A person exerts a horizontal force of 36 N on the end of a door 97cm wide. What is the magnitude of the torque if the force is exerted perpendicular to the door and What is the magnitude of the torque if the force is exerted at a 70.0 angle to the face of the door?
Thank you
torque=36N*.97m
torque=36N*.97m*cos20
To find the magnitude of the torque in both cases, we need to multiply the force by the distance from the axis of rotation (in this case, the hinges) to the point where the force is applied.
1. When the force is exerted perpendicular to the door:
- Given force, F = 36 N
- Distance, r = 97 cm = 0.97 m (converted to meters)
- The torque (τ) is calculated by the equation: τ = F * r
- Substituting the values: τ = 36 N * 0.97 m = 34.92 N·m
So, the magnitude of the torque when the force is exerted perpendicular to the door is 34.92 N·m.
2. When the force is exerted at a 70.0° angle to the face of the door:
- Again, given force, F = 36 N
- The distance from the force to the axis of rotation is perpendicular to the door and is still 97 cm = 0.97 m.
- But now we need to find the component of the force that is perpendicular to the door, which is given by F * sin(θ), where θ is the angle of 70.0°.
- Therefore, the perpendicular component of the force, F_perpendicular = 36 N * sin(70.0°) ≈ 34.65 N (rounded to two decimal places)
- The torque (τ) is still calculated by the equation: τ = F_perpendicular * r
- Substituting the values: τ = 34.65 N * 0.97 m = 33.66 N·m (rounded to two decimal places)
So, the magnitude of the torque when the force is exerted at a 70.0° angle to the face of the door is approximately 33.66 N·m.