Two blocks of mass 2.9 kg and 1.9 kg are suspended from a rigid support S by two inextensible wires each of length 1 meter, as shown in the figure. The upper wire has negligible mass and the lower wire has a uniform mass of 0.2 kg/m. The whole system of blocks, wires and support have an upward acceleration of 0.2 $m / s^{2}$ . Acceleration due to gravity is 9.8 $m / s^{2}$ . Then the difference in tension at the mid point's (in N) of the upper wire and the lower wire is

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Solution

Here,
$m_{1} = 1.9 k g$
$m_{2} = 2.9 k g$
The whole system of block, wires and support have an upward acceleration of $0.2 m / s^{2}$ .
So applying Newton's second law:
Let, $T =$ Tension at the mid point of the lower wire
$λ =$ Mass per unit length of wire
l $=$ Half-length of the wire $= 0.5 m$
$∴ T - (m_{1} + λ l) g = (m_{1} + λ l) a$
$T = (m_{1} + λ l) (a + g)$
$T = (1.9 + 0.2 \times 0.5) (0.2 + 9.8) = 20 N$

If, $T^{'} =$ Tension at the mid point of the upper wire
$∴ T^{'} = [m_{1} + (λ \times 1) + m_{2}] a + [m_{2} g + (λ \times 1) g + m_{1} g]$
$T^{'} = [m_{1} + (λ \times 1) + m_{2}] (a + g)$
$T^{'} = [1.9 + 0.2 \times 1 + 2.9] (9.8 + 0.2) T^{'} = 5 \times 10 = 50 N$

So, $T^{'} - T = 50 - 20 = 30 N$

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Two blocks of masses $2.9 kg$ and $1.9 kg$ are suspended from a rigid support by two inextensible wires each of length $1 m$ , the upper wire has negligible mass and the lower wire has a uniform mass of $0.2 kg/m$ . The whole system of block, wire and support has an upward acceleration of $0.2 {m/s}^{2}$ . (take $g = 9.8 m / s e c^{2}$ )