Torque & Equilibrium

Torque & Equilibrium

9 min read Updated Mar 26, 2026

You can’t open a door by pushing on the hinge. You push near the handle — as far from the hinge as possible — because that’s where your push has the most rotational effect. Try pushing 1 cm from the hinge and you’ll be amazed at how stiff the door feels. Push at the handle (~80 cm out) and it swings easily. Same door, same hinge, same push — totally different result.

That “rotational effect of a force” is torque. It’s the rotational version of force: where ordinary force makes things accelerate in a straight line, torque makes things spin around a pivot. The MCAT uses torque in seesaws, beams, levers, the human body (every joint!), and any time a problem mentions a hinge, a pivot, or a fulcrum.

The Torque Equation

The product rsinθr\sin\theta — or equivalently FsinθF\sin\theta times rr — is called the lever arm (the perpendicular distance from the pivot to the line of action of the force). You can read the formula two equivalent ways:

  1. τ=(rsinθ)×F\tau = (r\sin\theta) \times F = lever arm × full force.
  2. τ=r×(Fsinθ)\tau = r \times (F\sin\theta) = full distance × perpendicular component of force.

Both give the same answer. Use whichever matches your mental picture of the problem.

Sign Convention for Torque

Torque has a sign that tells you which way it’s spinning:

  • Counterclockwise (CCW) = positive (standard convention).
  • Clockwise (CW) = negative.

The MCAT may use either convention — what matters is consistency. Pick one for a given problem and stick with it.

Maximizing Torque

Since τ=rFsinθ\tau = rF\sin\theta, you maximize torque by making each factor as big as possible:

  • rr as large as possible (push far from the pivot).
  • FF as large as possible (push harder).
  • θ=90°\theta = 90° (push perpendicular to the lever arm, since sin90°=1\sin 90° = 1).

This is why door handles are placed far from hinges, why long wrenches loosen stubborn bolts, and why a bicycle pedal arm is a foot long, not an inch. All of them are torque-maximizing geometry.

Static Equilibrium

An object is in static equilibrium when it’s not moving and not rotating. That requires two conditions, not one:

The “two conditions, not one” point is what makes equilibrium problems trickier than basic Newton’s-law problems. You can balance forces perfectly and still get rotation; you can balance torques perfectly and still get sliding. You need both.

Interactive Seesaw Simulator

Predict First

A 10 kg mass sits 1 m to the left of a seesaw's fulcrum. A 5 kg mass is placed on the right side. Can the lighter mass balance the seesaw?

Place two weights on opposite sides of a fulcrum and watch the beam respond. Drag the weights directly on the beam or use the sliders below to adjust masses and distances. Notice how a heavy mass close to the fulcrum can perfectly balance a lighter mass farther away — that’s torque, where the product of mass and distance determines the rotational effect, not mass alone. Hit “Balance It!” to let the math find equilibrium for you.

Solving Equilibrium Problems

Same recipe every time:

  1. Draw a free-body diagram showing all forces and where they act on the object.
  2. Choose a pivot point. (Any point works — the math is the same. Smart choice = easier math.)
  3. Apply ΣF=0\Sigma F = 0 (usually as separate x and y component equations).
  4. Apply Στ=0\Sigma\tau = 0 about your chosen pivot.
  5. Solve the equations simultaneously.

Choosing the Pivot Point Strategically

You can calculate torques about any point — the answer comes out the same. But a smart choice eliminates unknown forces from the torque equation by putting the pivot right where one of those unknowns acts.

If you don’t know the force at a support, make that support your pivot. The torque from that force is automatically zero (because r=0r = 0 for any force at the pivot itself), and you never have to solve for it.

Torque in Biological Systems

The human body is full of torque problems. Every joint is a pivot point. Every muscle attaches to a bone at some distance from that joint. Bones act as lever arms.

Take your bicep. The bicep muscle attaches to the forearm only a few cm past the elbow joint, while the weight you’re holding might be 30+ cm out at your hand. That gives the bicep a short lever arm and the weight a long lever arm — so the bicep has to pull with a force several times bigger than the weight, just to hold things steady. (For a 5 kg dumbbell at the hand, the bicep typically pulls with ~15× that — like 750 N or so.)

This is why even moderately heavy gym weights can feel surprisingly hard: the muscle is at a mechanical disadvantage. It’s also why most muscle attachments throughout the body sacrifice mechanical advantage for speed — a small muscle contraction near the joint produces a large, fast motion at the end of the limb.

A 40 kg child sits 2 m from the pivot of a seesaw. Where must a 60 kg child sit to balance it? (Use g=10g = 10 m/s².)
Click to reveal answer

1.33 m from the pivot. For equilibrium: τ1=τ2\tau_1 = \tau_2m1gr1=m2gr2m_1 g r_1 = m_2 g r_2. gg cancels: (40)(2)=(60)(r2)(40)(2) = (60)(r_2)r2=80/60=1.33r_2 = 80/60 = 1.33 m. The heavier kid sits closer to the pivot.

A 500 N force is applied at the end of a 0.3 m wrench at a 60° angle to the wrench. What is the torque?
Click to reveal answer

τ130\tau \approx 130 N·m. τ=rFsinθ=(0.3)(500)(sin60°)=(0.3)(500)(0.866)130\tau = rF\sin\theta = (0.3)(500)(\sin 60°) = (0.3)(500)(0.866) \approx 130 N·m. If the force were perpendicular (90°), torque would be the maximum: (0.3)(500)(1)=150(0.3)(500)(1) = 150 N·m.

A horizontal 4 m beam of negligible mass rests on a pivot 1 m from one end. A 10 kg mass hangs from the long end (3 m from the pivot). What mass must be hung from the short end to balance the beam? (g=10g = 10 m/s²)
Click to reveal answer

30 kg. Equilibrium of torques: m1gr1=m2gr2m1(1)=(10)(3)m1=30m_1 g r_1 = m_2 g r_2 \Rightarrow m_1 (1) = (10)(3) \Rightarrow m_1 = 30 kg. The shorter lever arm needs proportionally more mass to balance.