A 3.0-m rod is pivoted about its left end. A force of 20 N is applied perpendicular to the rod at a distance of1.2 m from the pivot causing a ccw torque, and a force of 4.8 N is applied at the end of the rod 3.0 m from the pivot. The 4.8-N force is at an angle of 30o to the rod and causes a cw torque. What is the net torque about the pivot? (Take ccw as positive.)Select one:a.-16.8 N·mb.31.2 N·mc.-31.2 N·md.16.8 N·me.9.6 N·m
Question
A 3.0-m rod is pivoted about its left end. A force of 20 N is applied perpendicular to the rod at a distance of1.2 m from the pivot causing a ccw torque, and a force of 4.8 N is applied at the end of the rod 3.0 m from the pivot. The 4.8-N force is at an angle of 30o to the rod and causes a cw torque. What is the net torque about the pivot? (Take ccw as positive.)Select one:a.-16.8 N·mb.31.2 N·mc.-31.2 N·md.16.8 N·me.9.6 N·m
Solution
To solve this problem, we need to calculate the torque caused by each force and then add them up. Torque (τ) is calculated by the equation τ = rFsinθ, where r is the distance from the pivot point, F is the force, and θ is the angle between the force and the direction from the pivot point.
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For the 20 N force: The distance r = 1.2 m, the force F = 20 N, and the angle θ = 90° (since the force is perpendicular to the rod). So, τ1 = (1.2 m)(20 N)sin90° = 24 N·m. This torque is counterclockwise, so it is positive.
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For the 4.8 N force: The distance r = 3.0 m, the force F = 4.8 N, and the angle θ = 30°. So, τ2 = (3.0 m)(4.8 N)sin30° = 7.2 N·m. This torque is clockwise, so it is negative.
The net torque is the sum of these two torques: τ_net = τ1 + τ2 = 24 N·m - 7.2 N·m = 16.8 N·m.
So, the correct answer is d. 16.8 N·m.
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