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 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 = 2.10. The angle is 16 degrees.

the answer must be in 4 significant figures.

Explanation / Answer

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. If this helps then kindly rate 5-stars.

A baby elephant is stuck in a mud hole. To help pull it out, game keepers use a rope to apply a 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. 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 = 2.40. (Take ? = 18.0°.)

sol:

Component of FA perpendicular to FA = 0 because component ofany vector perpendicular to the vector = 0.
If we take components of FB and FC perpendicular to FA, then theycancel each other because FA and FB are on opposite sides of FA andmake same angle with FA.

Therefore, resultant force = sum of components of the three forcesalong FA.
Or resultant force = FA + F cos + F cos
                           = FA+ 2F cos
                           = FA + 2F cos 18.0°

It is given that resultant force = 2.4 FA
Therefore
FA + 2F cos 18.0° = 2.4 FA
         2F cos 18.0°= 1.4 FA
                     F/FA = 1.4/(2 cos 18.0°)
                              =  0.7/  0.951        

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