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
Ratio
F / FA = 1.32/( 2• cos θ)=
=1.32/2•cos22º = 0.71
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
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