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Torque worksheet

Grades 11–12 · CCSS HSG.SRT.C.8, HSA.CED.A.4

Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Solve for: Lever arm
- Torque (τ): 56 N·m
- Force (F): 180 N
- Angle between lever arm and force (θ): 78 °
- Show lever arm in: Meters (m)
Find: Lever arm length
Answer
2.
Problem 2.
- Solve for: Torque
- Force (F): 130 N
- Lever arm length (r): 16 cm
- Angle between lever arm and force (θ): 95 °
- Show torque in: Newton meters (N·m)
Find: Torque
Answer
3.
Problem 3.
- Solve for: Lever arm
- Torque (τ): 45 N·m
- Force (F): 200 N
- Angle between lever arm and force (θ): 72 °
- Show lever arm in: Millimeters (mm)
Find: Lever arm length
Answer
4.
Problem 4.
- Solve for: Torque
- Force (F): 132 N
- Lever arm length (r): 28 cm
- Angle between lever arm and force (θ): 90 °
- Show torque in: Pound-force feet (lbf·ft)
Find: Torque
Answer
5.
Problem 5.
- Solve for: Lever arm
- Torque (τ): 45 N·m
- Force (F): 115 N
- Angle between lever arm and force (θ): 90 °
- Show lever arm in: Meters (m)
Find: Lever arm length
Answer
6.
Problem 6.
- Solve for: Torque
- Force (F): 150 N
- Lever arm length (r): 24 cm
- Angle between lever arm and force (θ): 74 °
- Show torque in: Pound-force inches (lbf·in)
Find: Torque
Answer
7.
Problem 7.
- Solve for: Torque
- Force (F): 160 N
- Lever arm length (r): 26 cm
- Angle between lever arm and force (θ): 138 °
- Show torque in: Kilonewton meters (kN·m)
Find: Torque
Answer
8.
Problem 8.
- Solve for: Torque
- Force (F): 200 N
- Lever arm length (r): 22 cm
- Angle between lever arm and force (θ): 78 °
- Show torque in: Newton meters (N·m)
Find: Torque
Answer
9.
Problem 9.
- Solve for: Force
- Torque (τ): 40 N·m
- Lever arm length (r): 32 cm
- Angle between lever arm and force (θ): 72 °
- Show force in: Kilonewtons (kN)
Find: Force
Answer
10.
Problem 10.
- Solve for: Lever arm
- Torque (τ): 38 N·m
- Force (F): 170 N
- Angle between lever arm and force (θ): 85 °
- Show lever arm in: Millimeters (mm)
Find: Lever arm length
Answer

Answer key: Torque worksheet

1. Lever arm length: 0.32 m
1. Evaluate sin θ for θ = 78°
  $sin 78 °$ $= 0.9781476$
2. Work backward: divide the torque by force × sin θ to find the lever arm length needed
  $\frac{56}{180 \times 0.9781476}$ $= 0.318062 m$
2. Torque: 20.72 N·m
1. Change lever arm to m, the standard unit the formula uses (1 cm = 0.01 m)
  $16 cm \times 0.01$ $= 0.16 m$
2. Evaluate sin θ for θ = 95°
  $sin 95 °$ $= 0.9961947$
3. Multiply lever arm × force × sin θ. Only the square-on part of the push turns the object
  $0.16 \times 130 \times 0.9961947$ $= 20.72085 N·m$
3. Lever arm length: 236.58 mm
1. Evaluate sin θ for θ = 72°
  $sin 72 °$ $= 0.95105652$
2. Work backward: divide the torque by force × sin θ to find the lever arm length needed
  $\frac{45}{200 \times 0.95105652}$ $= 0.236579 m$
3. Change the answer to mm, the unit you picked: divide by 0.001
  $\frac{0.236579}{0.001}$ $= 236.579 mm$
4. Torque: 27.26 lbf·ft
1. Change lever arm to m, the standard unit the formula uses (1 cm = 0.01 m)
  $28 cm \times 0.01$ $= 0.28 m$
2. Evaluate sin θ for θ = 90°
  $sin 90 °$ $= 1$
3. Multiply lever arm × force × sin θ. Only the square-on part of the push turns the object
  $0.28 \times 132 \times 1$ $= 36.96 N·m$
4. Change the answer to lbf·ft, the unit you picked: divide by 1.355817948331
  $\frac{36.96}{1.355817948331}$ $= 27.260297 lbf·ft$
5. Lever arm length: 0.39 m
1. Evaluate sin θ for θ = 90°
  $sin 90 °$ $= 1$
2. Work backward: divide the torque by force × sin θ to find the lever arm length needed
  $\frac{45}{115 \times 1}$ $= 0.391304 m$
6. Torque: 306.28 lbf·in
1. Change lever arm to m, the standard unit the formula uses (1 cm = 0.01 m)
  $24 cm \times 0.01$ $= 0.24 m$
2. Evaluate sin θ for θ = 74°
  $sin 74 °$ $= 0.9612617$
3. Multiply lever arm × force × sin θ. Only the square-on part of the push turns the object
  $0.24 \times 150 \times 0.9612617$ $= 34.605421 N·m$
4. Change the answer to lbf·in, the unit you picked: divide by 0.112984829028
  $\frac{34.605421}{0.112984829028}$ $= 306.283785 lbf·in$
7. Torque: 0.03 kN·m
1. Change lever arm to m, the standard unit the formula uses (1 cm = 0.01 m)
  $26 cm \times 0.01$ $= 0.26 m$
2. Evaluate sin θ for θ = 138°
  $sin 138 °$ $= 0.66913061$
3. Multiply lever arm × force × sin θ. Only the square-on part of the push turns the object
  $0.26 \times 160 \times 0.66913061$ $= 27.835833 N·m$
4. Change the answer to kN·m, the unit you picked: divide by 1,000
  $\frac{27.835833}{1,000}$ $= 0.027836 kN·m$
8. Torque: 43.04 N·m
1. Change lever arm to m, the standard unit the formula uses (1 cm = 0.01 m)
  $22 cm \times 0.01$ $= 0.22 m$
2. Evaluate sin θ for θ = 78°
  $sin 78 °$ $= 0.9781476$
3. Multiply lever arm × force × sin θ. Only the square-on part of the push turns the object
  $0.22 \times 200 \times 0.9781476$ $= 43.038494 N·m$
9. Force: 0.13 kN
1. Change lever arm to m, the standard unit the formula uses (1 cm = 0.01 m)
  $32 cm \times 0.01$ $= 0.32 m$
2. Evaluate sin θ for θ = 72°
  $sin 72 °$ $= 0.95105652$
3. Work backward: divide the torque by lever arm × sin θ to find the force needed
  $\frac{40}{0.32 \times 0.95105652}$ $= 131.432778 N$
4. Change the answer to kN, the unit you picked: divide by 1,000
  $\frac{131.432778}{1,000}$ $= 0.131433 kN$
10. Lever arm length: 224.38 mm
1. Evaluate sin θ for θ = 85°
  $sin 85 °$ $= 0.9961947$
2. Work backward: divide the torque by force × sin θ to find the lever arm length needed
  $\frac{38}{170 \times 0.9961947}$ $= 0.224383 m$
3. Change the answer to mm, the unit you picked: divide by 0.001
  $\frac{0.224383}{0.001}$ $= 224.383258 mm$