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Center of Gravity

  1. Oct 10, 2004 #1
    Hi, I'm kind of stuck on this problem and maybe you guys can help me. If so I greatly apreciate it.

    An 850-N painter stands 1.20m from one end of a 3.00m scaffold supported at each end by a stepladder. The scaffold weighs 250-N and there is 40.0-N can of paint opposite the painter. How much force is exerted by each stepladder?
     
  2. jcsd
  3. Oct 10, 2004 #2
    im sorry.. the 40N can of paint is opposite the end of the painter and .50m away the that end
     
  4. Oct 11, 2004 #3

    HallsofIvy

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    Let F1 be the force exerted by the ladder at the "paint can" end and F2 be the force exerted by the ladder at the "painter" end. The total weight of painter, scaffold, and paint can is 850+ 250+ 40= 1140 N. Since the scaffold does not fall, we must have
    F1+ F2= 1140.

    That's one equation in two unknowns. To get another, consider the torque around some point. In fact, we can eliminate one of the unknowns immediately by choosing to calculate the torque about one of the endpoints. The torque about the endpoint on the "paint can end", taking positive to be counterclockwise, is:
    first the step ladder at that end is exerting force F1 but is distance 0 from the pivot: that contributes 0.
    The paint can has weight 40 N and is 0.5 m from that end: torque -20 Nm (negative becaue it acts clockwise).
    The scaffold has weight 250 N and we can take its weight at its center of gravity: the middle of the scaffold, 1.5 meters from the fulcrum: torque is -250(1.5)= -375 Nm.
    The painter has weigth 750 N and is 1.2 meters from his end so 3-1.2= 1.8 m from the fulcrum: torque is -750(1.8)= -1350 Nm.
    Finally, the other ladder is exerting F2 N upward 3 m from the fulcrum: torque is 3F2.

    Since the whole system is not rotating, the total torque (about any point and, in particular, about the fulcrum used here) must be 0:
    -20- 375-1350+ 3F2= 0 or 3F2= 1745 so F2= 581 2/3 N.
    From F1+ F2= 1140, F1= 1140- 581 2/3= 558 1/3 N.
     
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