Answer:
The tension in the cable B is approximately 8375.227 N
Explanation:
The given parameters are;
The orientation of one cable, B = Horizontal
The orientation of the other cable, A = 40° to the vertical
The horizontal speed with which the chair swings= 32 revolutions per minute
ω = 32×2×π/60 rad /s
The weight of the seat = 255 N
The weight of the person = 825 N
The acceleration due to gravity, g = 9.8 m/s²
Therefore, we have;
The centripetal force [tex]F_c = \left (m_p + m_s \right ) \times l \times \omega ^2[/tex], which gives;
[tex]F_c = \left (\dfrac{825 + 255}{9.8} \right ) \times 7.5 \times \left (\dfrac{32\times 2 \times \pi}{60} \right )^2 \approx 9281.457[/tex]
The weight of the person and the seat W = 825 + 255 = 1080 N = The vertical force component in the system
The vertical component of the cable A = The vertical force component in the system = 1080 N
Let [tex]T_A[/tex] represent the tension in A
The vertical component of the tension = [tex]T_A[/tex] × cos 40° = 1080
[tex]T_A[/tex] = 1080/(cos(40°)) ≈ 1409.84 N
The horizontal component of [tex]T_A[/tex] =
From ∑Fₓ = 0, given that the centripetal force is acting outwards and the tension in cable B and the horizontal component of [tex]T_A[/tex] are acting inwards, we have;
The centripetal force, [tex]F_c[/tex] = The tension in the cable B + The horizontal component of [tex]T_A[/tex]
9281.457 = The tension in the cable B + 906.23
The tension in the cable B = 9281.457 - 906.23 ≈ 8375.227 N.
