A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?

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Q: A force of 50 N is applied at a distance of 0.5 m from the pivot at an angle of 60 degrees. What is the torque?

Solution: Torque (τ) = F × r × sin(θ) = 50 N × 0.5 m × sin(60°) = 50 N × 0.5 m × (√3/2) = 43.3 Nm.

Steps: 8

Step 1: Identify the values given in the problem. We have a force (F) of 50 N, a distance (r) of 0.5 m from the pivot, and an angle (θ) of 60 degrees.
Step 2: Recall the formula for torque (τ), which is τ = F × r × sin(θ).
Step 3: Substitute the values into the formula. We have τ = 50 N × 0.5 m × sin(60°).
Step 4: Calculate sin(60°). The value of sin(60°) is √3/2.
Step 5: Substitute sin(60°) into the equation. Now we have τ = 50 N × 0.5 m × (√3/2).
Step 6: Calculate the multiplication. First, calculate 50 N × 0.5 m = 25 N·m.
Step 7: Now multiply 25 N·m by (√3/2). This gives us τ = 25 N·m × (√3/2).
Step 8: Calculate the final value. 25 N·m × (√3/2) = 25 N·m × 0.866 = 43.3 Nm.