The fishing pole below (F = 94 N , = 1.90 m) makes an angle of 20.0° with the horizontal. What is the torque exerted by the fish about an axis perpendicular to the page and passing through the fisher's hand?
i TRIED doing fdsinthetha=(1.90*94*sin20) this was not the right answer. pls help
sine20? is the fish pulling horizontally? If the fish is pulling vertically, it is sin70.
Not enough information is given. Needs 2 angles, not one, shown in image.
Torque is:
Force*cos[90-(20 degrees + x degrees)]*distance
To calculate the torque exerted by the fish about an axis perpendicular to the page and passing through the fisher's hand, you need to use the formula:
Torque = Force x Distance x sin(θ)
Where:
- Torque is the measure of force that can cause an object to rotate around an axis.
- Force (F) is the magnitude of the force applied.
- Distance (d) is the perpendicular distance from the axis of rotation to the line of action of the force.
- θ is the angle between the force vector and the line connecting the axis of rotation to the point of application of the force.
Plugging in the values you provided:
- Force (F) = 94 N
- Distance (d) = 1.90 m
- θ = 20.0°
First, convert the angle from degrees to radians:
θ (in radians) = θ (in degrees) x (π/180)
θ (in radians) = 20° x (π/180) ≈ 0.3491 radians
Now, substitute the values into the formula:
Torque = 94 N x 1.90 m x sin(0.3491)
Calculating the result:
Torque ≈ 94 N x 1.90 m x sin(0.3491)
Torque ≈ 94 N x 1.90 m x 0.3420
Torque ≈ 61.6296 N·m
Therefore, the torque exerted by the fish about an axis perpendicular to the page and passing through the fisher's hand is approximately 61.6296 N·m.
To calculate the torque exerted by the fish about an axis perpendicular to the page and passing through the fisher's hand, you need to use the formula:
Torque = force × perpendicular distance
where,
force is the force applied by the fishing pole (F),
perpendicular distance is the perpendicular distance between the axis of rotation and the line of action of the force (r),
and the angle between the force and the perpendicular distance is given by θ.
In this case, the force applied by the fishing pole is F = 94 N and the perpendicular distance is the horizontal distance from the axis of rotation to the point of application of force, which is given by the equation d = l × sin(θ).
Given:
F = 94 N,
l = 1.90 m, and
θ = 20.0°.
1. First, calculate the perpendicular distance (r) using the equation:
r = l × sin(θ)
r = 1.90 m × sin(20.0°)
r = 1.90 m × 0.3420
r ≈ 0.6498 m
2. Now, plug the values of force (F) and perpendicular distance (r) into the torque formula:
Torque = F × r
Torque = 94 N × 0.6498 m
Torque ≈ 61.0952 N·m
Therefore, the torque exerted by the fish about an axis perpendicular to the page and passing through the fisher's hand is approximately 61.0952 N·m.