Question

A mobile at the art museum has a 2.0 steel cat and a 4.0 steel dog suspended from a lightweight cable, as shown in the figure . It is found that = 29 when the centerrope is adjusted to be perfectly horizontal.

Answer 1

for the CAT point
(2*9.81)/ sin 150 = T1 / sin 90 = T2 / sin 120
39.24 = T1 = 1.1547 T2
T2 = 39.24 / 1.1547 = 33.983 N
Tension in the 1 st part of the cable = 39.24 N.................ans
Tension in the horizontal part of the cable = 33.983 N.................ans

For the DOG point
T2 = 33.983 N
wt of dog = 4*9.81 = 39.24 N
T3 = ( 39.24^2 + 33.983^2)^0.5 = 51.909 N tension in the third part of the cable ans

39.24 / sin A = 51.909 / sin 90 = 33.983 / sin B
sin A = 39.24 / 51.909 = 0.7559382765994337
A = 49.107 degrees angle of rope 3.....ans.........................