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Q20P

Expert-verifiedFound in: Page 346

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Question: A bowler holds a bowling ball (M = 7.2 Kg) in the palm of his hand (Figure 12-37). His upper arm is vertical; his lower arm (1.8 kg)** ** is horizontal. What is the magnitude of (a) the force of the biceps muscle on the lower arm and (b) the force between the bony structures at the elbow contact point?**

**Answer**

** **

- The magnitude of the force of the biceps muscle on the lower arm,$T=6.5\times {10}^{2}\text{N}$ .
- Force between the bony structures at the elbow contact point, $F=5.6\times {10}^{2}\text{N}$ .

** **

- Distance between the elbow point of contact and the lower arm CM, $D=0.15\text{m},d=0.04\text{m}$ .
- Distance between the elbow point of contact and the palm, L =0.33m .
- Mass of the ball,
*M =7.2 kg*. - Mass of the lower arm,
*m =1.8 kg*.

** **

**The ball is held in hands and balanced. It is neither moving linearly nor rotating about any pivot point. So, it is in static equilibrium condition. By selecting a pivot point and applying the static equilibrium conditions, you can write the torque equation in terms of force and distance. Solving this equation, you would get the unknown tension and force. **

Using the figure given in the problem and condition of static equilibrium, we can write torque and force equations as,

$\sum Torque=0\phantom{\rule{0ex}{0ex}}\left(d\times T\right)+\left(0\times F\right)-\left(D\times mg\right)-\left(L\times Mg\right)=0...........................\left(1\right)\phantom{\rule{0ex}{0ex}}\sum {F}_{x}=0\phantom{\rule{0ex}{0ex}}F+\left(M+m\right)g-T=0.........................\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}$

Substitute the values in equation 1, and we get,

$\left(0.04\times T\right)+0-\left(0.15\times 1.8\times 9.8\right)-\left(0.33\times 7.2\times 9.8\right)=0\phantom{\rule{0ex}{0ex}}T\times 0.04=25.9308\phantom{\rule{0ex}{0ex}}T=648\text{N}\phantom{\rule{0ex}{0ex}}$

Hence, the magnitude of the force of the bicep muscle on the lower arm is $6.5\times {10}^{2}\text{N}$ .

Substitute the values in equation 2, and we get,

$F+\left(7.2+1.8\right)\times 9.8-648=0\phantom{\rule{0ex}{0ex}}F+88.2-648=0\phantom{\rule{0ex}{0ex}}F\approx 560\text{N}\phantom{\rule{0ex}{0ex}}$

Hence, the force between the bony structures at the elbow contact point is* * $5.6\times {10}^{2}\text{N}$.

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