One block of 2·0 kg mass is placed on top of another block of 3·0 kg mass. The coefficient of static friction between the two blocks is 0·2. The bottom block is pulled with a horizontal force F such that both the blocks move together without slipping. Taking acceleration due to gravity as 10 m/s², the maximum value of the frictional force is :
50 N
30 N
4 N
10 N
Answer is Wrong!
Answer is Right!
This question was previously asked in
UPSC NDA-1 – 2023
– The maximum static friction force is given by $f_{s,max} = \mu_s N$.
– For the blocks to move together without slipping, the actual static friction force must be less than or equal to $f_{s,max}$. The question asks for the maximum value of the frictional force in this scenario, which is the maximum static friction force that can exist between the blocks.
Normal force on the top block (due to the bottom block) N = m₁g = 2.0 kg × 10 m/s² = 20 N.
Maximum static friction force $f_{s,max} = \mu_s N = 0.2 \times 20 N = 4 N$.
For the blocks to move together, the actual friction force on the top block is $f = m_1 a$, where ‘a’ is the acceleration of the system. As the pulling force F on the bottom block increases, ‘a’ increases, and thus the required friction force ‘f’ increases. Slipping occurs when the required ‘f’ exceeds $f_{s,max}$. The maximum value the frictional force can reach while preventing slipping is $f_{s,max}$.