Friction Force MCQ Level - 2 Solved MCQs Physics

During downward motion :
F = mg Sinθ - mg Cosθ
During upward motion :
2F = mg Sinθ + μmg Cosθ
Solving above two equations, we get m = 1/3 tanθ
The correct answer is: 

So block 'Q' is moving due to force while block 'P' due to friction.
Friction direction on both P + Q blocks as shown.

First block 'Q' will move and P will move with 'Q' so by FBD taking 'P' and 'Q' as system

When applied force is 4 N  then FBD

i.e., a_p = a_q = 0
4 kg block is moving due to friction and maximum friction force is 8 N
So acceleration 
Slipping will start at when Q has +ve acceleration equal to maximum acceleration of P i.e., 2 m/s² 

The correct answer is: Maximum acceleration of 4 kg block is 2 m/s²

Maximum friction 
Pseudo force = ma = 5N
acceleration due to the applied Pseudo force 
So ​required friction force is only 5N although maximum friction force available is 6N.
The correct answer is: 5 N

Direction of friction is such that it opposes the relative velocity.
The correct answer is: Block A (left), block B (right due to block A, right due to ground)

For chain to move with constant speed P needs to be equal to frictional forces on the chain. As the length of chain on the rough surface increases.
Hence the friction force f_k = µ_kN increases.
The correct answer is: the magnitude of P should increase with time

FBD of A and C

If acceleration of 'C' is a
For block 'A' N = 8 ma ...(1)
for block A to remain stationary with respect to block B,
8mg = μN (Limiting condition)   ...(2)
a = g/μ
and acceleration a can be written by the equation of system (A + B + C)

The correct answer is: 

Limiting condition for m to not slip in vertical downward direction,

Same normal force would acceleration M, thus 
Taking m + M as system

The correct answer is: 240 N

For smooth plane  a = g sinθ 
For rough plane, a' = g(sinθ - μcosθ)
∴ t' = nt

when θ = 45° ,  sinθ = cosθ = 1/√2
On solving we get, 
The correct answer is: 

Let m_A and m_B be the mass of blocks A and B respectively.
As the force F increases from 0 to µ_s m_Ag, the frictional force f on block A is such that f = F. When F = µ_sm_Ag, the frictional force f attains maximum value f = µ_smg.
As F is further increased to µ_s(m_A + m_B)g, the block A does not move. In this duration frictional force on block A remains constant at µ_sm_Ag.
As F further increased, system will start moving and kinetic friction (µx_Ag) will start acting on A (µ_s>µ_k). Hence C is correct choice.
The correct answer is:

Limiting friction between A and B = 60 N
Limiting friction between B and C = 90 N
Limiting friction between C and ground = 50 N
Since limiting friction is least between C and ground, slipping will occur at first between C and ground. This will occur when F = 50 N.
The correct answer is: 60 N