Question

A crate with a mass of 175.5 kg is suspended from the end of a uniform boom with a mass of 94.7 kg. The upper end of the boom is supported by a cable attached to the wall and the lower end by a pivot (marked X) on the same wall.
Calculate the tension in the cable.

Answer 1

The tension in the cable is mathematically given as

T =266.659 N

What is the tension in the cable.?

Generally, the equation for the angle of the boom with horizontal is  mathematically given as

A = tan-1(5 /10)

$A= 26.57 \textdegree$

The angle of cable with horizontal

B = tan-1(4 / 10)

B= 21.80 degrees

Hence

175.5 * cos26.57 + 94.7 *cos26.57 * 1/2 = T (sin(26.57+21.80)) * 1

T =266.659 N

In conclusion, the tension in the cable.

T =266.659 N

Read more about tension

https://brainly.com/question/12111847

#SPJ1