Respuesta :
Answer:
(a) The tension, [tex]T_{ACB}[/tex], in the support cable is approximately 1212.57 N
(b) The tension, [tex]T_{CD}[/tex], in the traction cable is approximately 166.3 N
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
[tex]\Sigma F_x = \Sigma F_y = 0[/tex]
(a) For [tex]\Sigma F_x = 0[/tex], we have;
[tex]\Sigma F_x[/tex] = [tex]T_{ACB}[/tex] × cos(10°) - ([tex]T_{CD}[/tex] × cos(30°) + [tex]T_{ACB}[/tex] × cos(30°)) = 0
Which gives;
[tex]T_{CD}[/tex] = 0.13716·[tex]T_{ACB}[/tex]
0.13716·[tex]T_{ACB}[/tex] - [tex]T_{CD}[/tex] = 0......................(1)
For [tex]\Sigma F_y = 0[/tex], we have;
[tex]\Sigma F_y[/tex] = [tex]T_{ACB}[/tex] × sin(10°) + ([tex]T_{CD}[/tex] × sin(30°) + [tex]T_{ACB}[/tex] × sin(30°)) - 900 = 0
0.674·[tex]T_{ACB}[/tex] + 0.5·[tex]T_{CD}[/tex] = 900............(2)
Multiplying equation (2) by 2 and adding to equation (1) gives;
1.48445·[tex]T_{ACB}[/tex] = 1800
The tension, [tex]T_{ACB}[/tex], in the support cable is therefore;
[tex]T_{ACB}[/tex] = 1212.57 N
(b) The tension, [tex]T_{CD}[/tex] ,in the traction cable, CD, is given as follows
From, [tex]T_{CD}[/tex] = 0.13716·[tex]T_{ACB}[/tex], we have;
[tex]T_{CD}[/tex] = 0.13716 × 1212.57 ≈ 166.3 N
