- #1

sinequanon

- 6

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## Homework Statement

A 15 meter beam jutting out of the side of a building is held by a hinge (at the wall) and a cable at 10 meters from the wall. The angle between the beam and the cable is 60 degrees and the mass of the beam is 250 kg. If a 1000 N object is located on the beam, 7 meters from the wall, what is the reacting force

**R**from the hinge and at what angle is it applied? Assume the system is in equilibrium.

## Homework Equations

1. ΣF

_{x}= R

_{x}-

*T*cosΘ = 0

2. ΣF

_{y}= R

_{y}+

*T*sinΘ - F

_{object}- F

_{beam}= 0

3.

*T*sinΘ(d

_{cable}) - F

_{beam}(d

_{beam}) - F

_{object}(d

_{object})

*Use 10 m/s

^{2}for the value of gravitational acceleration.

## The Attempt at a Solution

Alright, so I just wanted to double check to see if I'm actually doing this correctly.

First I substitute into the third equation in order to find the cable tension.

*T*sin60(10 m) - (2500 N)(7.5 m) - (1000 N)(7 m) = 0

*T*= 2973.44

Then, I would substitute the

*T*value into the other equations.

ΣF

_{x}= R

_{x}- 2973.44cos60 = 0

R

_{x}= 1486.72 N

ΣF

_{y}= R

_{y}+ 2973.44sin60 - 1000 - 2500 = 0

R

_{y}= 924.925 N

From here, it appears to be a simple matter of using the Pythagorean Theorem and then just using inverse cosine to find Θ

_{R}.

R = [tex]\sqrt{1486.72^2 + 924.925^2}[/tex] = 1750.95 N

cos

^{-1}Θ = 1486.72/1750.95

Θ = 31.88°

I was hoping someone would be able to double check to see if my understanding of this matter is correct or otherwise. I was also wondering if someone could tell if my final answer

*R*should be positive or negative, as that is one thing I haven't a clue about.