a body weighing 2000 N is suspended with a chain AB. It is pulled by a horizontal force 320 N as shown in figure. Find the force in the chain & lateral displacement.
Answers
Answer:
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Therefore the Force in the chain is 2025 N and the lateral displacement is 0.32 m.
Given:
Weight of the body = W = 2000 N
Length of the chain = L = 2 m
Horizontal Force applied on the body = F = 320 N
To Find:
The force in the chain and the Lateral displacement.
Solution:
The given question can be easily solved using the below approach.
Given that,
Weight of the body = W = 2000 N
Length of the chain = L = 2 m
Horizontal Force applied on the body = F = 320 N
Let Force in the chain be 'x'.
The angle between the chain and horizontal = θ
As the body is in equilibrium we can use 'Lami's Theorem'.
By Lami's Theorem,
⇒ 2000 / Sin ( 180 - θ ) = 320 / Sin ( 90 + θ ) = F / Sin 90
Case 1:
⇒ 2000 / Sin ( 180 - θ ) = 320 / Sin ( 90 + θ ) { Sin ( 180 - θ ) = Sin θ; Sin ( 90 + θ )= Cos θ }
⇒ 2000 / Sin θ = 320 / Cos θ
⇒ Sin θ / Cos θ = 2000 / 320
⇒ Tan θ = 6.25
⇒ θ = Tan ⁻¹ ( 6.25 ) = 80.9° ≅ 81°
Angle between Force in the chain and horizontal = 81°
Case 2:
⇒ 2000 / Sin ( 180 - θ ) = F / Sin 90
⇒ 2000 / Sin θ = F / 1
⇒ F = 2000 / Sin 81 = 2024.93 ≅ 2025 N
Hence the force in the chain = F = 2025 N
From the figure,
Cos θ = x / 2 ⇒ x = 2Cos θ
⇒ x = 2 Cos 80.9 = 0.3163 ≅ 0.32
Lateral Displacement of the chain = x = 0.32 m
Therefore the Force in the chain is 2025 N and the lateral displacement is 0.32 m.
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