Two blocks A 2kg and B 5kg rest one over the other on a smooth inclined plane the coefficient of static and dynamic friction between A and B is the same and equal to 0.6 the maximum horizontal force F that can be applied to B in order that both A and B do not have any relative motion is
1) 42 N
2) 42kgf
3) 5.4 kgf
4) 1.2 N
Answers
Answer:
1.2kgf.
Explanation:
The mass of block A from the question is taken as m1= 2 kg and the mass of the block B is m₂ =5 kg.
Also the co-efficient of friction is give as μ
is 0.6.
The frictional force for A due to B will be the force required to stop the body from slipping. Let that force be Fx.
So, we will get Fx= μ m₂.
Fx = (0.6) (5).
Fx= 3 kgf.
The acceleration will be calculated by for the bodies as:-
Acceleration will be a = Fx / m₂.
a = 3 / 5.
a = 0.6 m/s^2.
Now, to get the maximum force for stopping the relative motion of the bodies will be F + Fx.
F(max) + Fx = m(12) x a
And the value of mass for 1 due to 2 will be 5+2= 7 kg.
So, the value of maximum force is:-
Fmax= (7 x 0.6) - 3
Fmax= 4.2 - 3.
Fmax will be 1.2 kgf.
So, the force required is 1.2 kgf.
Answer:
42N
Explanation:
Now both should move with same acceleration. So,
Fr=ma=2a
F−Fr=5a
a = mμg/m = μg = 6m/s2
F - 0.6×2 × 10 = 5 × 6 = 42N