# How do we decide which tension forces get included in the force- and torque-balance equations?

Consider a static equilibrium problem like the following:

The uniform boom shown below weighs 700 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?

To get the correct answer, we can define the hinge as the origin and set up the appropriate force- and torque-balance equations. The correct equations include the tension in the cable (one variable) as well as the tension in the boom (which is the negative counterpart to the force on the hinge; two variables). Then $$\sum F_x = 0 \\ \sum F_y = 0 \\ \sum \tau = 0$$ is a square system.

But why don't we include the tension in the wall between the hinge and the point where the cable is attached? If we do include it, we have two more variables; can we come up with two more equations so that the system remains solvable?

• You are talking about force on boom due to hinge, I.e the normal force due to hinge . Why would tension act between a wall and a boom – Anusha Aug 5 at 10:52