A deep-sea fisherman struggles to bring in a big salmon (see figure below). Find the force each hand must exert on the pole if the tension in the line is 159 N. Neglect the weight of the pole and assume that the force exerted by the left hand is directed along the horizontal. (Let θ = 53.50 and L = 1.10 m.)
left hand
magnitude
direction above positive x axis
right hand
magnitude
direction
To find the force exerted by each hand on the pole, we can first find the horizontal and vertical components of the tension force in the line.
The tension force in the line can be broken down into its horizontal and vertical components using trigonometry.
The horizontal component of the tension force is given by:
F_horizontal = Tension * cos(θ)
where Tension is the magnitude of the tension force in the line and θ is the angle between the tension force and the positive x-axis.
Substituting the given values, we have:
F_horizontal = 159 N * cos(53.50°)
Next, we can find the vertical component of the tension force using:
F_vertical = Tension * sin(θ)
Again, substituting the given values:
F_vertical = 159 N * sin(53.50°)
The force exerted by the left hand, which is directed along the horizontal, is equal to the horizontal component of the tension force.
Therefore, the force exerted by the left hand is:
Left hand magnitude = F_horizontal = 159 N * cos(53.50°)
The force exerted by the right hand, which is directed in the opposite direction of the vertical component, is equal to the negative of the vertical component of the tension force.
Therefore, the force exerted by the right hand is:
Right hand magnitude = -F_vertical = - (159 N * sin(53.50°))
The left hand magnitude is in the positive x-axis direction, while the right hand magnitude is in the negative y-axis direction. The direction is given as above the positive x-axis for the left hand. The direction for the right hand would be below the negative y-axis.