A man opens a 1.14-m wide door by pushing on it with a force of 49.5 N directed perpendicular to its surface.

(a) What magnitude torque does he apply about an axis through the hinges if the force is applied at the center of the door?
N · m

(b) What magnitude torque does he apply at the edge farthest from the hinges?
N · m

To calculate the torque applied by the man, we need to know the distance between the point of application of the force and the axis of rotation. In this case, the axis of rotation is through the hinges of the door.

(a) When the force is applied at the center of the door, the distance between the point of application and the axis of rotation is half the width of the door, which is 1.14 m / 2 = 0.57 m.

Torque (τ) can be calculated using the formula:
τ = r * F * sin(θ)

Where:
- τ is the torque
- r is the distance between the point of application of the force and the axis of rotation
- F is the magnitude of the force
- θ is the angle between the force vector and the lever arm (which is perpendicular to the surface of the door in this case)

Since the force is applied perpendicular to the surface of the door, θ = 90 degrees.

Plugging the values into the formula:
τ = 0.57 m * 49.5 N * sin(90 degrees)
τ = 0.57 m * 49.5 N * 1
τ = 28.215 N · m

Therefore, the magnitude of the torque applied at the center of the door is 28.215 N · m.

(b) When the force is applied at the edge farthest from the hinges, the distance between the point of application and the axis of rotation is the width of the door, which is 1.14 m.

Using the same formula as above, with the new value of the distance (r):
τ = 1.14 m * 49.5 N * sin(90 degrees)
τ = 1.14 m * 49.5 N * 1
τ = 55.83 N · m

Therefore, the magnitude of the torque applied at the edge farthest from the hinges is 55.83 N · m.