what's the tension in the cable? what is the force exerted on the beam by the wall?

Question

uniform beam weighing 254N stucks out from a vertical wall.
lightweight cable connects the end of the beam to the wall, making an angle of 65 between the beam and the cable.

Plz dont just give answers, show how you did it
thank you in advance

Answer 1

Here's how to get an answer:

Consider the torques summed at the junction of the beam and the wall:

The beam
-254*L/2
where L is the length of the beam
this is negative since it is downward

the cable (assume the cable is weightless)
T*L*sin(q)
where T is the tension
and q is the angle
this is positive since it is upward

sum and set to zero
T*L*sin(q)-254*L/2=0
L divides out
q=65
T=254/(sin(65)*2)

T=140 N

Now that we know the tension, look at a FBD of the beam and sum all of the horizontal forces acting on the beam:

the wall F
positive is to the right
tension
-T*cos(q)
sum and set equal to zero to find
F=T*cos(q)
=140*cos(65)
=59.2N

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