Question

A 2.00 m long horizontal uniform beam is supported by a wire as shown in the figure. The wire makes an angle of 20.0 degrees with the beam. A 40.0 kg signboard is attached to the beam 1.40 m from the wall . What is the tension in the string if the beam has 20.0 kg mass?

Answer 1

The beam alone is exerting a normal force of (20 x 9.8)/2 = 98N. in the wire.
Then deal with the ball. It is positioned 1.4m. from the wall, so is 0.6m. from the wire.
The weight of the ball is (40 x 9.8) = 392N.
392/((1.4/0.6) + 1) = 117.6N. normal force is on the wall, so (392 - 117.6) = 274.4N. is on the wire.
Total normal force = (274.4 + 98) = 372.4N. normal is on the wire.
Now sketch a right triange, the "vertical" side is 372.4N., the angle opposite is 20 degrees. Find the "length" of the hypotenuse.
372.4/(Sin 20) = 1,089N., nearest answer is B.

Just to explain the derivation of the normal force of the ball on the wire end.
The (1.4/0.6) is a ratio of 2.33:1. You add the 2 sides of the ratio, and that gives a "no. of parts" that you need to divide the force by.
The answer is 1 part. You just subtract that from the 392N. total, to obtain the normal force component on the wire end.