Answer
(a) The tension in the cable is 658 N.
(b) The vertical component of the force exerted on the beam at the wall is 95 N.
The horizontal component of the force exerted on the beam at the wall is 526 N.
Work Step by Step
(a) We can find the angle $\theta$ that the cable makes with the beam.
$sin(\theta) = \frac{3.00~m}{5.00~m}$
$\theta = arcsin(0.60)$
$\theta = 36.9^{\circ}$
Let's consider an axis of rotation where the beam is attached to the wall. Let $T$ be the tension in the cable.
$\sum \tau = 0$
$T~sin(\theta)~(4.00~m) -(190~N)(2.00~m) - (300~N)(4.00~m)= 0$
$T~sin(\theta)~(4.00~m) = (190~N)(2.00~m) + (300~N)(4.00~m)$
$T = \frac{(190~N)(2.00~m) + (300~N)(4.00~m)}{(4.00~m)~sin(\theta)}$
$T = \frac{(190~N)(2.00~m) + (300~N)(4.00~m)}{(4.00~m)~sin(36.0^{\circ})}$
$T = 658~N$
The tension in the cable is 658 N.
(b) Let $F_{w,y}$ be the vertical component of the force exerted on the beam at the wall. Let the upward direction be positive.
$\sum F_y = 0$
$F_{w,y}+T~sin(\theta)-190~N-300~N = 0$
$F_{w,y} = 190~N + 300~N - T~sin(\theta)$
$F_{w,y} = 190~N + 300~N - (658~N)~sin(36.9^{\circ})$
$F_{w,y} = 95~N$
The vertical component of the force exerted on the beam at the wall is 95 N.
Let $F_{w,x}$ be the horizontal component of the force exerted on the beam at the wall. Let the direction to the right be positive.
$\sum F_x = 0$
$F_{w,x}-T~cos(\theta) = 0$
$F_{w,x}=T~cos(\theta)$
$F_{w,x}=(658~N)~cos(36.9^{\circ})$
$F_{w,x} = 526~N$
The horizontal component of the force exerted on the beam at the wall is 526 N.
