Torque & Rotational Equilibrium Simulator

A uniform beam rests on a pivot at its center, with a weight hanging on each side. Tick marks on the beam are 1 m apart, out to 5 m on each side. Because the beam is uniform and pivoted at its middle, its own weight acts at the pivot and adds no torque, so only the two hanging weights matter. The simulator uses $g = 9.8$ m/s².

The panel lists the CCW torque from the left weight, the CW torque from the right weight and the net torque, with counterclockwise counted as positive, all in N·m to one decimal place. The verdict reads “Balanced — Στ = 0”, “Rotates CCW — left side drops” or “Rotates CW — right side drops”. The beam tilts toward the side that wins, more for a larger net torque. Treat the tilt as an indicator only: a real beam with a nonzero net torque would keep rotating until something stopped it.

Setting a mass to 0 removes that weight; setting a distance to 0 hangs it directly on the pivot, where it makes no torque however heavy it is. Try doubling one mass and then finding the distance on the same side that restores balance: it should be half the original distance.

Torque measures how effectively a force turns an object about an axis: $$\tau = r_\perp F = rF\sin\theta$$ where $r$ is the distance from the axis to where the force acts and $\theta$ is the angle between $\vec{r}$ and $\vec{F}$. A hanging weight pulls straight down on a level beam, so $\theta = 90°$ and $\tau = mgr$.

A rigid object is in rotational equilibrium when the torques about any axis add to zero: $$\sum\tau = 0 \quad\Rightarrow\quad m_L g\, r_L = m_R g\, r_R$$ Because $g$ appears on both sides, it cancels: balance depends only on $m_L r_L = m_R r_R$.

Full static equilibrium also needs $\sum F = 0$. Here the pivot pushes up with a force equal to the total weight of the beam and both masses. That support force acts at the axis, so it produces no torque about the pivot, which is why choosing the pivot as the axis is usually the fastest route.

If the torques do not cancel, Newton's second law in rotational form, $\alpha = \sum\tau / I$, says the beam gains angular acceleration in the direction of the net torque.

A 6 kg mass hangs 2.5 m to the left of the pivot. Where must a 5 kg mass hang on the right for the beam to balance?

Left torque. $\tau_L = (6)(9.8)(2.5) = 147$ N·m, counterclockwise.

Balance condition. $(5)(9.8)\, r_R = 147$, so $r_R = 147/49 = 3.0$ m. Using the shortcut, $r_R = (6)(2.5)/5 = 3.0$ m. The lighter mass must sit farther out.

Off balance. If the 5 kg mass is moved to 2.0 m, $\tau_R = (5)(9.8)(2.0) = 98$ N·m, and the net torque is $147 - 98 = +49$ N·m, counterclockwise, so the left side drops.

Pivot force. In the balanced case, ignoring the beam's own mass, $\sum F = 0$ means the pivot pushes up with $(6 + 5)(9.8) = 107.8$ N.

To check in the simulator, set m_L = 6 kg, r_L = 2.5 m, m_R = 5 kg and r_R = 3.0 m. Both torques read 147.0 N·m and the verdict says balanced. Slide r_R to 2.0 m and the panel shows 98.0 N·m on the right, a net of +49.0 N·m, and “Rotates CCW — left side drops”.

Rotational equilibrium is Topic 5.5 in Unit 5, Torque and Rotational Dynamics, and it builds directly on Topic 5.3, Torque. Exam problems include seesaws, shelves held by a cable, planks resting on two supports and ladders against walls. In each, choose an axis where an unknown force acts so that force drops out of the torque equation, then use $\sum F = 0$ to find what remains.