# 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

#### Solution

• 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.

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

$= -6.58 Nm$