A massive uniform bar produces torque because its own weight acts at its center of mass. When additional forces act
on the bar, the net torque about the pivot is the signed sum of the bar-weight torque and all applied-force torques.
1. Torque from an applied force
For a force \(F_i\) applied at position \(x_i\), with pivot at \(x_p\), the signed lever arm is
\[
r_i=x_i-x_p.
\]
If the force makes angle \(\theta_i\) with the bar, the signed torque is
\[
\tau_i=(x_i-x_p)F_i\sin\theta_i.
\]
Only the perpendicular component \(F_i\sin\theta_i\) contributes to torque.
2. Torque from the bar’s own weight
For a homogeneous bar, the center of mass is at the midpoint:
\[
x_{\mathrm{cm}}=\frac{L}{2}.
\]
The weight of the bar is
\[
W=Mg.
\]
If the bar angle is \(\phi\) from the horizontal, gravity acts downward, so its angle relative to the bar is
\[
\theta_g=-90^\circ-\phi.
\]
Therefore, the torque from the bar’s own weight is
\[
\tau_g=(L/2-x_p)Mg\sin(-90^\circ-\phi).
\]
3. Sign convention
The calculator uses this sign convention:
\[
\tau>0 \Rightarrow \text{counterclockwise},
\qquad
\tau<0 \Rightarrow \text{clockwise}.
\]
For a horizontal bar pivoted at the left end, the center of mass is to the right of the pivot and gravity pulls
downward. This produces a clockwise torque, so the torque is negative.
4. Net torque
The net torque is the signed total:
\[
\tau_{\mathrm{net}}=\tau_g+\sum_i\tau_i.
\]
If \(\tau_{\mathrm{net}}\neq0\), the bar has a rotational acceleration tendency about the pivot. If
\(\tau_{\mathrm{net}}=0\), the bar is in rotational equilibrium.
5. Rotational equilibrium
Rotational equilibrium requires
\[
\tau_{\mathrm{net}}=0.
\]
This means
\[
\tau_g+\sum_i\tau_i=0.
\]
Clockwise and counterclockwise torque contributions cancel.
6. Solving for a missing balancing force
If one applied force \(F_u\) is unknown, then
\[
\tau_g+\tau_{\mathrm{known}}+(x_u-x_p)F_u\sin\theta_u=0.
\]
Solving for \(F_u\),
\[
F_u=
\frac{-(\tau_g+\tau_{\mathrm{known}})}
{(x_u-x_p)\sin\theta_u}.
\]
A balancing force is possible only if its lever arm and perpendicular component are not zero.
7. Worked example
A \(4.0\ \mathrm{kg}\) uniform bar has length \(1.2\ \mathrm{m}\). It is pivoted at the left end and held
horizontally.
\[
M=4.0\ \mathrm{kg},
\qquad
L=1.2\ \mathrm{m},
\qquad
x_p=0.
\]
Its center of mass is
\[
x_{\mathrm{cm}}=\frac{L}{2}=0.6\ \mathrm{m}.
\]
Its weight is
\[
W=Mg=(4.0)(9.80665)=39.2266\ \mathrm{N}.
\]
The bar is horizontal, so \(\phi=0^\circ\). Gravity is downward, so
\[
\theta_g=-90^\circ.
\]
The torque due to the bar’s weight is
\[
\tau_g=(0.6)(39.2266)\sin(-90^\circ).
\]
\[
\tau_g=-23.5\ \mathrm{N\,m}.
\]
The magnitude is \(23.5\ \mathrm{N\,m}\), and the negative sign means clockwise:
\[
\boxed{|\tau_g|=23.5\ \mathrm{N\,m}\ \text{clockwise}}.
\]
Formula summary
| Concept |
Formula |
Use |
| Bar center of mass |
\(x_{\mathrm{cm}}=L/2\) |
Location where the weight of a homogeneous bar acts. |
| Bar weight |
\(W=Mg\) |
Force of gravity on the bar. |
| Bar-weight torque |
\(\tau_g=(L/2-x_p)Mg\sin(-90^\circ-\phi)\) |
Torque caused by the bar’s own weight. |
| Applied-force torque |
\(\tau_i=(x_i-x_p)F_i\sin\theta_i\) |
Torque from one applied force. |
| Net torque |
\(\tau_{\mathrm{net}}=\tau_g+\sum_i\tau_i\) |
Total rotational effect about the pivot. |
| Rotational equilibrium |
\(\tau_{\mathrm{net}}=0\) |
Clockwise and counterclockwise torques cancel. |
| Balancing force |
\(F_u=\frac{-(\tau_g+\tau_{\mathrm{known}})}{(x_u-x_p)\sin\theta_u}\) |
Force needed to make the net torque zero. |
Key idea: a massive uniform bar is not torque-free. Its weight acts at \(L/2\), so changing the pivot position changes
the gravitational torque.
