Question

Five weightless rods of equal length are joined together so as to from a rhombus ABCD with one diagonal BD. If
a weight W be attached to C and the system be suspended from A, show that the thrust in the rod BD is equal to
W / 3​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Question:

Five weightless rods of equal length are joined together so as to from a rhombus ABCD with one diagonal BD. If a weight W be attached to C and the system be suspended from A, show that the thrust in the rod BD is equal to W /√3​.

Answer:

Let the five rods AB, BC, CD, DA, and BD form the rhombus ABCD and the diagonal BD. Since the system is suspended from A and there's a weight W attached to C, AC is going to be vertical and consequently BD horizontal.

Suppose that the rod AB or AD makes an angle θ with the horizontal through A. let us give a small symmetrical displacement such C that θ changes to $\theta + \delta \theta$ . Since A is fixed, we measure the depth of C, the point  of application of W, below A. Now AC is 2a sin θ where a is the length of a rod and hence the work done W during the tiny displacement is W δ(2a sin θ )

Point C moves down and hence the sign. The length BD is 2a cos θ and also the work done by the thrust T within the rod AD is T δ(2a cos θ).

By the principle of virtual work,

T δ(2a cos θ) + W δ(2a sin θ) = 0

2a δθ(W cos θ - T sin θ) = 0

Since, δθ can't be equal to zero,

∴ W cos θ - T sin θ = 0

T = W cos θ/ sin θ = W cot θ

In equilibrium position, θ = 60°

Hence, T = W cot 60° = W × 1/√3

T = W/√3

Hence proved.

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