A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force F A → , as part a of the drawing shows. By itself, however, force F A → is insufficient. Therefore, two additional forces F B → and F C → 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 / F A when k = 2.00 .
A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force F A → , as part a of the drawing shows. By itself, however, force F A → is insufficient. Therefore, two additional forces F B → and F C → 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 / F A when k = 2.00 .
Solution Summary: The author explains how the x and y components of a vector will cancel out each other, and their horizontal components will add up.
A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a force
F
A
→
, as part a of the drawing shows. By itself, however, force
F
A
→
is insufficient. Therefore, two additional forces
F
B
→
and
F
C
→
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
/
F
A
when
k
=
2.00
.
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