How would I go about solving for the magnitude of the force the pivot exerts on the bar?

Assuming the bar Is uniform in density.

My idea was: since the system is in both translational and rotational equilibrium.

$$T\sin\theta = F$$ (Pivot on Bar)

Solving for $$T$$

let M = mass of bar

let m = mass of block

$$T\cos\theta*l = Mg*\frac{1}{2} + mg$$

$$T = \frac{Mg\frac{1}{2} + mg}{\cos\theta*l}$$

Then

$$F$$ (Pivot on Bar) = $$T\sin\theta$$

Although when I attempt the examples I don't receive the right answer.

thanks for any help out there.

• Which member, horizontal or vertical, member is the “bar” – Bob D Oct 2 '19 at 3:38
• @BobD the horizontal bar holding the block mass – Bdyce Oct 2 '19 at 3:39
• Ok. Have you taken a class in statics? This is a statics problem – Bob D Oct 2 '19 at 4:34
• @BobD I am taking that class right now, I just can't seem to find the solution for this. – Bdyce Oct 2 '19 at 4:35
• OK, see my answer – Bob D Oct 2 '19 at 4:45