A and B are two pegs separated by 13 cm. A body of 169 kgwt is suspended by the thread of 17 cm connecting to A & B,such that two segments of strings are perpendicular. Then tensions in shorter and longer parts of string are..
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Answered by
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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.
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.
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kvnmurty:
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Answered by
13
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
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
Attachments:
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