A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?

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Q: A force of 20 N is applied at an angle of 60 degrees to the lever arm of length 2 m. What is the torque about the pivot?

Solution: Torque (τ) = F × r × sin(θ) = 20 N × 2 m × sin(60°) = 20 × 2 × (√3/2) = 20√3 Nm ≈ 34.64 Nm.

Steps: 10

Step 1: Identify the force applied, which is 20 N.
Step 2: Identify the length of the lever arm, which is 2 m.
Step 3: Identify the angle at which the force is applied, which is 60 degrees.
Step 4: Use the formula for torque: Torque (τ) = Force (F) × Length of lever arm (r) × sin(angle (θ)).
Step 5: Substitute the values into the formula: τ = 20 N × 2 m × sin(60°).
Step 6: Calculate sin(60°), which is √3/2.
Step 7: Substitute sin(60°) into the equation: τ = 20 N × 2 m × (√3/2).
Step 8: Simplify the equation: τ = 20 × 2 × (√3/2).
Step 9: The 2 in the numerator and denominator cancels out: τ = 20√3 Nm.
Step 10: Calculate the approximate value of 20√3, which is about 34.64 Nm.