A person exerts a horizontal force of 42 N on the end of a door 96 cm wide. What is the magnitude of the torque if the force is exerted (a) perpendicular to the door and (b) at a 60.0° angle to the face of the door?

The magnitude of the force is \(F = 42\;{\rm{N}}\).

The width of the door is \(r = 96\;{\rm{cm}} = 0.96\;{\rm{m}}\).

Torque is defined as the measure of the force that causes an object to rotate about an axis. It is also called the rotational equivalent of linear force.

Mathematically, torque is represented as follow:

\(\begin{align}\vec \tau &= \vec r \times \vec F\\\left| \tau \right| &= rF\sin \theta \end{align}\) … (i)

Here, F is the applied force, and r is the perpendicular distance between the point about which the torque is calculated and the point of application of force.

When the force exerted is perpendicular to the door, the angle is \(\theta = {90^{\rm{o}}}\).

From equation (i),

\(\begin{align}\tau & = rF\sin {90^{\rm{o}}}\\ &= \left( {0.96\;{\rm{m}}} \right)\left( {42\;{\rm{N}}} \right)\\ &= 40.32\;{\rm{m}}\;{\rm{N}}\end{align}\)

When the force exerted is at \(\theta = {60^{\rm{o}}}\) to the face of the door,

\(\begin{align}\tau &= rF\sin {60^{\rm{o}}}\\ &= \left( {0.96\;{\rm{m}}} \right)\left( {42\;{\rm{N}}} \right)\frac{{\sqrt 3 }}{2}\\ \approx 35\;{\rm{m}}\;{\rm{N}}\end{align}\)