Question

A light bar AD is suspended from a cable BE and supports a 20-kg block at C. The ends A and D of the bar are in contact with frictionless vertical walls. Determine the tension in cable BE and the reactions at A and D.

Answer 1

Tension in cable BE= 196.2 N

Reactions A and D both are 73.575 N

Step-by-step explanation:

The free body diagram is as attached sketch. At equilibrium, sum of forces along y axis will be 0 hence

`T_(BE)-W=0` hence

`T_(BE)=W=20*9.81=196.2 N`

Therefore, tension in the cable,
`T_(BE)=196.2 N`

Taking moments about point A, with clockwise moments as positive while anticlockwise moments as negative then

`196.2* 0.125- 196.2* 0.2+ D_x* 0.2=0`

`24.525-39.24+0.2D_x=0`

`D_x=73.575 N`

Similarly,

`A_x-D_y=0`

`A_x=73.575 N`

Therefore, both reactions at A and D are 73.575 N