A3.00-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 30.0o to the rod and causes a cw torque. what is the net torque about the pivot? (take ccw as positive.)

a. 26.4 n·m

b. 4.68 n·m

c. 8.58 n·m

d. -8.58 n·m

e. -16.4 n·m

The net torque about the pivot is and the answer is 'c'

c.

Explanation:

The torque is the force apply in a distance so it is the moment so depends on the way to be put it the signs so:

Now to find the net Torque is the summation of both torques

work: w = fd (given)

work done by chet   =50*1 = 50 j

work done by mika =40*2 = 80 j

work done by sara =30*1.5 = 45 j

work done by bill =60*0.5 = 30 j

students in order from the greatest amount of work done to the least is mika> chet> sara> bill

hence option (b) is correct.

b.

explanation:

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