Question

$\mathbf{2}\mathbf{.}\mathbf{00}\mathit{m}$Question: The fishing pole in Figure P10.28 makes an angle of$\mathbf{20}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$ with the horizontal. What is the torque exerted by the fish about an axis perpendicular to the page and passing through the angler’s hand if the fish pulls on the fishing line with a force$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{100}\mathbf{ }\mathit{N}$at an angle$\mathbf{37}\mathbf{.}{\mathbf{0}}^{\mathbf{o}}$ below the horizontal? The force is applied at a point from the angler’s hands.

Short answer

The torque exerted by the fish about an axis perpendicular to the page and passing through the angler’s hand if the fish pulls on the fishing line with a force$\overrightarrow{\mathbf{F}}=100 \text{N}$ at an angle$37.0°$ below the horizontal is $\tau =168 \text{N}·\text{m}$.

Step-by-step solution

Defining the torque formula

The torque acting on an object can be defined by$\tau =rF\sin \theta$

where F is the magnitude of force, r is the distance, and$\theta$ is the angle between the force and magnitude.

Using the torque formula to find the torque exerted by the fish

Consider the given figure.

Hence, from the figure, the angle can be calculated as $\left(\theta =20°+37°=57°\right)$.

Thus, the torque can be

$\begin{array}{rcl}\tau & =& \left(2\right)\left(100\right)\left(\sin 57°\right)\\ & =& 168 \text{N}·\text{m}\end{array}$

Therefore, the torque exerted by fish about an axis perpendicular to the page and passing through the angler’s hand is $\tau =168 \text{N}·\text{m}$.