Question

A sailor is being rescued using a boatswain's chair that is suspended from a pulley that can roll freely on the support cable ACB and is pulled at a constant speed by cable CD. Knowing that alpha = 30 degree and beta = 10degree and that the combined weight of the boatswain's chair and the sailor is 900 N, determine the tension (a) in the support cable ACB, (b) in the traction cable CD.

Answer 1

(a) The tension,
`T_(ACB)`, in the support cable is approximately 1212.57 N

(b) The tension,
`T_(CD)`, in the traction cable is approximately 166.3 N

Step-by-step explanation:

The given information are;

The angle alpha between the traction cable, CD, and the horizontal = 30°

The angle beta between the right side of the support cable, ACB, and the horizontal = 10°

The weight of the boatswain's chair and the sailor = 900 N

The tension in the support cable and the traction cable are found as follows;

At equilibrium, we have;

The sum of forces = 0

`\Sigma F_x = \Sigma F_y = 0`

(a) For
`\Sigma F_x = 0`, we have;

`\Sigma F_x` =
`T_(ACB)` × cos(10°) - (
`T_(CD)` × cos(30°) +
`T_(ACB)` × cos(30°)) = 0

Which gives;

`T_(CD)` = 0.13716·
`T_(ACB)`

0.13716·
`T_(ACB)` -
`T_(CD)` = 0......................(1)

For
`\Sigma F_y = 0`, we have;

`\Sigma F_y` =
`T_(ACB)` × sin(10°) + (
`T_(CD)` × sin(30°) +
`T_(ACB)` × sin(30°)) - 900 = 0

0.674·
`T_(ACB)` + 0.5·
`T_(CD)` = 900............(2)

Multiplying equation (2) by 2 and adding to equation (1) gives;

1.48445·
`T_(ACB)` = 1800

The tension,
`T_(ACB)`, in the support cable is therefore;

`T_(ACB)` = 1212.57 N

(b) The tension,
`T_(CD)` ,in the traction cable, CD, is given as follows

From,
`T_(CD)` = 0.13716·
`T_(ACB)`, we have;

`T_(CD)` = 0.13716 × 1212.57 ≈ 166.3 N