A uniform 1.0-N meter stick is suspended horizontally by vertical strings attached at each end.A 2.0-N weight is suspended from the 10-cm position on the stick,another 2.0-N weight is suspended from the 50 cm position,and a 3.0-N weight is suspended from the 60 cm position.What is the tension in the string attached at the 100-cm end of the stick?
A)1.9 N
B)3.0 N
C)3.5 N
D)4.0 N

Question

Quizplus · Accepted Answer

The torque produced by each weight must be balanced by the torque produced by the tension in the strings. Since the stick is symmetric, the tension in each string is the same.

Taking the pivot point at the left end of the stick, we can write:

(2.0 N)(0.1 m) + (2.0 N)(0.5 m) + (3.0 N)(0.6 m) = T(1.0 m)

T = (2.0 N)(0.1 m + 0.5 m + 0.6 m + 1.0 m)/1.0 m
T = 3.5 N

Therefore, the tension in the string attached at the 100-cm end of the stick is 3.5 N, which corresponds to choice C.

A Uniform 1