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Standard XIII

Physics

Tension in a String

Three blocks ...

Question

Three blocks are connected by a light inextensible string as shown in the figure and placed on a horizontal table. If $m_{1} = 10 k g, m_{2} = 20 k g$ and $m_{3} = 30 k g$ , find the tension $T_{1}$ given that $T_{3} = 60 N$

zero

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45 N

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30 N

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10 N

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Solution

The correct option is D $10 N$
The FBD of the mass $m_{1}$ is as shown

Let $a$ be common the acceleration of the system, which is given by
$a = \frac{(Supporting Force-Opposing Force)}{Total mass} = \frac{T_{3}}{m_{1} + m_{2} + m_{3}}$
$= \frac{60}{10 + 20 + 30} = 1 m / s^{2}$
From the FBD of $m_{1}$ , we have
$T_{1} = m_{1} a = 10 \times 1 = 10 N$

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Three blocks of mass $M_{1}$ , $M_{2}$ & $M_{3}$ are connected by massless strings as shown on a frictionless table. Where $T_{1}$ is the tension in the string between block $M_{1}$ and $M_{2}$ , $T_{2}$ is the tension in the string between block $M_{2}$ and $M_{3}$ . They're pulled with a force $F = 40 N$ . If $M_{1}$ =10 kg, $M_{2}$ = 6 kg and $M_{3}$ = 4 kg, then tension $T_{2}$ will be:

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Three blocks are connected by a light inextensible string as shown in the figure and placed on a horizontal table. If m1=10 kg,m2=20 kg and m3=30 kg, find the tension T1 given that T3=60 N

The correct option is D 10 NThe FBD of the mass m1 is as shown Let a be common the acceleration of the system, which is given by a=(Supporting Force-Opposing Force)Total mass=T3m1+m2+m3 =6010+20+30=1 m/s2 From the FBD of m1, we have T1=m1a=10×1=10 N

Three blocks are connected by a light inextensible string as shown in the figure and placed on a horizontal table. If $m_{1} = 10 k g, m_{2} = 20 k g$ and $m_{3} = 30 k g$ , find the tension $T_{1}$ given that $T_{3} = 60 N$