- #1

jybe

- 41

- 1

## 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}= √(FHx

^{2}+ FHy

^{2})

F

_{H}= 2744 N

tan(theta) = (1372/2376.4)

theta = 30 degrees from horizontal