A,B are two pegs seperated by 13cm.A body of 169 Kgwt is suspended by thread of 17cm connecting toA,B.such that two segments of strings are perpendicular then tensions in shorter and longer parts of strings having are
Answers
Explanation:
See diagram.
CD = L SinФ = (17 - L) CosФ --- (1)
=> tan Ф = 17/L - 1
=> L = 17/(1+tanФ) -- (2)
AB = L CosФ + (17 - L) SinФ = 13 cm --- (3)
- (1) * cosФ + (3) * sinФ =>
(17 -L) cos²Ф + (17-L) sin²Ф = 13 sinФ
=> 17 - L = 13 sinФ cm
=> 17 - 17/(1+tanФ) = 13 sinФ
=> 17 tanФ = 13 (1+tanФ) sinФ
=> 17 = 13 (cosФ + sinФ)
=> 17/13 = √2 Sin (π/4 + Ф)
=> Ф = Sin⁻¹ (17/13√2) - π/4 = 22.62°
=> sinФ = 0.384 cosФ = 0.923
=> L = 12.0 cm nearly
Writing the equations of static Equilibrium for concurrent forces:
T1 / Sin(180°-Ф) = T2 / Sin(90°+Ф) = 169 / Sin90°
=> T1 /sinФ = T2 / CosФ = 169
The tension forces in the smaller and larger parts of the string are :
=> T1 = 64.896 kg wt. T2 = 155.987 kg wt
= 65kg wt. =156kg wrt
See diagram.
CD = L SinФ = (17 - L) CosФ --- (1)
=> tan Ф = 17/L - 1
=> L = 17/(1+tanФ) -- (2)
AB = L CosФ + (17 - L) SinФ = 13 cm --- (3)
- (1) * cosФ + (3) * sinФ =>
(17 -L) cos²Ф + (17-L) sin²Ф = 13 sinФ
=> 17 - L = 13 sinФ cm
=> 17 - 17/(1+tanФ) = 13 sinФ
=> 17 tanФ = 13 (1+tanФ) sinФ
=> 17 = 13 (cosФ + sinФ)
=> 17/13 = √2 Sin (π/4 + Ф)
=> Ф = Sin⁻¹ (17/13√2) - π/4 = 22.62°
=> sinФ = 0.384 cosФ = 0.923
=> L = 12.0 cm nearly
Writing the equations of static Equilibrium for concurrent forces:
T1 / Sin(180°-Ф) = T2 / Sin(90°+Ф) = 169 / Sin90°
=> T1 /sinФ = T2 / CosФ = 169
The tension forces in the smaller and larger parts of the string are :
=> T1 = 64.896 kg wt. T2 = 155.987 kg wt
respectively.
