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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

Respuesta :

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 (Ï„) 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:

[tex]\tau_{1} = \tau_{2}[/tex]

[tex]F_{1}.r_{1}.sin(\theta) = F_{2}.r_{2}.sin(\theta)[/tex]

Assuming part (1) is related to unknown force:

[tex]F = \frac{F_{2}.r_{2}.sin(\theta}{r_{1}.sin(\theta) }[/tex]

Replacing the corresponding values:

[tex]F = \frac{18*1.8*sin(30)}{3*sin(30)}[/tex]

[tex]F = \frac{18*1.8}{3}[/tex]

F = 10.8

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

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