
A constant force F is applied in a horizontal direction as shown. The contact force between M and m is N and between m and M’ is N’ then:
(A) N=N’
(B) N>N’
(C)N’>N
(D) cannot be determined from the given data

Answer
486.6k+ views
Hint: Here we have three bodies of given masses and placed close to each other in contact. The floor is smooth and there is no scope of any frictional force present. A constant force F acts on the first body and we need to find the force between the masses. Newton’s third law of motion can be used here.
Complete step by step answer:
Since F acts on the M and this is constant, if we see the whole system then the total mass of the system is (M+m+M’)
Applying Newton’s second law we get, F=(M+m+M’)a
\[a=\dfrac{F}{M+m+M'}\]
Between M and m contact force is N, applying Newton’s third law
\[N=\dfrac{(m+M')F}{M+m+M'}\]-----(1)
Between m and M’ contact force is N’, applying Newton’s third law \[N'=\dfrac{M'F}{M+m+M'}\]----(2)
Comparing the above two it is visible that numerator of (1) is greater than the numerator of (2) while the denominators are same, so, N>N’
Hence, the correct option is (B).
Note:Newton’s second law states that force applied on a body is equal to the product of the mass of the body and acceleration produced in the body. If the external force is zero, then the acceleration of the body is zero. Newton’s third law states that forces always occur in pairs.
Complete step by step answer:

Since F acts on the M and this is constant, if we see the whole system then the total mass of the system is (M+m+M’)
Applying Newton’s second law we get, F=(M+m+M’)a
\[a=\dfrac{F}{M+m+M'}\]
Between M and m contact force is N, applying Newton’s third law
\[N=\dfrac{(m+M')F}{M+m+M'}\]-----(1)
Between m and M’ contact force is N’, applying Newton’s third law \[N'=\dfrac{M'F}{M+m+M'}\]----(2)
Comparing the above two it is visible that numerator of (1) is greater than the numerator of (2) while the denominators are same, so, N>N’
Hence, the correct option is (B).
Note:Newton’s second law states that force applied on a body is equal to the product of the mass of the body and acceleration produced in the body. If the external force is zero, then the acceleration of the body is zero. Newton’s third law states that forces always occur in pairs.
Recently Updated Pages
Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Accountancy: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Trending doubts
State and prove Bernoullis theorem class 11 physics CBSE

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

Write the differences between monocot plants and dicot class 11 biology CBSE

Why is steel more elastic than rubber class 11 physics CBSE

Explain why a There is no atmosphere on the moon b class 11 physics CBSE

State the laws of reflection of light
