
Three masses ${m_1},{m_2}$ and ${m_3}$ are connected with weightless strings in succession and are placed on a frictionless table. If mass ${m_3}$ is dragged with a force T. The tension in the string between ${m_2}$ and ${m_3}$ is:
(1) $\dfrac{{{m_2}}}{{{m_1} + {m_2} + {m_3}}}T$
(2) $\dfrac{{{m_3}}}{{{m_1} + {m_2} + {m_3}}}T$
(3) $\dfrac{{{m_1} + {m_2}}}{{{m_1} + {m_2} + {m_3}}}T$
(4) $\dfrac{{{m_2} + {m_3}}}{{{m_1} + {m_2} + {m_3}}}T$
Answer
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Hint: To find the solution of the given question apply the concept of Newton’s second law of motion which establishes a relation between force ‘F’ and the acceleration ‘a’. According to Newton's second law of motion the rate of change of momentum of a body is directly proportional to the force applied, and the change in momentum takes place in the direction of the applied force.
Formula Used: $F = ma$
Complete answer:
Create a rough diagram of the three masses ${m_1},{m_2}$ and ${m_3}$ connected with a weightless string in succession which is placed on a frictionless table for a better understanding of the forces acting on it.
Newton’s second law of motion is defined as the acceleration of an object produced by a net force which is directly proportional to the magnitude of the net force, in the same direction to that of the net force and is inversely proportional to the mass of the object.
Mathematically it is given as,
$ \Rightarrow F \propto ma$
$ \Rightarrow F = kma$
Where ‘k’ is denoted as the constant of proportionality, it is equal to 1 when the values are taken in SI units.
Thus, the final expression is given as, $F = ma$
$ \Rightarrow a = \dfrac{F}{m}$
$ \Rightarrow a = \dfrac{T}{{{m_1} + {m_2} + {m_3}}}$
$T - {T_1} = {m_3}a$ --(1)
${T_1} - {T_2} = {m_2}a$ --(2)
${T_2} = {m_1}a$ --(3)
Put equation (3) in (2)
${T_1} - {m_1}a = {m_2}a$
${T_1} = {m_2}a + {m_1}a$ --(4)
Put value of ‘a’ in equation (4)
$ \Rightarrow {T_1} = \dfrac{{{m_2}T}}{{{m_1} + {m_2} + {m_3}}} + \dfrac{{{m_1}T}}{{{m_1} + {m_2} + {m_3}}}$
$ \Rightarrow {T_1} = \dfrac{{\left( {{m_2} + {m_1}} \right)T}}{{{m_1} + {m_2} + {m_3}}}$
Hence, option (3) is the correct option.
Note:
Newton's second law of motion describes the relationship between an object's mass and the amount of force needed to accelerate it. This means the more mass an object has, the more force is needed to accelerate it, and greater is the force, more would be the object's acceleration.
Formula Used: $F = ma$
Complete answer:
Create a rough diagram of the three masses ${m_1},{m_2}$ and ${m_3}$ connected with a weightless string in succession which is placed on a frictionless table for a better understanding of the forces acting on it.
Newton’s second law of motion is defined as the acceleration of an object produced by a net force which is directly proportional to the magnitude of the net force, in the same direction to that of the net force and is inversely proportional to the mass of the object.
Mathematically it is given as,
$ \Rightarrow F \propto ma$
$ \Rightarrow F = kma$
Where ‘k’ is denoted as the constant of proportionality, it is equal to 1 when the values are taken in SI units.
Thus, the final expression is given as, $F = ma$
$ \Rightarrow a = \dfrac{F}{m}$
$ \Rightarrow a = \dfrac{T}{{{m_1} + {m_2} + {m_3}}}$
$T - {T_1} = {m_3}a$ --(1)
${T_1} - {T_2} = {m_2}a$ --(2)
${T_2} = {m_1}a$ --(3)
Put equation (3) in (2)
${T_1} - {m_1}a = {m_2}a$
${T_1} = {m_2}a + {m_1}a$ --(4)
Put value of ‘a’ in equation (4)
$ \Rightarrow {T_1} = \dfrac{{{m_2}T}}{{{m_1} + {m_2} + {m_3}}} + \dfrac{{{m_1}T}}{{{m_1} + {m_2} + {m_3}}}$
$ \Rightarrow {T_1} = \dfrac{{\left( {{m_2} + {m_1}} \right)T}}{{{m_1} + {m_2} + {m_3}}}$
Hence, option (3) is the correct option.
Note:
Newton's second law of motion describes the relationship between an object's mass and the amount of force needed to accelerate it. This means the more mass an object has, the more force is needed to accelerate it, and greater is the force, more would be the object's acceleration.
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