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Finding the resultant force

  1. May 7, 2007 #1
    1. The problem statement, all variables and given/known data
    The question im a bit confused about is this:

    A block is dragged across the ground by a light cable,attached to a winch. Th block is initally at rest.

    Information given:
    Volume: 6m³
    Density 2875 kg/m³
    Coefficent of friction between block and the ground: 0.4
    Diameter: 1.4m

    Find the force that must be applied to the block to overcome friction and cause it to accelerate at a rate of 0.4m/s²

    3. The attempt at a solution

    So I've first got to find the mass right? m=D.V= 2875*6= 17,250 KG

    Then I can find its weight. w=m.g = 17,250*9.81= 169,222.5 N

    Now I need to find the force acting down on the block right? Fa= m.a
    =17,250*0.4= 6900N

    And the frictional force acting Frictionalforce= μ*w = 0.4*169,222.5= 67,689N

    Then the total force should equal: Tforce=√Frictional force²+Fa²= √67689²+6900²= 68,039.77308 right?

    Have I gone wrong somewhere?

    Many thanks
  2. jcsd
  3. May 7, 2007 #2
    You're doing OK until you add forces.
    Newton's seond law states:
    [tex]\Sigma F = ma[/tex]
    Sigma means add up the forces algebraicly. Squaring stuff is when you have forces acting along x- and y- axes.
  4. May 7, 2007 #3
    So F - friction force = ma
    where F=? Frictional force=67689 ma= (17250*0.4)=6900

    So F - 67689 = 6900?

    Many thanks
  5. May 7, 2007 #4

    Don't really understand what you have written here. But if you solve your last line for F, you and I will get very similar answers.
  6. May 7, 2007 #5
    Would that mean F= 6900+67689=74589? Right?
    Many thanks
  7. May 7, 2007 #6
    The number looks good to me--just slap on units and a direction, and check significant figures, and that's it.
  8. May 7, 2007 #7
    Out of interest, how would I go about finding out the angle?
    Many thanks
  9. May 7, 2007 #8
    Everything done solved for the x- component of the force. No information was given about the y-component.
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