Chapter 9, Problem 11TP

To determine

(a)

The force which is greater and the reason for the two forces to be different.

Answer to Problem 11TP

The force on bicep is $943\text{N}$ and force of the dumbbell is $89\text{N}$. The force on the bicep is nearly $11$ times the force of dumbbell and it is due to the torque exerted on the hand.

Explanation of Solution

Given:

The weight of the dumbbell is $w_1=20\text{lb}$.

Formula used:

The relation between the mass and weight of forearm is,

  $w=mg$

The relation between the force and the weight of the bicep using the second condition of equilibrium in which the net torque is zero is,

  $\begin{array}{c}{r}_1{F}_{\text{B}}=\left(r_2{w}_2\right)+\left(r_3{w}_1\right)\\ F_{\text{B}}=\frac{\left(r_2{w}_2\right)+\left(r_3{w}_1\right)}{r_1}\end{array}$

The free body diagram for the hand is shown below.

  

  Figure-(1)

From the above figure, all the three forces are acting perpendicular at the pivot.

Let mass of the forearm be $m_2=2.5\text{kg}$.

Calculation:

The weight of the forearm is calculated as,

  $\begin{array}{c}w=mg\\ w_1=\left(2.5\text{kg}\right)\left(9.8\text{m}/{\text{s}}^{\text{2}}\right)\\ =24.5\text{N}\end{array}$

The force on the bicep is calculated as,

  $\begin{array}{c}{F}_{\text{B}}=\frac{\left(r_2{w}_2\right)+\left(r_3{w}_1\right)}{r_1}\\ =\frac{\left(0.16\text{m}\right)\left(24.5\text{N}\right)+\left(0.38\text{m}\right)\left(88.96\text{N}\right)}{0.04\text{m}}\\ =\frac{\left(3.92\text{N}\cdot \text{m}\right)+\left(33.81\text{N}\cdot \text{m}\right)}{0.04\text{m}}\\ =\frac{37.73\text{N}\cdot \text{m}}{0.04\text{m}}\end{array}$

Solve further,

  $\begin{array}{c}{F}_{\text{B}}=943.25\text{N}\\ \simeq 943\text{N}\end{array}$

The force on dumbbell is calculated as,

  $\begin{array}{c}{w}_2=\left(20\text{lb}\times \frac{4.448\text{N}}{1\text{lb}}\right)\\ =88.96\text{N}\\ \simeq \text{89}\text{N}\end{array}$

Compare the force of the dumbbell with the force on the bicep.

  $\begin{array}{c}\frac{w_1}{w_2}=\frac{943\text{N}}{89\text{N}}\\ =10.59\\ \simeq 11\\ w_1\simeq 11w_2\end{array}$

Conclusion:

Thus, the force on bicep is $943\text{N}$ and force of the dumbbell is $89\text{N}$. The force on the bicep is nearly $11$ times the force of dumbbell and it is due to the torque exerted on the hand.

To determine

(b)

Whether the force on the bicep muscles changes or if it remains the same if the weight is curled towards the body.

Answer to Problem 11TP

The force on bicep is $667\text{N}$ for the hand bent at $45°$ and force on bicep is $943\text{N}$ if the hand is held perpendicular.

Explanation of Solution

Given:

The bicep makes an angle of $\theta =45°$

Formula used:

The relation between the force and weight of the bicep is,

  $F_{\text{B}}=\frac{\left(r_2{w}_2\sin \theta \right)+\left(r_3{w}_1\sin \theta \right)}{r_1}$

Torque is generated by the weight of the forearm and the weight of the dumbbell.

Calculation:

The force on the bicep is calculated as,

  $\begin{array}{c}{F}_{\text{B}}=\frac{\left(r_2{w}_2\sin \theta \right)+\left(r_3{w}_1\sin \theta \right)}{r_1}\\ =\frac{\left(0.16\text{m}\right)\left(24.5\text{N}\right)\left(\sin 45°\right)+\left(0.38\text{m}\right)\left(88.96\text{N}\right)\left(\sin 45°\right)}{0.04\text{m}}\\ =\frac{\left(2.77\text{N}\cdot \text{m}\right)+\left(23.91\text{N}\cdot \text{m}\right)}{0.04\text{m}}\\ =\frac{26.68\text{N}\cdot \text{m}}{0.04\text{m}}\end{array}$

Solve further,

  $F_{\text{B}}=667\text{N}$

Conclusion:

Thus, the force on the bicep is $667\text{N}$ for the hand bent at $45°$ and force on the bicep is $943\text{N}$ if the hand is held perpendicular.