A100 kg uniform beam is attached to a vertical wall at one end and is supported by a cable at the other end. calculate the magnitude of the tension in the wire if the angle between the cable and the horizontal is θ = 45°.

The magnitude of the tension in the wire supporting the beam is 692.96 N.

The given parameters;

mass of the beam = 100 kg

The magnitude of the tension in the wire as follows;

T = mg x sinθ

where;

m is the mass of the beam being supportedg is acceleration due to gravity

T = 100 x 9.8 x sin(45)

T = 692.96 N

Thus, the magnitude of the tension in the wire supporting the beam is 692.96 N.

Learn more here:link

First, draw a diagram (unless it is provided).
Then draw an FBD (free-body diagram for the beam)
Consider the conditions of equilibrium ( ∑ F = 0, ∑ M = 0 ).
Solve for unknowns.

Assuming the wall support is hinged (else insufficient information to solve).

We denote the beam ABC of length 2L, with a weight of 981 N acting at its CG at B.
The cable is attached at C at 45 degrees with the horizontal, with a tension of T newtons.

Sum vertical forces: we cannot sum vertical forces without knowing the vertical reaction at the wall support A.

Next, take moments about A (counter-clockwise positive)
T*(2Lsin(45)-981L=0
solve for T
T=981L/(2Lsin(45)=981/sqrt(2)=693.7 N