Question

32. Three
se blocks A, B, C of masses M1, M2, and M3, are connected by means
of two identical wires each of length L, radius r and Young's modulus Y
as shown in the figure. Blocks A and B are lying on a smooth horizontal
surface and the string connecting blocks B and C passes over a light
frictionless fixed pulley. When released, the system starts accelerating
uniformly. Calculate the longitudinal strain produced in the wire
connecting Block A and B.
M]
M2
M₃​

Answer 1

Answer:

the longitudinal strain produced in the wire  connecting Block A and B is

$Strain = \frac{M_1(M_3 - M_1 - M_2)g}{\pi r^2 Y(M_1 + M_2 + M_3)}$

Explanation:

As shown in the figure let say the system is moving with acceleration "a" so we have

$M_3g - T_1 = M_3a$

on the other side of the string we have

$T_1 - (M_1 + M_2)g = (M_1 + M_2) a$

now we have

$(M_3 - M_1 - M_2)g = (M_1 + M_2 + M_3) a$

so we have

$a = \frac{M_3 - M_1 - M_2}{M_1 + M_2 + M_3} g$

so from above equation of acceleration of the blocks we can find the tension in the string between A and B

$T = M_1 a$

$T = \frac{M_1(M_3 - M_1 - M_2)}{M_1 + M_2 + M_3} g$

so strain in the string is given as

$Strain = \frac{T}{AY}$

$Strain = \frac{M_1(M_3 - M_1 - M_2)g}{\pi r^2 Y(M_1 + M_2 + M_3)}$

#Learn

Topic : Strain

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