A uniform horizontal beam with a length of 8m and weigh of 200N is attached to a wall by a pin connection ..Its far end is supported by the cable tha makes an angle of 53.0 with the beam. If 600N person stands 2m from the wall , find the tension in the cable as well as the magnitude and direction of the force exerted by the wall on the beam.

The tension in the cable is approximately 1440.0 N

Explanation:

The length of the beam = 8 m

The weight of the beam = 200 N

The angle made by the beam support cable = 53°

The weight of the person = 600 N

The position where the person stands = 2 m

The forces acting are the weight of the cable acting at the center of the beam and the weight of the person acting at 2 m from the wall

Therefore, we have;

Sum of moments = 0, which gives;

2000 × 4 + 2 × 600 - 8 × = 0

2000 × 4 + 2 × 600  = 8 ×

= 9200/8 = 1150 N

The tension in the cable, T = The component of the force in the cable

The vertical component of the tension   = T×sin(53)

Therefore;

T =  /(sin(53)) = 1150/(sin(53 degrees)) = 1439.96 ≈ 1440 N

The tension in the cable, T is approximately 1440 N.

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