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When a tensile or compressive load 'P' i...

When a tensile or compressive load `'P'` is applied to rod or cable, its length changes. The change length `x` which, for an elastic material is proportional to the force Hook' law).
`P alpha x` or `P = kx`
The above equation is similar to the equation of spring. For a rod of length `L`, area `A` and young modulue `Y`. the extension `x`can be expressed as.
`x = (PL)/(AY)` or `P = (AY)/(x)`, hence `K = (AY)/(L)`
Thus rod or cables attached to lift can be treated as springs. The energy stored in rod is called stra energy & equal to `(1)/(2)Px`. The loads placed or dropped on the floor of lift cause stresses in the cable and can be evaluted by spring analogy. If the cable of lift cause stresses and load is placed dropped, then maximum extension in cable can be calculated by enerfy conservation.
Four identical rods of geometry as described in problem `(2)` are attached with lift. If weight of the lift cage is `1000N`, and elastic limit of each rod is taken as `9 xx 10^(4)N//m^(2)` then the number of persons it can carry safety is equal to. `(g = 10m//sec^(2))`, assume average mass of a person as `50 kg` and lift moves with constant speed)

A

`7`

B

`26`

C

`24`

D

`25`

Text Solution

Verified by Experts

The correct Answer is:
B
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When a tensile or compressive load 'P' is applied to rod or cable, its length changes. The change length x which, for an elastic material is proportional to the force Hook' law). P alpha x or P = kx The above equation is similar to the equation of spring. For a rod of length L , area A and young modulue Y . the extension x can be expressed as. x = (PL)/(AY) or P = (AY)/(x) , hence K = (AY)/(L) Thus rod or cables attached to lift can be treated as springs. The energy stored in rod is called stra energy & equal to (1)/(2)Px . The loads placed or dropped on the floor of lift cause stresses in the cable and can be evaluted by spring analogy. If the cable of lift cause stresses and load is placed dropped, then maximum extension in cable can be calculated by enerfy conservation. In the above problem, the maximum stress developed in the rod is equal to - (N//m^(2))

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When a tensile or compressive load 'P' is applied to rod or cable, its length changes. The change length x which, for an elastic material is proportional to the force Hook' law). P alpha x or P = kx The above equation is similar to the equation of spring. For a rod of length L , area A and young modulue Y . the extension x can be expressed as. x = (PL)/(AY) or P = (AY)/(x) , hence K = (AY)/(L) Thus rod or cables attached to lift can be treated as springs. The energy stored in rod is called stra energy & equal to (1)/(2)Px . The loads placed or dropped on the floor of lift cause stresses in the cable and can be evaluted by spring analogy. If the cable of lift cause stresses and load is placed dropped, then maximum extension in cable can be calculated by enerfy conservation. It two rods of same length (4m) and cross section areas 2 cm^(2) and 4 cm^(2) with same young modulus 2 xx 10^(10)N//m^(2) are attached one after the other with mass 600 kg then angular frequency is-

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