#### Answer

(a) The required horizontal force is 9.70 N
(b) The tension in the wire is 265 N

#### Work Step by Step

We can find the angle $\theta$ that the wire makes with the vertical.
$sin(\theta) = \frac{0.15 ~m}{4.0 ~m}$
$\theta = sin^{-1}( \frac{0.15 ~m}{4.0 ~m}) = 2.1^{\circ}$
(a) The vertical component of tension $T_y$ is equal to the weight $mg$.
$\frac{T_x}{T_y} = tan(\theta)$
$T_x = T_y~tan(\theta)$
$T_x = mg~tan(\theta)$
$T_x = (27~kg)(9.80~m/s^2)~tan(2.1^{\circ})$
$T_x = 9.70 ~N$
The required horizontal force will be equal to the horizontal component of the tension which is 9.70 N
(b) $\frac{T_x}{T} = sin(\theta)$
$T = \frac{T_x}{sin(\theta)}$
$T = \frac{9.70~N}{sin(2.1^{\circ})}$
$T = 265~N$
The tension in the wire is 265 N