Question

Four weightless rods form a rhombus PQRS with smooth at the joints. Another weightless rod joins the midpoints E and F of PQ and PS respectively. The system is suspended from P and a weight 2W is atta
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Answer 1

Answer:

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Answer 2

Answer:

The force exerted by the rods on the load is

F1 = 2Ften cos α, while the force exerted on the spring is F2 = 2Ften sin α (see. Fig. 48).

According t0 Hooke's law, F2 = (1.5l - 2l sin α) k , where k is the rigidity of the spring.

As a result, In order to determine the period of small oscillations, we must determine the force ΔF acting on the load for a small change Δh in the height of the load relative to the equilibrium position h0 = 2l cos α0.

We obtain Consequently, since Δh = - 2l sin α0 Δα,

we find that The period of small oscillations of the load can be found from the formula where in is the mass of the load determined from the equilibrium condition

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