Torque Calculations to Know for AP Physics 1 (2025)

Key Concepts

1. Torque Formula

   • Vector Form:
     τ⃗=r⃗×F⃗

     • τ⃗: Torque (vector), indicating the rotational effect of a force about a pivot.

     • r⃗: Lever arm vector (distance from pivot to the point of force application).

     • F⃗: Force vector.

   • Scalar Form:
     τ = rF sin⁡θ

     • τ: Magnitude of torque.

     • r: Magnitude of the lever arm (distance in meters or feet).

     • F: Magnitude of the applied force (in Newtons or pounds).

     • θ: Angle between r⃗ and F⃗.

Key Details:

• Maximum Torque: When θ = 90°, sin⁡(90^∘) = 1, so τ = rF.

• No Torque: When θ = 0° or 180°, sin(0°) = 0, so τ = 0

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2. Units of Torque

   • SI Unit: Newton-meters (N·m).

     • Example: A 10 N force applied 2 m from the pivot produces τ = 20 N·m

   • Imperial Unit: Foot-pounds (ft·lb).

     • Example: A 5 lb force applied 3 ft from the pivot produces τ = 15 ft·lb

3. Net Torque (Στ)

   • Net Torque Formula:
     Στ=τ1​+τ2​+⋯+τn​

   • Sign Convention:

     1. Positive (+): Counterclockwise rotation.

     2. Negative (−): Clockwise rotation.

4. Equilibrium Condition

   • Rotational equilibrium occurs when: Στ=0

     • The object is either at rest or rotates at constant angular velocity.

5. Relationship Between Torque and Angular Acceleration

   • Torque is related to angular acceleration as: τ = Iα

     • I: Moment of inertia (kg · m2), a measure of the object's resistance to rotational acceleration.

     • α: Angular acceleration (rad/s²).

Sample Problems

Problem 1: Torque Due to a Force

A 2 m-long beam is hinged at one end and held horizontally by a vertical force of 100 N applied at its free end.

Calculate the torque due to this force.

Solution:

Given:

r = 2m, F = 100N, θ = 90^∘

Formula:

τ = rF sin θ

Calculation:

sin(90^∘)=1

τ = 2⋅100⋅1 = 200N·m

Answer: The torque is 200N·m

Problem 2: Torque in Equilibrium

A 3 m-long uniform rod weighing 150 N is hinged at one end and held at a 45^∘ angle to the horizontal by a rope tied 2 m from the hinge. The rope is perpendicular to the rod.

1. Calculate the torque due to the rod’s weight about the hinge.

2. Determine the tension in the rope if the system is in equilibrium.

Solution:

1. Torque Due to the Rod’s Weight:

   • The rod’s weight acts at its center of mass, located 1.5 m from the hinge.

   • Given:
     r_W = 1.5m, W =150N, θ=45^∘, sin(45^∘)=0.707

   • Formula:
     τ_W​ = r_W​Wsinθ

   • Calculation:
     τ_W ​= 1.5⋅150⋅0.707 = 159.1N·m

2. Torque Due to the Rope’s Tension:

   • Torque from the rope:
     τ_T​=r_T​Tsin(90^∘), and sin⁡(90^∘)=1

τ_T​=2T

3. Equilibrium Condition:

   • For rotational equilibrium:
     Στ=0, so τ_T=τ_W​
     2T=159.1
     T=159.1/2​=79.55N

Answers:

1. Torque due to the rod’s weight: 159.1N·m.

2. Tension in the rope: 79.55N