Torque and Static Equilibrium: Find the Perpendicular Lever Arm

Nablio · 40 minPublished October 4, 2026

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The same push can turn a bar strongly, weakly or not at all. The difference is the force’s line of action relative to the point about which you calculate torque. This guide connects that geometry to the three equations needed to keep a rigid object at rest in a plane.

Before you start: know force diagrams, signed components, basic algebra and sine/cosine. No calculus is needed. By the end, you should be able to choose a useful pivot, find a perpendicular lever arm, assign a torque sign and check both force balance and torque balance.

Model: all examples are idealized classroom systems in an inertial reference frame, where a particle with no net force stays at rest or moves in a straight line at constant speed. Bars are rigid, forces lie in one plane and supports behave exactly as stated. We ignore deformation, strength limits and failure. These calculations do not assess whether a real support, cable or structure is safe.

1. Distinguish force from torque

A force describes an interaction and is measured in newtons (N). Its net value determines the acceleration of the centre of mass. Torque describes the turning effect of a force about a specified point or axis. It depends on where the force acts as well as on its direction and magnitude.

F: force magnitude, in N. Its signed components are F_x and F_y.
r: distance from the chosen point O to the point of application, in metres (m). The position components x and y are measured from O.
θ: the angle between the position vector from O to the force application point and the force vector, from 0° to 180°.
d_⊥: shortest distance from O to the force’s line of action, in m. This distance is never negative.
τ_O: signed torque about the axis through O perpendicular to the page, in newton-metres (N·m). Here counterclockwise is positive and clockwise is negative.

Torque and energy share dimensions, but describe different quantities. Write torque in N·m rather than J. A nonzero force can have zero torque about one point, and a force can have different torques about different points. Always name the point.

2. Measure to the line of action

Extend the force arrow into an infinite straight line: its line of action. Draw the shortest segment from O to that line. The segment meets it at a right angle and has length d_⊥. It need not end at the point where the force touches the object.

$d_{\perp}=r\sin\theta$

$|\tau_O|=Fd_{\perp}=rF\sin\theta$

The vertical bars mean magnitude. Assign the sign separately from the turning tendency. If the line of action passes through O, the perpendicular distance is zero, even if the force acts far from O. For fixed r and F, a 90° angle gives the largest torque magnitude.

For a horizontal bar extending right from O, only the vertical component of the force produces torque about O. More generally, in rightward x and upward y coordinates:

$\tau_O=xF_y-yF_x$

This component formula includes the sign. Do not calculate it and then add another minus sign for clockwise rotation. Likewise, use either Fd_⊥ or rF sin θ; multiplying by sin θ in both places would count the geometry twice.

O is left of the force application point A on a horizontal bar. The force points 35 degrees above the rightward direction. A perpendicular segment from O ends at Q on the backward extension of the force line. — Figure 1. OA is r. OQ is d⊥. The right-angle mark is at Q, on the force’s line of action. The force arrow has the same direction everywhere on that dashed line.

3. Worked example: an angled push

A 40 N force acts at the end of an ideal wrench, 0.30 m to the right of O. It points 35° above the rightward horizontal. Find its torque about O. This example concerns the torque of this force only; it does not claim the wrench is in equilibrium.

Find the distance and the sign

The position vector points right, so the angle between it and the force is 35°. The force tends to turn the wrench counterclockwise: the sign is positive.

$d_{\perp}=(0.30\,\mathrm m)\sin35^\circ$

$d_{\perp}\approx0.172\,\mathrm m$

$\tau_O=+(40\,\mathrm N)(0.172073\,\mathrm m)$

$\tau_O\approx+6.9\,\mathrm{N\cdot m}$

Check a second way

Resolve the force: F_y = 40 sin 35° ≈ 22.94 N. The horizontal component acts along a line through O and contributes no torque.

$\tau_O=(0.30\,\mathrm m)(22.943\,\mathrm N)$

$\tau_O\approx+6.9\,\mathrm{N\cdot m}$

If the same force were perpendicular to the wrench and upward, its torque would be +12 N·m. If it pointed horizontally right, its torque would be zero. If it pointed 35° below horizontal, the magnitude would stay 6.9 N·m but the sign would be negative. Keep unrounded values during calculation; round the final answer.

4. Choose a pivot and write all three balances

An object in static equilibrium is at rest and remains at rest in the chosen inertial frame. In this planar rigid-body model, use all three conditions:

$\sum F_x=0$

$\sum F_y=0$

$\sum \tau_O=0$

The symbol Σ means “add all contributions.” Include all external forces on the selected object. A zero net force alone does not rule out a turning effect. A zero torque about one chosen point alone does not rule out a net force. The conditions describe balance; initial rest is also needed to call the situation static.

A pivot is a choice in the calculation

O may be a real hinge, a contact point or any convenient fixed point in the diagram. Choosing O at the intersection of unknown force lines removes their torque terms, because each lever arm is zero. Those forces still belong in the force equations. When the net force is zero, the net torque is the same about every origin; once both conditions hold, changing the pivot changes no physical answer.

A practical order

Isolate the body and draw every external force at its point or line of application. Do not draw a force and its components as three separate forces.
Mark the force application positions, axes and positive turning sense. Put a uniform bar’s weight at its midpoint; a nonuniform bar needs its specified centre of mass.
Choose a pivot that removes useful unknowns. Write one signed torque equation, then both force equations.
Solve, restore units and substitute into all balances. A cable must pull with nonnegative tension; a simple contact can push but cannot pull. A negative result may reverse an assumed direction or invalidate a one-way support model.

We use ideal pins or hinges that exert horizontal and vertical forces but no separate applied couple. A built-in support that can exert its own moment would require a different force diagram.

5. Worked example: balance a bar

An ideal rigid bar of negligible weight is 1.20 m long. A frictionless pin at x = 0.40 m supports it. A 12 N force acts vertically downward at x = 0.10 m. A second downward force P acts at x = 1.00 m. Coordinates are measured from the left end. Find P and the pin’s vertical force for equilibrium.

A horizontal 1.20 metre bar has a 12 newton downward force at x 0.10 m, a pin at x 0.40 m, and downward force P at x 1.00 m. Distances from the pin are 0.30 m to the first force and 0.60 m to P. — Figure 2. Distances in the torque equation are measured from the chosen pin, not from the left end.

Use the pin to remove the support force

About the pin, the 12 N force turns counterclockwise and P turns clockwise. Their perpendicular lever arms are 0.30 m and 0.60 m.

$+(12)(0.30)-P(0.60)=0$

$P=6.0\,\mathrm N$

Each product in the first equation is a torque in N·m, with distances in metres and forces in newtons. The larger lever arm needs a smaller force to produce the matching torque magnitude.

Return to force balance

Let R_y be the upward pin reaction. There are no other horizontal forces, so R_x = 0.

$R_y-12-6=0$

$R_y=18\,\mathrm N$

Check about a different point

About the left end, the support has a nonzero lever arm. Substitute every force:

$\begin{aligned}\sum\tau_{\rm left}&=-12(0.10)\\&\quad+18(0.40)\\&\quad-6(1.00)=0\end{aligned}$

The torques are −1.2 + 7.2 − 6.0 = 0 N·m. The pin was convenient, not compulsory. Forgetting its force after moving the pivot would spoil the check.

6. Worked example: a hinge and an angled cable

An ideal uniform horizontal bar is 2.00 m long and weighs 60 N. It is pinned at its left end O. A 100 N downward load acts 1.50 m from O. A light cable at the right end pulls up and left at 30° above the leftward horizontal. Find the cable tension T and the hinge-force components H_x and H_y. The cable and hinge are ideal constraints; no strength or safety conclusion is implied.

A 2 metre bar is pinned at O on the left. Weight 60 N acts downward at x 1 m and load 100 N at x 1.5 m. A cable tension T acts up-left at the right endpoint, 30 degrees above leftward horizontal. Hx points right and Hy up at O. — Figure 3. The cable angle is measured from the leftward horizontal. Its upward component is T sin 30° and its leftward component has magnitude T cos 30°. Force-arrow lengths are schematic.

Balance turning about the hinge

The bar’s weight acts at its midpoint, 1.00 m from O. The hinge forces pass through O. The cable’s horizontal component also has zero torque about O; only its upward component contributes.

$\begin{aligned}2.00T\sin30^\circ&=60(1.00)\\&\quad+100(1.50)\end{aligned}$

$2.00T\sin30^\circ=210\,\mathrm{N\cdot m}$

$T=210\,\mathrm N$

The cable force makes 150° with the rightward position vector. Using sin 150° gives the same result as sin 30°. Its perpendicular lever arm is 2.00 sin 30° = 1.00 m, so Fd_⊥ also gives 210 N·m.

Find the hinge force from both component equations

$H_x-T\cos30^\circ=0$

$H_x\approx182\,\mathrm N\quad\text{right}$

$H_y+T\sin30^\circ-60-100=0$

$H_y=55\,\mathrm N\quad\text{up}$

The tension is not equal to the 160 N total weight. Its upward component is only 105 N; the hinge supplies the remaining 55 N. Its leftward component is balanced by the rightward hinge force.

Check all three balances

Horizontal forces: +181.865 − 181.865 = 0 N.
Vertical forces: +55 + 105 − 60 − 100 = 0 N.
Torques about O: +210 − 60 − 150 = 0 N·m.

The positive tension agrees with a taut cable pulling in the stated direction. A negative hinge component would mean the hinge force points opposite the arrow we initially chose; an ideal pin can act in either direction.

7. Six questions to try

Use the same sign convention throughout: right and up positive, counterclockwise torque positive. In questions 1–2, the position vector points right from O. Give torque units and explain its sign. Try each question before opening its hint; full solutions are in the next section.

1. Check the geometry before multiplying

A 25 N upward force is applied 0.16 m to the right of O. Find its torque. Compare it with a 25 N force applied 0.40 m to the right of O but pointing horizontally right. Which produces the larger torque magnitude?

Hint for question 1

Find the shortest distance to each force line. A longer distance to the application point does not guarantee a larger lever arm.

2. Use an obtuse angle

A force of 18 N acts at x = 0.50 m, y = 0. Its direction is 150° counterclockwise from the positive x-axis. Find d_⊥ and the signed torque about O. Does its leftward component make the torque negative?

Hint for question 2

Sketch the up-left force. Use sin 150° for the magnitude, then check xF_y − yF_x.

3. Test whether zero net force is enough

Two forces act on a horizontal bar: 10 N upward at x = −0.30 m and 10 N downward at x = +0.30 m. Their application points have y = 0. Find the net force and torque about x = 0. Could these two forces alone maintain static equilibrium?

On a horizontal bar, a 10 N upward force acts 0.30 m left of O and a 10 N downward force acts 0.30 m right of O. Both tend to turn the bar clockwise. — Figure 4. Equal and opposite forces can act along different lines.

Hint for question 3

The forces cancel as vectors. Check the turning tendency of each separately.

4. Choose a pivot, then check another one

A bar of negligible weight is pinned at x = 0.50 m. A 14 N downward force acts at x = 0.20 m, and a downward force Q acts at x = 1.10 m. The application points all lie on the horizontal bar. Find Q and the vertical pin reaction. Check torque about x = 0.

Hint for question 4

Relative to the pin, the downward forces have lever arms 0.30 m and 0.60 m on opposite sides.

5. Diagnose an incomplete equilibrium calculation

A uniform horizontal bar has length 2.00 m and weight 40 N. Apart from its weight, its only force is an upward force F at x = 1.50 m from the left end. There is no hinge or other support. A student chooses the left end as O, sets net torque to zero and claims the bar is in static equilibrium. Find their F, test the net force and explain the error.

Hint for question 5

Choosing a point for calculating torque does not create a physical support or a new force there.

6. Change the cable angle

Return to the hinge-and-cable example with the same bar and load positions. Change the cable angle to 45° above the leftward horizontal. Find T, H_x and H_y. Which component of T must remain the same? Explain what happens to the required tension as the angle approaches 0° from above, within this ideal model.

Hint for question 6

Torque balance fixes T sin α, where α is the angle above the leftward horizontal. Write that product before solving for T.

8. Solutions and checks

Solution 1: a longer radius can give zero torque

$\begin{aligned}\tau_O&=+(25)(0.16)\\&=+4.0\,\mathrm{N\cdot m}\end{aligned}$

For the horizontal force, the line of action passes through O. Thus d⊥ = 0 and τO = 0. The first force has the larger turning effect.

Solution 2: the vertical component fixes the sign here

$d_{\perp}=0.50\sin150^\circ=0.25\,\mathrm m$

$\begin{aligned}\tau_O&=+(18)(0.25)\\&=+4.5\,\mathrm{N\cdot m}\end{aligned}$

F_y = +9.0 N. With x = +0.50 m and y = 0, xF_y − yF_x = +4.5 N·m. The leftward component contributes no torque about O.

Solution 3: a pure turning effect

$\sum F_y=10-10=0$

$\begin{aligned}\sum\tau_O&=(-0.30)(10)\\&\quad+(0.30)(-10)\end{aligned}$

$\sum\tau_O=-6.0\,\mathrm{N\cdot m}$

Both torques are clockwise. These equal, opposite forces on separated lines form a couple: zero resultant force but nonzero torque. Because the net force is zero, shifting the torque origin cannot remove the −6.0 N·m result. These forces alone cannot maintain static equilibrium.

Solution 4: include the pin in the second check

$14(0.30)-Q(0.60)=0$

$Q=7.0\,\mathrm N$

$R_y=14+7=21\,\mathrm N$

The horizontal reaction is zero. About x = 0, all distances must be measured from that new origin:

$\begin{aligned}\sum\tau_{x=0}&=-14(0.20)\\&\quad+21(0.50)\\&\quad-7(1.10)=0\end{aligned}$

The terms are −2.8 + 10.5 − 7.7 = 0 N·m. Both force and torque balances pass.

Solution 5: a zero moment about one point is insufficient

$1.50F-40(1.00)=0$

$F=26.666\ldots\,\mathrm N$

$\sum F_y=26.666\ldots-40$

$\sum F_y=-13.333\ldots\,\mathrm N$

There is a downward resultant, so the bar is not in static equilibrium. Setting F = 40 N would balance the force but would give +20 N·m about the left end. No magnitude of that one upward force at the stated position satisfies both conditions. The arbitrarily chosen torque origin supplies no reaction force.

Solution 6: torque fixes the upward cable component

$T\sin\alpha=105\,\mathrm N$

$T=\frac{105}{\sin45^\circ}\approx148.5\,\mathrm N$

$H_x=T\cos45^\circ=105\,\mathrm N$

$H_y=160-105=55\,\mathrm N$

H_x points right and H_y up. Compared with 30°, the required tension and horizontal hinge force decrease, but the upward cable component and vertical hinge force stay the same. For 0° < α ≤ 90°, T = 105/sin α. As α approaches 0° from above, the ideal required tension grows without bound. At exactly 0°, a horizontal cable supplies zero counterclockwise torque and no finite tension can balance the stated loads. This is a limit of the ideal equilibrium arrangement, not a safe-load recommendation.

9. What to work through with a tutor

Start with a torque or equilibrium question you’re stuck on, or a basic concept you’d like to understand better. You can build the force diagram and first equations with your tutor. Ask the tutor to work through these decisions with you:

Check which object you isolated and identify the interaction behind each force.
Have you draw the force’s line of action and its perpendicular lever arm before using a formula.
Compare two pivot choices and show where an unknown force disappears from torque balance but remains in force balance.
Ask you to predict the effect of moving a load or changing a cable angle, then check the prediction with equations.
Review one new question you solve yourself, focusing on the first unsupported step rather than only the final number.

Explore Nablio mechanics tutoring. Use your own course outline and permitted practice material to choose the next problem.

10. Next steps and sources

If drawing the external forces was difficult, start with free-body diagrams and friction. For deciding which physical principle fits a moving system, continue with work, energy or momentum.

The explanations, numerical examples and diagrams in this guide are original. For standard definitions and a wider treatment, see:

Course context: Concordia’s PHYS 204 calendar description includes statics and rotational motion. McGill’s official PHYS 101 catalogue PDF describes introductory mechanics and rotation without calculus. Those broad descriptions support topic relevance; this guide does not establish a current section’s exact syllabus, exam coverage or teaching order. Consult your course outline. Nablio is independent of these universities.

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