Question

A horizontal, 8 m boom weighing 400 N is hinged at the wall, as shown in Fig. 5.19. A cable is attached at a point 4.3 m
away from the wall, and a 1150 N is attached to the right end. What is the tension in the cable?

Answer 1

We can solve this problem using torques. The torque is the perpendicular to the lever component of force times distance from the point of rotation to the force:

$$\tau=F_\bot\cdot d.$$

Something is in equilibrium (does not move, break or rotate) when the net (total) torque is zero, i.e.

$$\Sigma\tau=0.$$

Torques counter-clockwise are positive and they are negative when directed clockwise.

Now choose a point around which we will calculate the torques. Let it be the point O:

The force of gravity, if we deal with a uniform boom, is applied at the center of the rod.

$$-mg\frac{d}{2}+T_yd_1-Fd=0,$$

$$-400\cdot\frac{8}{2}+T_y\cdot4.3-1150\cdot8=0,$$

$$T_y=639.7\text{ N}.$$

If you know the angle between the cable and the rod or the wall, simply use sine or cosine correspondingly to calculate the tension $T$.