Answer
(a) The tension in each 35.0-cm wire is 152 N.
The tension in the 18.0-cm wire is 294 N.
(b) Each ball pushes with a force of 40.0 N.
Work Step by Step
(a) We can find the angle $\theta$ that each 35.0-cm wire makes with the vertical.
$sin(\theta) = \frac{0.125~m}{0.475~m}$
$\theta = arcsin(\frac{0.125~m}{0.475~m})$
$\theta = 15.26^{\circ}$
The vertical component of the tension in each 35.0-cm wire is equal in magnitude to the weight of one ball. We can find the tension $T$ in each 35.0-cm wire.
$T~cos(15.26^{\circ}) = mg$
$T = \frac{mg}{cos(15.26^{\circ})}$
$T = \frac{(15.0~kg)(9.80~m/s^2)}{cos(15.26^{\circ})}$
$T = 152~N$
The tension in the 18.0-cm wire is equal to the sum of the vertical component of the tension in each 35.0-cm wire, which is equal to the sum of the weight of each ball. Therefore, the tension in the 18.0-cm wire is $2mg = (2)(15.0~kg)(9.80~m/s^2)$, which is equal to 294 N.
(b) The normal force $F_N$ exerted by each ball is equal in magnitude to the horizontal component of the tension in each 35.0-cm wire.
$F_N = T~sin(15.26^{\circ})$
$F_N = (152~N)~sin(15.26^{\circ})$
$F_N = 40.0~N$
Each ball pushes with a force of 40.0 N.
