Question

For what value of \[{{m}_{1}}\] will \[8Kg\] mass be at rest?

Answer 1

Hint: In such type of question first we have to break the possible number of forces main forces acting in such numerical are weight and tension then we apply the Newton’s second law of motion to solve each mass by seeing the direction of all forces acting on it and required results are obtained.

Complete step by step solution:
Since in the given figure of pulleys masses are attached to the strings so two forces acts on them:-
1. Weight of the body vertically downwards.
2. Tension in the string in upward direction.
Since pulleys are massless so their mass is not considered during calculation.
The given figure block of \[8Kg\] is attached to the string with a pulley.
Its Weight \[=8g\](acts Vertically Downwards direction)
and Tension in the string \[{{T}_{1}}\] (Upwards direction).
Since the string to this mass is the same, the same tension acts on the whole string.
According to the question, the body of mass \[8Kg\] is at rest so no acceleration is acting on this body.
So According to Newton’s second Law of motion we can write,
\[\begin{align}
  & {{T}_{1}}=8g \\
 & \therefore {{T}_{1}}=80N \\
\end{align}\].
So the tension in the above is \[80N\].
For the second pulley two masses are attached to it and this system has tension \[2T\] acting in downward direction which balances the tension \[{{T}_{1}}\] acting in upward direction in the upper pulley.
So in equilibrium we can write,
\[\begin{align}
  & {{T}_{1}}=2T \\
 & \Rightarrow T=\dfrac{{{T}_{1}}}{2} \\
 & \therefore T=40N \\
\end{align}\]
So the tension on the strings in the lower system is \[40N\] when the mass of the upper pulley is at rest.
For \[5Kg\] body, acceleration is acting downward direction,
so according to Newton’s Law of motion we can write,
\[5g-T=5a\]
Since value of \[T=40N\]is given put in above equation we get,
\[\begin{align}
  & 50-40=5a \\
 & \Rightarrow 10=5a \\
 & \therefore a=2\dfrac{m}{{{s}^{2}}} \\
\end{align}\]
So we get the acceleration of \[5Kg\] body, since the lower pulley has the same value of acceleration for both the masses.
Now for \[{{m}_{1}}Kg\] body acceleration is acting upward direction,
so according to Newton’s Law of motion we can write,
\[T-{{m}_{1}}g={{m}_{1}}a\]
Value of Tension \[T=40N\] and acceleration \[a=2\dfrac{m}{{{s}^{2}}}\] is already calculated put in above equation we get,
\[\begin{align}
  & 40-10{{m}_{1}}=2{{m}_{1}} \\
 & \Rightarrow 40=12{{m}_{1}} \\
 & \therefore {{m}_{1}}=\dfrac{40}{12}=\dfrac{10}{3}Kg \\
\end{align}\]
So the mass of \[{{m}_{1}}\] when \[8Kg\] mass is at rest is \[\dfrac{10}{3}Kg\].

Note: According to Newton’s second law rate of change in momentum is directly proportional to external force applied, here in this type of numerical we will use the derived form of Newton’s second law which is represented as external force is equal to the product of mass and acceleration of the body.