Question
NEXT Question 3 Your answer is partially correct. Try again. One end of a meter stick is pinned to a table, so the stick can rotate free the stick in such a way that the net torque is zero. The f ly in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to The second force has a magnitude of 6.00 N and acts at a 59, 19 angle with respect to the length of the stick. Where along th rst force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. at a 59.1° angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned the tolerance is + /-2% Click if you would like to Show Work for this question: Open Show Work LINK TO TEXT Question Attempts: 1 of 5 used SAVE FOR LATER maple www.maplesoft.com All Rights R A Division of
Explanation / Answer
Note that the net torque = 0.
Also, given a force, the momen arm r, and the angle,
T = F r sin (A)
Thus,
T1 - T2 = 0
Here,
T1 = 2.00 N * 1 m = 2.00 mN
The other one is
T2 = 6.00 N * r * sin 59.1 = 5.148 r
Thus,
2.00 - 5.148r = 0
Thus,
r = 0.388 m [ANSWER]
