A 1330 N uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end.?

Question

There is a 50 degree angle between the horizontal and the wire and a 30 degree angle between the horizontal and the beam.
Calculate the magnitude of the tension in the wire. Calculate the magnitude of the horizontal and vertical components of the force that the wall exerts on the left end of the beam.

Answer 1

Ok, look, I have made a graphic of your problem, and I was thinking about it, but I think I have it know, if it's not correct, please let me know ok.

I will tell you the forces on the beam.

There is a reaction from the wall to the beam, the weight and the tension.

Now, the tension is making and angle of 80º with the beam, why ?, because the beam makes 30º with the horizontal, and 50º with the tension and the groud, look, I used geometry for that.

Now, to find the tension let's apply torques, I am gonna use the perpendicular componentes of the weight and the tension, and apply torques respect to the wall.
You don't consider the force that the wall exerts to the beam

So : T*cos(80)*L = 1330*cos(60)*L/2

T = 1914 Newtons

To find the reaction, the force that the wall exerts on the left, you just have to draw a perpendicular force to the wall.
For this let's apply torques respect the center of mass of the beam, so you don't consider the weight.
Perpendicular component of the force exerts by the wall :

R*cos(30)*L/2 = 1914*cos(80)*L/2

R = 384 Newtons

The componentes : Rx = 384*sen(30) = 192

Ry = 384*cos(30) = 332.5 Newtons

Hope that might help you

Answer 2

Forces in y = Tsin50 - 1330/2*sin30 = 0
T = 434 N in the wire.
At the wall:
Forces in x = 1330/2*cos30 = 575.9 N (horiz. force)
Forces in y = 665*sin30 = 332.5 N (vert. force)

Answer 3

Equations:

Sum Fx :
T cos50 + Whoriz = 0

Sum Fy:
T sin50 + Wvert - mg = 0

Sum Moments:
(T sin20)L - (mg cos30) L/2 = 0
or
T = (mg/2) (cos30 / sin20)

Horizontal:
Whoriz = T cos50

Vertical:
Wvert = mg - T sin50