What is the tension force for this system in rotational equilibrium?

[Np14]

Homework Statement

See picture

Relevant Equations

Torque = Fnet x L

The system is in rotational equilibrium and therefore experiences no net torque, meaning all individual torques must add to zero.

τ_NET = 0 = FF_Tsin(θ)L - F_gL - F_g(L/2)

τ_NET = 0 = F_Tsin30°(0.6?) - (0.5)(9.8)(0.6) - (2.0)(0.6/2)

My only problem (I think) is figuring out what the length L is for the cord holding the spring scale.

 

[kuruman]

View attachment 241037

The system is in rotational equilibrium and therefore experiences no net torque, meaning all individual torques must add to zero.

τ_NET = 0 = FF_Tsin(θ)L - F_gL - F_g(L/2)

τ_NET = 0 = F_Tsin30°(0.6?) - (0.5)(9.8)(0.6) - (2.0)(0.6/2)

My only problem (I think) is figuring out what the length L is for the cord holding the spring scale.

Is L the length of the cord or some other length?

 

[haruspex]

My only problem (I think) is figuring out what the length L is for the cord holding the spring scale

I think you mean that your problem is finding the moment arm for that force. You seem to be clear that L is the length of the beam.

There are essentially three ways to find the moments due to forces at oblique angles, all leading to the same answer, as you can check.
1. Use the line and magnitude of the force as is, and set the moment arm as the perpendicular distance from the line of the force to the axis.
2. Take the moment arm as the distance from the point of application of the force to the axis, but only use that component of the force which is at right angles to that moment arm.
3. Use the force as in 1 and the moment arm as in 2, but multiply their product by the sine of the angle between them.

 

[Np14]

EDIT:

τNET = 0 = FFTsin(θ)L - FgL - Fg(L/2)

EDIT: My bad, there should only be one F for the first expression (didn't use two in my calculations anyways)

Is L the length of the cord or some other length?

I see what you mean. The length should be some constant times L, I just don't know what it should be.

1. Use the line and magnitude of the force as is, and set the moment arm as the perpendicular distance from the line of the force to the axis.

Using the first way,
sin30 x 0.6 = 0.3

I guess I just typed the equations into the calculator wrong because I got 29 N now which is the correct answer. Anyways, thanks for the clarification.