A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2Mg force. In this situation, tension in the string is x/5 Mg for x=__________. Neglect mass of the string and friction of the block (bigger mass) with ice slab.
(Given g= acceleration due to gravity)
Fig. Hanging mass
A hanging mass M is connected to a four times bigger mass by using a string-pulley arrangement, as shown in the figure. The bigger mass is placed on a horizontal ice-slab and being pulled by 2Mg force. In this situation, tension in the string is x/5 Mg for x=__________. Neglect mass of the string and friction of the block (bigger mass) with ice slab.
(Given g= acceleration due to gravity)
Correct Answer: 6
Solution and Explanation
\(a = \frac{Mg}{4M + M }= \frac{g}{5}\)
(in upward direction)
\(T = M ( g + \frac{g}{5} ) = \frac{6Mg}{5}\)
⇒ x = 6
Therefore , tension in the string is \(\frac{x}{5}\) Mg for x= 6
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Concepts Used:
Newtons Laws of Motion
Newton’s First Law of Motion:
Newton’s 1st law states that a body at rest or uniform motion will continue to be at rest or uniform motion until and unless a net external force acts on it.
Newton’s Second Law of Motion:
Newton’s 2nd law states that the acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the object’s mass.
Mathematically, we express the second law of motion as follows:
Newton’s Third Law of Motion:
Newton’s 3rd law states that there is an equal and opposite reaction for every action.