A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a

Question

A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply force A, as part (a) of the drawing shows. By itself, however, force A is insufficient. Therefore, two additional forces B and C are applied, as in part (b) of the drawing. Each of these additional forces has the same magnitude F and is oriented at ? = 21.5° away from A. The magnitude of the resultant force acting on the elephant in part (b) of the drawing is six times that in part a. Find the ratio F / FA.
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Section 1.8 Addition of Vectors by Means of Components

Explanation / Answer

    The x component of force acting on elephant in part b is                  F x   =   FA cos + F B cos + F C cos                             = 3 F cos       The y component of force acting on elephant in part b is                  F y   = F A   sin + FB sin - FC sin                            = 0 + 0 = 0      The resultant force acting on elephant in part b is                  F' = F x   = 3 F cos (21.5 )                                  = 2.791 F The magnitude of the resultant force acting on the elephant in part (b) is six times that in part a                   FA = F' / 6                          = 2.791 F / 6                  The ratio F/ F A   = 6 / 2.791                                               = 2.1495                          

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