# Torque Self-Test: Net Torque

The pivot point is at O.

\(F_1 = 10 N\), and is at a distance of 0.25 m from O, where \(\theta = 80^\circ\)

\(F_2 = 7.0 N\), acting perpendicular to the object, at a distance of 1.25 m from O

\(F_3 = 12 N\), is 0.60 m from O, and acts at \(\theta = 40 ^\circ\) from the horizontal

Find the total (net) torque on the object.

The net total torque on the object is:

- A. -1.7 N m
- B. -15.8 N m
- C. -6.6 N m
- D. 18 N m
- E. 61 N m

- A. No. Remember, counter clockwise torques are + and clockwise are -
- B. No. Remember, counter clockwise torques are + and clockwise are -
**C. Correct**- D. No. Remember, counter clockwise torques are + and clockwise are -
- E. No. Remember, counter clockwise torques are + and clockwise are -

As with forces, individual torques can add to give the net torque.

In this question, we have three separate forces. All three forces act at a distance from the pivot point, O. This gives rise to three individual torques, each one caused by a separate force. Note that we have forces acting in different directions, and so may cause different directions of torque. Let's take the forces causing a counter-clockwise motion of the object to have positive torque. Clockwise motions will be taken to have negative torques.

Letting \(\tau_1\) be the torque caused by \(F_1\), similarly with \(\tau _ 2\) and \(\tau _3\), we can see that \(F_1\) and \(F_2\) both cause torques in the negative direction (ie. clockwise), while \(F_3\) causes a positive torque (ie. counter-clockwise). We need to consider this when summing the total torque, \(\tau \) on this object.

\(\sum \tau = \tau_1 + \tau_2 + \tau_3\)

Let's take each torque separately, beginning with the one caused by \(F_1\)

\(\tau _1 = r_1 F_1 \sin(\theta_1)\)

\( = (0.25) \cdot (10) \cdot \sin (80) = - 2.46\)

The torque caused by \(F_2\),

\(\tau_2 = r_2 \cdot F_2 \cdot \sin(\theta _2)\)

\(= (1.25) \cdot (7.0) \cdot \sin (90) = -8.75\)

And by \(F_3\),

\(\tau_3 = r_3 \cdot F_3 \cdot \sin ( \theta_3)\)

\(= (0.6) \cdot (12) \cdot \sin (40) = 4.63\)

Summing up \(\tau_1\),\(\tau_2\), and \(\tau _3\), we get the total, or net torque.

\(\tau = -2.46 + (-8.75) + 4.63\)

\(= -6.58 Nm\)