Torque measures the turning effect of a force about a pivot point. For a massless bar, the bar itself has no weight
contribution, so the net torque comes only from the applied forces.
1. Torque from one force
If a force \(F\) is applied at position \(x_i\) on a bar with pivot at \(x_p\), the lever arm is
\[
r_i=x_i-x_p.
\]
If the force makes angle \(\theta_i\) with the positive direction of the bar, then the signed torque is
\[
\tau_i=r_iF_i\sin\theta_i.
\]
Only the force component perpendicular to the bar produces torque:
\[
F_{\perp}=F_i\sin\theta_i.
\]
2. Sign convention
This calculator uses the common convention:
\[
\tau>0 \Rightarrow \text{counterclockwise},
\qquad
\tau<0 \Rightarrow \text{clockwise}.
\]
The sign comes automatically from the lever arm and the force angle. For example, an upward force to the right of
the pivot gives a positive torque, while a downward force to the right of the pivot gives a negative torque.
3. Net torque
The net torque is the signed sum of all individual torques:
\[
\tau_{\mathrm{net}}=\sum_i \tau_i.
\]
Written out,
\[
\tau_{\mathrm{net}}
=
\sum_i (x_i-x_p)F_i\sin\theta_i.
\]
4. Rotational equilibrium
A bar is in rotational equilibrium when the net torque about the pivot is zero:
\[
\tau_{\mathrm{net}}=0.
\]
This does not mean that no forces act. It means that the clockwise and counterclockwise torque effects cancel.
5. Solving a missing balancing force
Suppose one force \(F_u\) is unknown. Let the sum of all known torques be \(\tau_{\mathrm{known}}\). For equilibrium,
\[
\tau_{\mathrm{known}}+(x_u-x_p)F_u\sin\theta_u=0.
\]
Solving for the unknown force gives
\[
F_u=
\frac{-\tau_{\mathrm{known}}}{(x_u-x_p)\sin\theta_u}.
\]
This only works if the unknown force has a nonzero lever arm and a nonzero perpendicular component.
6. Worked example
A \(2.0\ \mathrm{m}\) bar is pivoted at its center, so
\[
x_p=1.0\ \mathrm{m}.
\]
Force 1 is \(15\ \mathrm{N}\) downward at \(x_1=0.6\ \mathrm{m}\), so
\[
r_1=x_1-x_p=0.6-1.0=-0.4\ \mathrm{m},
\qquad
\theta_1=-90^\circ.
\]
Its torque is
\[
\tau_1=(-0.4)(15)\sin(-90^\circ)=+6.0\ \mathrm{N\,m}.
\]
Force 2 is \(5\ \mathrm{N}\) downward at \(x_2=1.6\ \mathrm{m}\), so
\[
r_2=1.6-1.0=0.6\ \mathrm{m},
\qquad
\theta_2=-90^\circ.
\]
Its torque is
\[
\tau_2=(0.6)(5)\sin(-90^\circ)=-3.0\ \mathrm{N\,m}.
\]
Therefore,
\[
\tau_{\mathrm{net}}=6.0-3.0=3.0\ \mathrm{N\,m}.
\]
Since the net torque is positive, the rotation tendency is counterclockwise:
\[
\boxed{\tau_{\mathrm{net}}=3.0\ \mathrm{N\,m}\ \text{counterclockwise}}.
\]
Formula summary
| Concept |
Formula |
Use |
| Lever arm |
\(r_i=x_i-x_p\) |
Signed distance from pivot to force position. |
| Perpendicular force component |
\(F_{\perp}=F_i\sin\theta_i\) |
Only this component creates torque. |
| Individual torque |
\(\tau_i=(x_i-x_p)F_i\sin\theta_i\) |
Signed torque from one applied force. |
| Net torque |
\(\tau_{\mathrm{net}}=\sum_i\tau_i\) |
Add all signed torques. |
| Rotational equilibrium |
\(\tau_{\mathrm{net}}=0\) |
Clockwise and counterclockwise effects cancel. |
| Balancing force |
\(F_u=\frac{-\tau_{\mathrm{known}}}{(x_u-x_p)\sin\theta_u}\) |
Find the force needed to make the net torque zero. |
Key idea: torque depends on force, distance from the pivot, and angle. A large force produces no torque if it acts
through the pivot or parallel to the bar.
