A baby elephant is stuck in a mud hole. To help pull it out, game keepersuse 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 FBand Fc are applied, as in part b of the drawing. Each of these additionalforces has the same magnitude F. The magnitude of the resultant forceacting on the elephant in part b of the drawing is k times larger thanthat in part a. Find the ratio F/FA when k = 1.61.(a)F/FA=Number iFB23° F23UnitsFC

Answered: e a rope to apply a force FA, as part a…

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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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 = 1.61.
(a)
F/FA=Number i
FB
23° F
23
Units
FC

Expert Solution

Step 1

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 cosθ + F cosθ

= FA+ 2F cosθ

= FA + 2F cos 23.0°

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