Question

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 30.0o 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

Answer 1

Answer:

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

c. $T_{net}=8.58$

Explanation:

$T=F*d$

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:

$T_1=F_1*d_1$

$T_1=7.8N*1.6m=12.48N*m$

$T_2=F_2*d_2$

$T_2=2.60N**cos(30)*3.0m$

$T_2= - 3.9 N*m$

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

$T_{net}=T_1+T_2$

$T_{net}=12.48N-3.9N=8.58N$