then the math is actually pretty easy. The table is usually a 50/50 balance. You need to figure out how much weight is shifted to the other side. Take your angle in degrees (0 for flat, 90 for vertical) multiply by 0.0111... and then multiply half the weight by that to find how much is shifted. Example:
    1100 / 2 = 550, that's your normal weight of each side.
    40 degrees * 0.0111... = 0.444... * 550 = 244 shifted (306/794).

    Again this won't take into account the fact that the legs add weight to the outsides, or if the table is not symmetric.
  10. Mar 18, 2009 #9
    this gave me some trouble 2 would it be sin(40)/(550)=1/x i used law of sines
    then x would be (550)/(sin40) oh i see then 244 shifted to the other side that makes sense so it woul be what u said above.
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