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A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force FA, as part a of the drawing shows. By itself, however, force FA is insufficient. Therefore, two additional forces FB and FC are applied, as in part b of the drawing. Each of these additional forces has the same magnitude F. The magnitude of the resultant force acting on the elephant in part b of the drawing is k times larger than that in part a. Find the ratio F/FA when k = 2 cos = 25

  • Physics -

    As the forces FB and FC are equal the horizontal component of force is given as FB = FC = F•cos θ,
    θ =22º
    The net force applied on the elephant is
    Fnet = FA + 2 F•cos θ ........ (1) ( there are two forces on either side of FA)
    Given that Fnet is k times FA
    Fnet = k• FA where k = 2.32•FA
    Fnet = 2.32• Fa .............. (2)
    Plug in (2) in (1)
    2.32 •FA = FA + 2• F• cos θ
    2.32• FA - FA = 2• F•cos θ
    (2.32 - 1) • FA = 2 F cos θ
    1.32• FA= 2• F• cos θ
    F / FA = 1.32/( 2• cos θ)=
    =1.32/2•cos22º = 0.71

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