Question

As illustrated in Fig. 3.33, consider an athlete performing flexion/extension exercises of the lower arm to strengthen the biceps muscles. The athlete is holding the weight of W1 ¼ 150 N in his hand, and the weight of his lower arm is W2 ¼ 20 N. As measured from the elbow joint at point O, the center of gravity of the lower arm (point A) is located at a distance a ¼ 7:5 cm and the center of gravity of the weight held in the hand is located at a distance b ¼ 32

Answer 1

Missing Question Data:

Some part of the question along with a related diagram was missing. I have attached it here for convenience.

Data:

$W_{1}\;=\;150\;N\\\\W_{2}\;=\;20\;N\\\\a\;=\;7.5\;cm\;=\;0.075\;m\\\\b\;=\;32\;cm\;=\;0.32\;m\\\\\theta_{1}\;=\;0^{o}\\\\\theta_{2}\;=\;30^{o}\\\\\theta_{3}\;=\;60^{o}$

Answer with Explanation:

To find the net Momentum, we can use the following formula,

$M_{NET}\;=\;\sum M\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; (Taking\;clock-wise\;direction\;as\;positive)$

$M_{NET}\;(\theta_{1}\;=\;0^{o})\;=\;W_{1}\;*\;b\;+\;W_{2}\;*a\\\\M_{NET}\;(\theta_{1}\;=\;0^{o})\;\;=\;150\;*\;0.32\;+\;20\;*0.075\\\\M_{NET}\;(\theta_{1}\;=\;0^{o})\;\;=\;49.5\;N-m$

The value of net Momentum at any angle $\theta$ can be obtain by the formula,

$M_{NET}\;(\theta)\;=M_{NET}\;(0^{o})\;cos(\theta)\\\\\therefore\;M_{NET}\;(\theta_{2}\;=\;30^{o})\;=M_{NET}\;(0^{o})\;cos(\theta_{2})\\\\M_{NET}\;(\theta_{2}\;=\;30^{o})\;=\;49.5\;*\;cos(30^{o})\\\\\M_{NET}\;(\theta_{2}\;=\;30^{o})\;=\;42.9\;N-m$

Similarly,

$M_{NET}\;(\theta)\;=M_{NET}\;(0^{o})\;cos(\theta)\\\\\therefore\;M_{NET}\;(\theta_{3}\;=\;60^{o})\;=M_{NET}\;(0^{o})\;cos(\theta_{3})\\\\M_{NET}\;(\theta_{3}\;=\;60^{o})\;=\;49.5\;*\;cos(60^{o})\\\\\M_{NET}\;(\theta_{3}\;=\;60^{o})\;=\;24.8\;N-m$