MCAT content library / Chemistry & physics / Torque and levers: solve a static-equilibrium problem
Choose a useful pivot, retain every external force, and use perpendicular moment arms to solve an original forearm-and-load model.
About 14 minutes, including the worked examples and checks.
Useful foundations
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Concept: a force can translate and turn an object
The same force can have different turning effects depending on where it acts. A pull close to a hinge may require a much greater force than a pull farther away. Torque describes this turning effect about a chosen pivot. Solving a static-equilibrium problem requires both a force balance and a torque balance, within the mechanics scope of AAMC Category 4A.1
For a force of magnitude F applied at position r from the pivot, the torque magnitude is τ = rF sinθ = Fd⊥. Here θ is the angle between the position vector and the force, and d⊥ is the shortest perpendicular distance from the pivot to the force’s line of action. The distance along the object is the moment arm only when the force is perpendicular to that object.2
Choose counterclockwise torque as positive and clockwise torque as negative for a planar diagram. A force directed through the pivot has zero moment arm even if its magnitude is large. Torque is expressed in N·m. Although that dimension also occurs in energy, a torque is a turning effect, not an amount of energy.
For a rigid object at rest, ΣFx = 0, ΣFy = 0 and Στ = 0. One condition cannot replace another. Equal opposite forces at different positions can sum to zero force while producing a nonzero turning effect. Conversely, a force through the pivot contributes no torque about that pivot but still belongs in the force balance.3
Pattern: draw every external force on one system
Use the following original simplified model of a horizontal forearm together with a held load. Treat them as one rigid system. The elbow is an ideal pin that can exert horizontal and vertical forces but no applied torque couple. The tendon supplies one force. Ignore other contacts and assume the system remains at rest. These are model assumptions, not a description of all forces in a real limb.
| Quantity | Value | Location or direction |
|---|---|---|
| Forearm weight | 12 N | Downward; center of mass at 0.150 m |
| Held-load weight | 28 N | Downward at 0.300 m |
| Tendon attachment | 0.060 m | Measured from the elbow |
| Tendon direction | 30° | Above the horizontal toward the elbow |
| Elbow force | Rₓ and Rᵧ | Acts at the pivot |
Because the forearm and held load are one system, the contact force between them is internal and is not added again. The two weights, tendon pull and joint force are external. If you instead isolated the forearm alone, you would replace the held-load weight with the contact force exerted by that stationary load. Mixing both descriptions would double count the load.4
Rule: use a pivot that removes an unknown torque
Taking torques about the elbow makes both joint-force components contribute zero torque. This does not make them zero forces. The tendon position vector points right while its force points up and left: their angle is 150°, whose sine equals sin30° = 0.500. Thus its perpendicular moment arm is 0.060 × 0.500 = 0.030 m.
The tendon tends to turn the system counterclockwise, while both weights tend to turn it clockwise. Torque balance gives 0.030T − 12(0.150) − 28(0.300) = 0. The total clockwise torque is 1.80 + 8.40 = 10.20 N·m, so T = 340 N. Using the full 0.060 m as the tendon moment arm would incorrectly halve this answer.
Now check translation. With right and up positive, the tendon components are Tx = −340 cos30° ≈ −294 N and Ty = 340 sin30° = 170 N. Horizontal balance gives Rx ≈ +294 N. Vertical balance gives Ry + 170 − 12 − 28 = 0, hence Ry = −130 N: the joint force on this system points downward in that component.
A downward joint component is compatible with equilibrium because the tendon’s upward component exceeds the total weight. The system exerts an opposite force on the joint, but that reaction acts on another object and is not an additional force in this system’s diagram. Choosing a different pivot would change the individual torque terms; satisfying both force and torque balance would give the same physical answer.
Application: separate location from total weight
In a uniform gravitational field, the combined center of mass can be located by weighting each position by its mass, or equivalently by its weight. Here xCM = [12(0.150) + 28(0.300)]/(12 + 28) = 0.255 m. A single downward 40 N force at 0.255 m reproduces the two gravitational torques. Placing that same 40 N at the hand would not.5
If only the 28 N load moves to 0.200 m, the clockwise torque becomes 12(0.150) + 28(0.200) = 7.40 N·m. With the tendon geometry unchanged, T becomes 7.40/0.030 ≈ 247 N. The load’s torque falls by one third, but the total torque does not, because the forearm’s weight remains at its original location.
Alternatively, keep both weights in place and change the tendon to pull vertically at the same 0.060 m attachment. Its moment arm doubles, so T falls from 340 N to 170 N. The required upward tendon component is still 170 N; it is now the entire tendon force. A change in angle can therefore change total force without changing the vertical component demanded by torque balance.
Application: interpret mechanical advantage carefully
For an ideal weightless lever with only an input force, a load force and a pivot reaction, torque balance gives Fout/Fin = din/dout. This output-to-input force ratio is mechanical advantage. Both distances are perpendicular moment arms. An input arm of 0.40 m and output arm of 0.10 m give mechanical advantage 4: an input 25 N can balance an output 100 N.6
In a small ideal movement with forces along their respective displacements, moving the input 4 cm moves the output 1 cm. Input and output work are each 1 J; force multiplication does not create energy. The forearm model includes its own weight as an extra load, so its force ratio cannot be obtained by silently discarding that torque. Before using a lever shortcut, check the system boundary, every external load, the perpendicular distances and the stated support model.
ACTIVE PRACTICE · MECHANICAL REASONING
Find the Tendon Force From Torque Balance
Use the original horizontal forearm model. The forearm weighs 12 N at 0.150 m; the held load is 28 N initially at 0.300 m. The tendon attaches 0.060 m from the elbow and pulls toward the elbow, at an initial 30° above horizontal. The elbow acts as a pin without a torque couple.
Clockwise weight torque: N·m. Effective tendon moment arm: m.
Required tendon force: N. Joint reaction components: Rx = N, Ry = N.
Using the elbow as pivot removes joint-force torque but not joint forces. Rx is right-positive, Ry up-positive. Tendon force points up and left. The two force-balance equations are checked after torque balance.
Check your reasoning
Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.
Check 1
A force acts at a point 0.080 m from a pivot on a straight horizontal rod. Its line of action makes an acute 30° angle with the rod. What perpendicular moment arm should multiply the force magnitude in the torque calculation? Use sin 30° =0.500 and cos 30° ≈0.866.
A. 0.020 m
Review this choice. This halves the required moment arm a second time. The distance is 0.080 sin 30° =0.040 m.
B. 0.040 m
Correct. The perpendicular distance is r sinθ, giving 0.080×0.500 =0.040 m.
C. 0.069 m
Review this choice. This uses r cos 30°. That projection runs along the force direction instead of giving the shortest distance to its line of action.
D. 0.080 m
Review this choice. The full attachment distance would be the moment arm for a perpendicular force, not for the supplied 30° angle.
The principle to carry forward
Use the perpendicular distance to the force line, or multiply the attachment distance by the sine of the included angle.
Check 2
A weightless lever is in static equilibrium about an ideal pin. Its only other forces are an input and a 60 N load, which create opposing torques. Their perpendicular moment arms are 0.240 m and 0.080 m, respectively. What input force and output-to-input mechanical advantage are required?
A. Input 180 N; mechanical advantage⅓.
Review this choice. This reverses the moment-arm ratio. The longer input arm reduces the required force.
B. Input 60 N; mechanical advantage 1.
Review this choice. Equal force magnitudes would balance only if the perpendicular arms were equal.
C. Input 20 N; mechanical advantage 3.
Correct. F_in(0.240)=60(0.080), so F_in=20 N. The force ratio 60/20 equals 3.
D. Input 6.67 N; mechanical advantage 9.
Review this choice. The arm ratio is 3, not its square. Torque depends linearly on the moment arm.
The principle to carry forward
With the stated three-force model, mechanical advantage is the input moment arm divided by the output moment arm.
Check 3
A horizontal uniform bar has weight 20 N acting 0.200 m from a pin at its left end. A vertical cable pulls upward 0.100 m from that pin. A separate 30 N load moves from 0.400 m to 0.200 m from the pin. The system is stationary, and the pin exerts no torque couple. What is the new cable tension?
A. 80 N
Review this choice. The original tension is 160 N, but halving only the load’s distance does not halve the bar’s unchanged 4 N·m torque.
B. 100 N
Correct. The new downward torque is 20(0.200)+30(0.200)=10 N·m. Dividing by the cable’s 0.100 m moment arm gives 100 N.
C. 120 N
Review this choice. The new downward torque is 10 N·m, not 12 N·m. Keep the two loads and their stated locations separate.
D. 160 N
Review this choice. This is the initial value: [20(0.200)+30(0.400)]/0.100. Moving the load inward reduces its torque.
The principle to carry forward
Update only the changed load’s torque, then retain every other term in the equilibrium equation.
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Sources
1. AAMC: What’s on the MCAT Exam? — Category 4A
2. OpenStax University Physics Volume 1: Torque
3. OpenStax University Physics Volume 1: Conditions for Static Equilibrium
4. OpenStax University Physics Volume 1: Examples of Static Equilibrium