- #1

Jbreezy

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Hi I was trying to see if I did this right.

A shop owner wants to hang a sign of mass 200 kg which is supported by which is supported by a uniform 155 N beam. What is the tension in the guy wires and the horizontal and vertical forces that the wall applies to the beam? The length of the beam is 400 cm.

The diagram shows that the angle between the guy wire and the beam is 65 degrees. It also shows that the wall to center of mass for the sign is 350 cm. I hope that gives a good picture.

2. Homework Equations

Mb = mass of beam

r = distance

g = gravity

Ms = mass of sign

t = torque

T = tension

3. The Attempt at a Solution

OK so sum the forces in each direction

Fy = Fu +Tsin(θ) - Mbg - Msg = 0

Fx = Fh - Tcos(θ) = 0

We will return back to the equations after we torque about the beam where it meets the wall.

Ʃt = Tr1 - Mbr2 +Msgr3

r1 = 4tan(65) 4 is in meters it is the length of the beam. Tan(65) is to find the lever arm.

r2 = 2 meters

r3 = 3.5 meters this is from the wall to the signs center of mass. The diagram shows this.

We solve for Tension and get

T = (Mbr2 +Msgr3) / (4tan(65))

T = (155N x 2) + (200kg x 9.8 m/s^2 x 3.5 m) /(4tan(65))

T = 835 N

Then put this back into the summation of forces to get the values for the Fh and Fu

I got .. Fh = 353.2 using Fh = Tcos(65)

And I got Fu = 1357 using Mb + Msg -Tsin(65)

When I plugged my values back into the summation of forces equations with the values I got

.00000071 for summation in the y direction

I got 2.29 E-8 for summation in the y direction.

This is close to 0 but is it OK? Thanks much.