A force of 40 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 40 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) = 40 N × 2 m × sin(60°) = 40 × 2 × (√3/2) = 40√3 Nm ≈ 34.64 Nm.

Steps: 8

Step 1: Identify the values given in the problem. We have a force (F) of 40 N, a length of the lever arm (r) of 2 m, and an angle (θ) of 60 degrees.
Step 2: Understand the formula for torque (τ). The formula is τ = F × r × sin(θ).
Step 3: Calculate sin(60 degrees). The value of sin(60 degrees) is √3/2.
Step 4: Substitute the values into the torque formula. We have τ = 40 N × 2 m × sin(60 degrees).
Step 5: Replace sin(60 degrees) with its value. Now the equation looks like τ = 40 N × 2 m × (√3/2).
Step 6: Simplify the equation. The 2 in the numerator and denominator cancels out, so we have τ = 40 N × √3.
Step 7: Calculate the final torque. This gives us τ = 40√3 Nm.
Step 8: If needed, approximate the value of √3. √3 is approximately 1.732, so τ ≈ 40 × 1.732 Nm ≈ 34.64 Nm.