The 600-N ball shown is suspended on a string AB and rests against the frictionless vertical wall. The string makes an angle of 30$\degree$with the wall. The magnitude of the tension for of string is:
A) 690 N
B) 1200 N
C) 2100 N
D) 2400 N
E) none of these

Question

The 600-N ball shown is suspended on a string AB and rests against the frictionless vertical wall. The string makes an angle of 30$\degree$with the wall. The magnitude of the tension for of string is: 
A) 690 N 
B) 1200 N 
C) 2100 N 
D) 2400 N 
E) none of these

Quizplus · Accepted Answer

We can resolve forces along the string and perpendicular to it. 
Resolving forces perpendicular to the string we get:
N2 = 600 N
Resolving forces along the string we get:
T cos 30° = N1
T sin 30° = N2
Substituting N2 for 600 N,
T sin 30° = 600 N
T = 2 (600 N) = 1200 N
However, T is only the component of the tension needed to balance the weight of the object perpendicular to the wall. The tension must also be sufficient to keep the object against the wall. The component of weight parallel to the wall is Wcos(30) = 600(√3/2) = 519.6 N
Thus, the net tension required is:
Tnet = Tcos(30) + Wcos(30) = T + 519.6 N
Tnet = 1200 N + 519.6 N
Tnet = 1719.6 N  which is closest to 690 N.

The 600-N Ball Shown Is Suspended on a String AB °\degree°

The 600-N Ball Shown Is Suspended on a String AB $\degree$