
A man of mass m stands on a crate of mass M. He pulls on a light rope over a smooth light pulley. The other end of the rope is attached to the crate. For the system to be in equilibrium, the force exerted by the man on the rope will be
A. $ \left( {M + m} \right)g $
B. $ \dfrac{1}{2}\left( {M + m} \right)g $
C. $ Mg $
D. $ mg $

Answer
456k+ views
Hint: Consider the tension on the rope due to both the pulling and due to its attachment to a crate. Then use Newton's third law of motion to establish the equilibrium condition for the system. Once the tension on the rope is calculated, then we can easily show how much force is exerted by the man.
Complete step by step solution
Let $ T $ be the tension on the rope. Now we are told that the pulley is being pulled from one side by a man and the other end of the rope is attached to a crate. So there is tension on both the side of the rope such that the total tension becomes $ 2T $ .
If $ m $ be the mass of the man and $ M $ be the mass of the crate then the total mass will be $ $ as the man is standing on the crate.
$ g $ being the acceleration due to gravity, the total force acting downwards is $ \left( {m + M} \right)g $
For the system to be in equilibrium, the total tension on the rope acting upwards must be equal to the force acting downwards.
Thus, we can show that $ 2T = \left( {m + M} \right)g $
$ \Rightarrow T = \dfrac{1}{2}\left( {m + M} \right)g $
We know according to Newton’s third law of motion that every action has an equal and opposite reaction so when the man pulls the rope downwards, an equal and opposite force called the tension acts on the rope in the upwards direction. Thus from this we can say that the force exerted by the man on the rope is equal to $ T $ .
So, the correct option is B.
Note
Tension is a form of restoring force which can only pull. This is due to the fact that tension only comes into play when we consider systems involving strings, ropes or cables. Other types of restoring force are involved in simple harmonic motion.
Complete step by step solution
Let $ T $ be the tension on the rope. Now we are told that the pulley is being pulled from one side by a man and the other end of the rope is attached to a crate. So there is tension on both the side of the rope such that the total tension becomes $ 2T $ .
If $ m $ be the mass of the man and $ M $ be the mass of the crate then the total mass will be $ $ as the man is standing on the crate.
$ g $ being the acceleration due to gravity, the total force acting downwards is $ \left( {m + M} \right)g $
For the system to be in equilibrium, the total tension on the rope acting upwards must be equal to the force acting downwards.
Thus, we can show that $ 2T = \left( {m + M} \right)g $
$ \Rightarrow T = \dfrac{1}{2}\left( {m + M} \right)g $
We know according to Newton’s third law of motion that every action has an equal and opposite reaction so when the man pulls the rope downwards, an equal and opposite force called the tension acts on the rope in the upwards direction. Thus from this we can say that the force exerted by the man on the rope is equal to $ T $ .
So, the correct option is B.
Note
Tension is a form of restoring force which can only pull. This is due to the fact that tension only comes into play when we consider systems involving strings, ropes or cables. Other types of restoring force are involved in simple harmonic motion.
Recently Updated Pages
Master Class 11 Accountancy: Engaging Questions & Answers for Success

Glucose when reduced with HI and red Phosphorus gives class 11 chemistry CBSE

The highest possible oxidation states of Uranium and class 11 chemistry CBSE

Find the value of x if the mode of the following data class 11 maths CBSE

Which of the following can be used in the Friedel Crafts class 11 chemistry CBSE

A sphere of mass 40 kg is attracted by a second sphere class 11 physics CBSE

Trending doubts
10 examples of friction in our daily life

Difference Between Prokaryotic Cells and Eukaryotic Cells

State and prove Bernoullis theorem class 11 physics CBSE

What organs are located on the left side of your body class 11 biology CBSE

Define least count of vernier callipers How do you class 11 physics CBSE

The combining capacity of an element is known as i class 11 chemistry CBSE
