The magnitude of a torque describes how strongly it turns an object, in the same way that the magnitude of a force describes how hard it pushes on something. When you turn a door knob, the magnitude is how hard you twist the knob. When you push on a door, the magnitude is the product of how hard you push and the distance from the hinges.
Torque is produced by applying a force at some distance from center of rotation. The magnitude \(T\) is found by multiplying the applied force \(F\) by the distance \(d\) between the force and the axis of rotation measured perpendicularly. The perpendicular distance \(d\) between the force and the axis of rotation is called the moment arm, or the lever arm. Note that if the moment arm is largest when the force is at a right angle with the shank of the wrench.
\begin{equation*}
T = F d = F r \sin\theta
\end{equation*}
Since the torque is the product of the force’s magnitude and the perpendicular distance, the greater the force, the greater the torque, but the closer force is to the axis of rotation, the smaller the moment.
Figure2.5.2.Moment Arm
The direction of a torque is the direction that the torque tends to rotate the object about its axis. A clockwise torque is used to tighten ordinary nuts and screws, while a counterclockwise torque loosens them.
Example2.5.3.Torque when force is perpendicular.
A mechanic applies a force of \(\N{400}\) perpendicular to the handle of a \(\cm{60}\) long wrench in order to loosen a bolt as shown in the left diagram of Figure 2.5.2 .
Determine the torque applied to the bolt.
Answer.
\(T = \Nm{240}\text{,}\) in the clockwise direction.
Solution.
Since the applied force is perpendicular to the handle, the moment arm \(d\) is simply the length of the handle, \(r.\)
\begin{align*}
F \amp= \N{400}\\
r \amp = \cm{60} = \cm{60} \times \left[ \frac{\m{1}}{\cm{100}} \right] = \m{0.60}\\
T \amp= F d = F r\\
\amp = (\N{400}) (\m{0.60})\\
\amp = \Nm{240}
\end{align*}
Therefore, the applied torque will be \(\Nm{240}\text{,}\) in the clockwise direction.
Example2.5.4.Torque when force is applied at an angle.
A mechanic applies a force of \(\N{400}\) at an angle of 60 degrees relative to the \(\cm{60}\) long wrench to loosen a bolt, as shown in the right diagram of Figure 2.5.2.
Determine the torque applied the bolt.
Answer.
\(T = \Nm{207.8}\text{,}\) in the clockwise direction.
Solution.
Since the applied force is not perpendicular to the handle, the moment arm \(d\) is less than the length of the handle, \(r.\)
\begin{equation*}
d = r \sin \theta
\end{equation*}
\begin{align*}
F \amp= \N{400}\\
r \amp = \cm{60} = \cm{60} \times \left[ \frac{\m{1}}{\cm{100}} \right] = \m{0.60}\\
T \amp= F d = F r \sin \theta\\
\amp = (\N{400}) (\m{0.60}) \sin \ang{60}\\
\amp = \Nm{207.8}
\end{align*}
Therefore, the applied torque will be \(\Nm{207.8}\text{,}\) in the clockwise direction.
Note that this torque is less than the value found in the previous example, because applying the force at an angle is less effective than applying the force perpendicular to the handle.
