A 3.00-m rod is pivoted about its left end.A force of 7.80 N is applied perpendicular to the rod at a distance of 1.60 m from the pivot causing a ccw torque,and a force of 2.60 N is applied at the end of the rod 3.00 m from the pivot.The 2.60-N force is at an angle of 30.0^o to the rod and causes a cw torque.What is the net torque about the pivot? (Take ccw as positive. ) A)16.4 N·m B)-8.58 N·m C)8.58 N·m D)-16.4 N·m E)4.68 N·m

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Quizplus · Accepted Answer

The net torque is calculated by taking the sum of individual torques, considering their direction (ccw as positive, cw as negative). The torque caused by the 7.80 N force is $7.80 \, \text{N} \times 1.60 \, \text{m} = 12.48 \, \text{N}\cdot\text{m}$ (ccw). The torque caused by the 2.60 N force is $2.60 \, \text{N} \times 3.00 \, \text{m} \times \cos(30^\circ) = 2.60 \times 3.00 \times \sqrt{3}/2 \approx 6.75 \, \text{N}\cdot\text{m}$ (cw). The net torque is $12.48 \, \text{N}\cdot\text{m} - 6.75 \, \text{N}\cdot\text{m} = 5.73 \, \text{N}\cdot\text{m}$. However, none of the provided options match this calculation, suggesting there may have been a misunderstanding in the question's details or in the calculation process. Given the options, the closest approach to a logical answer based on the provided calculations would be to reevaluate the calculation or the interpretation of the question.

A 300-M Rod Is Pivoted About Its Left End