Torque and Equilibrium

TL;DR

Torque is a twisting force that makes things rotate. For an object to be in equilibrium, both the net force and the net torque acting on it must be zero. This means it's either staying still or moving at a constant velocity without spinning.

1. The Mental Model

Imagine trying to open a stubborn jar lid or tighten a bolt. You're not just pushing on it; you're applying a twist. This twisting action is torque, and for the lid or bolt to stop moving, all the twists must balance out.

2. The Core Material

What is Torque?

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Torque measures how effectively a force causes rotation around a pivot point (also called a fulcrum or axis of rotation). It depends on three things: the size of the force, the distance from the pivot to where the force is applied (the "lever arm"), and the angle at which the force is applied. The formula for torque ($\tau$) is:

$\tau = rF \sin(\theta)$

Where:
* $r$ is the lever arm (distance from pivot to force).
* $F$ is the magnitude of the force.
* $\theta$ is the angle between the lever arm and the force vector.

The units for torque are Newton-meters (Nm). By convention, counter-clockwise torques are usually positive, and clockwise torques are negative. Think about pushing a door: pushing far from the hinges with a perpendicular push is most effective.

What is Equilibrium?

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An object is in equilibrium if its state of motion isn't changing. This means it's either perfectly still (static equilibrium) or moving at a constant velocity without accelerating or rotating (dynamic equilibrium). For equilibrium to occur, two conditions must be met:

1. Translational Equilibrium: The net force acting on the object is zero. This means all forces pushing or pulling it cancel out, so it won't move linearly.
   $\Sigma F_x = 0$
   $\Sigma F_y = 0$
2. Rotational Equilibrium: The net torque acting on the object is zero. This means all twisting forces cancel out, so it won't rotate or will rotate at a constant angular velocity.
   $\Sigma \tau = 0$

When solving problems, you often pick a convenient pivot point to calculate torques. If an object is in equilibrium, the net torque is zero regardless of which point you choose as the pivot. Picking a pivot where one or more unknown forces act can simplify calculations, as those forces won't produce torque about that point (since $r=0$).

How to Approach Equilibrium Problems

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graph TD
    A["Start: Object in Equilibrium?"] --> B{"Identify all forces acting on the object"};
    B --> C{"Choose a convenient pivot point"};
    C --> D{"Calculate torque for each force (rFsinθ)"};
    D --> E{"Assign direction to torques (CCW+, CW-)"};
    E --> F{"Set sum of forces = 0 (ΣF=0)"};
    F --> G{"Set sum of torques = 0 (Στ=0)"};
    G --> H["Solve the system of equations"];
    H --> I["End: Unknown forces/distances found"];

3. Worked Example

Let's say you have a uniform plank 4 meters long, weighing 200 N. It's supported at two points: one support is at the left end (0 m), and the other is 1 meter from the right end (at 3 m). A 500 N child sits 1 meter from the left end (at 1 m). We want to find the forces exerted by the two supports on the plank.

1. Draw a Free Body Diagram: Imagine the plank.

   • Weight of plank (200 N) acts at its center (2 m mark).
   • Child's weight (500 N) acts at 1 m mark.
   • Support force 1 ($F_1$) acts at 0 m mark, upwards.
   • Support force 2 ($F_2$) acts at 3 m mark, upwards.

2. Apply Translational Equilibrium:
   All upward forces must equal all downward forces.
   $F_1 + F_2 - 200 \text{ N} - 500 \text{ N} = 0$
   $F_1 + F_2 = 700 \text{ N}$ (Equation 1)

3. Apply Rotational Equilibrium:
   Let's pick the left end (0 m, where $F_1$ acts) as our pivot point. This means $F_1$ produces no torque, simplifying things.

   • Child's weight (500 N) at 1 m: $\tau_C = -(500 \text{ N}) \times (1 \text{ m})$ (clockwise, so negative)
   • Plank's weight (200 N) at 2 m: $\tau_P = -(200 \text{ N}) \times (2 \text{ m})$ (clockwise, so negative)
   • Support force 2 ($F_2$) at 3 m: $\tau_{F2} = +(F_2) \times (3 \text{ m})$ (counter-clockwise, so positive)

   Sum of torques = 0:
   $-500 \text{ N} \cdot 1 \text{ m} - 200 \text{ N} \cdot 2 \text{ m} + F_2 \cdot 3 \text{ m} = 0$
   $-500 \text{ Nm} - 400 \text{ Nm} + 3F_2 \text{ m} = 0$
   $-900 \text{ Nm} + 3F_2 \text{ m} = 0$
   $3F_2 = 900 \text{ N}$
   $F_2 = 300 \text{ N}$

4. Solve for the remaining unknown:
   Substitute $F_2 = 300 \text{ N}$ back into Equation 1:
   $F_1 + 300 \text{ N} = 700 \text{ N}$
   $F_1 = 400 \text{ N}$

So, the support at the left end exerts a force of 400 N, and the support 1 meter from the right end exerts a force of 300 N.

4. Key Takeaways

• Torque is a twisting force that causes rotation, calculated as the product of the force, lever arm, and sine of the angle between them.
• The direction of torque is important: counter-clockwise is usually positive, and clockwise is negative.
• For an object to be in equilibrium, both the net force and the net torque acting on it must be zero.
• Translational equilibrium means $\Sigma F_x = 0$ and $\Sigma F_y = 0$.
• Rotational equilibrium means $\Sigma \tau = 0$.
• When solving equilibrium problems, strategically choosing the pivot point can significantly simplify your calculations.
• You can choose any point as the pivot for calculating torques; if the object is in equilibrium, the net torque about any point will be zero.

Common Mistakes to Avoid:
- Forgetting to assign a sign (positive/negative) to your torques based on their direction of rotation.
- Not using the perpendicular distance (lever arm) or the correct angle for torque calculations.
- Confusing the location of a force with the pivot point; a force applied at the pivot produces zero torque.
- Ignoring the weight of the object itself, which acts at its center of mass for uniform objects.

5. Now Try It

A 6-meter-long uniform ladder weighing 150 N leans against a frictionless wall at an angle of 60 degrees with the ground. The ground is rough (meaning it provides friction). A 700 N person stands 2 meters up the ladder from the bottom. Determine the horizontal force exerted by the wall on the ladder and the magnitude of the normal force and friction force exerted by the ground on the ladder. What success looks like: You'll have three force magnitudes, one for the wall, and two for the ground (normal and friction).