Question

3. A very light bamboo fishing rod 3.0 m long is secured to a boat at the bottom end. It is

held in equilibrium by an 18 N horizontal force while a fish pulls on a fishing line

attached to the rod shown below. How much force F does the fishing line exert on the

rod? (3)

18 N

pivot

30°

1.8 m

3.0 in

Answer 1

The image in the attachment describes the situation of the fishing rod.

Answer: F = 10.8 N

Explanation: The image shows a fishing rod attached to an axis. To stay in equilibrium, Torque must be equal for the force of magnitude 18N and for the unknow force.

Torque (τ) is a measure of a force's tendency to cause rotation and, in physics, defined as:

τ = F.r.sin(θ)

F is the force acting on the object;

r is distance between where the torque is measured to where the force is applied;

θ is the angle between F and r;

For the fishing rod:

`\tau_(1) = \tau_(2)`

`F_(1).r_(1).sin(\theta) = F_(2).r_(2).sin(\theta)`

Assuming part (1) is related to unknown force:

`F = (F_(2).r_(2).sin(\theta)/(r_(1).sin(\theta) )`

Replacing the corresponding values:

`F = (18*1.8*sin(30))/(3*sin(30))`

`F = (18*1.8)/(3)`

F = 10.8

The fishing line exert on the the rod a force of 10.8N.