Chapter 8, Problem 38P

(a)

To determine

To find: The torque that is needed.

Answer to Problem 38P

The force value is $7.81\text{N}\cdot \text{m}$ .

Explanation of Solution

Given information:

The forearm in accelerates a $3.6\text{kg}$ ball at $7.0{\text{m/s}}^{\text{2}}$ by means of the triceps muscle, as shown..

Formula used:

  $I=m{r}_1^2$

Here,

  $I$ is the moment of inertia,

  $m$ is the mass,

  $r$ is the distance.

Calculation:

Mass of ball $m=3.6\text{kg}$

Acceleration of ball $a=7{\text{m/s}}^{\text{2}}$

Distance of the ball from the joint $r_1=30\text{cm}=\text{0}\text{.3}\text{m}$

Distance between triceps muscles $r_2=2.5\text{cm}=\text{0}\text{.025}\text{m}$

The moment of inertial of the ball $I=m{r}_1^2$

The angular acceleration of the arm-ball system $\alpha =\frac{a}{r_1}$

Hence torque needed $\tau =I\alpha$

  $\begin{array}{l}=m{r}_1^2\times \frac{a}{r_1}\\ =m{r}_{1}a\\ =\left(3.6\times 0.31\times 7\right)\text{N}\cdot \text{m}\\ =7.81\text{N}\cdot \text{m}\end{array}$

Conclusion:

The force value is $7.81\text{N}\cdot \text{m}$ .

(b)

To determine

To find: the force that must be exerted by the triceps muscle.

Answer to Problem 38P

The value is $312.4\text{N}$ .

Explanation of Solution

Given information:

The forearm in accelerates a $3.6\text{kg}$ ball at $7.0{\text{m/s}}^{\text{2}}$ by means of the triceps muscle, as shown

Formula used:

  $F=\frac{\tau }{r_2}$

Here,

  $F$ is the force,

  $\tau$ is the torque,

  $r_2$ is the distance.

Calculation:

As the force applied by the triceps muscle is perpendicular to the torque,

  $\begin{array}{l}F=\frac{\tau }{r_2}\\ \text{or}F=\frac{7.81}{0.025}\text{N}\\ F=312.4\text{N}\end{array}$

Conclusion:

The value is $312.4\text{N}$ .