Force Acting on Hinge: 2744 N Torque on Beam Homework: Find Force on Hinge

[jybe]

Homework Statement

"A beam of mass M = 280kg and length L = 2.2m is attached to a wall with a hinge (at a 90 degree angle to the wall, sticking out horizontally), and is supported at the other end by a wire making an angle of 30 degrees with the horizontal beam.

What is the force acting on the hinge?"

NOW...this isn't actually what I'm trying to figure out. I'll do the work for the question below, but what I really want to know is how my work would change if the beam was not at a right angle to the wall, but raised up, maybe at an angle of 20 degrees from the horizontal/70 degrees from the wall. Say the wire is now at an angle of 30 degrees from the horizontal but not the beam.

How would I account for the beam's weight in the Fy equation at the start? Would it still be mg?
And in the torque equation, I can't wrap my mind around how it would change, as far as both the wire and beam parts are concerned. If anyone could help me I would really appreciate it

Homework Equations

F_x = 0
0 = F_Hx - Tcos(theta) (Hx is the x compenent of the hinge force)

F_y = 0
0 = F_Hy + Tsin(theta) - mg

Net torque = 0
0 = Tsin(theta)*L - mg*(L/2)

T = (mg / 2*sin(theta))

The Attempt at a Solution

[/B]
F_Hx = Tcos(theta)
F_Hx = (mg/2sin(theta))*cos(theta)
F_Hx = 2376.4 N

F_Hy = mg - Tsin(theta)
F_Hy = mg - (mg/2sin(theta))*sin(theta)
F_Hy = 1372 N

F_H = √(FHx² + FHy²)
F_H = 2744 N

tan(theta) = (1372/2376.4)
theta = 30 degrees from horizontal

 

[haruspex]

how my work would change if the beam was not at a right angle to the wall,

You correctly posted the three statics equations, horizontal balance, vertical balance and torque balance.
Studying the details of those, which terms change if the beam is not horizontal?

 

[jybe]

You correctly posted the three statics equations, horizontal balance, vertical balance and torque balance.
Studying the details of those, which terms change if the beam is not horizontal?

I think the F_Hx changes, and also, I think the torque from the weight of the beam changes, but is it still mg, or is it mg multiplied by an angle? I'm just having a really hard time making sense of what changes and what doesn't.

 

[jybe]

Anyone?

 

[haruspex]

Sorry for the delay. Sometimes the system doesn't give me an alert when there is a reply.

the torque from the weight of the beam changes,

Yes. How does one determine the torque from a force about a given point?

I think the FHx changes

Its value might change, but not its appearance in the equations.

 

[jybe]

Yes. How does one determine the torque from a force about a given point?

The torque from the weight of the beam will be mg*length*cos(theta)

But then how do you express the torque on the beam from the wire now?

 

[haruspex]

The torque from the weight of the beam will be mg*length*cos(theta)

You have two different angles, no? One for the wire and one for the beam.