Consider the situation shown in the figure. Suppose a small electric field E exists in the space in the vertically upward direction and the upper block carries a positive charge Q on its top surface. The friction coefficient between the two blocks is µ but the floor is smooth. What maximum horizontal force F can be applied without disturbing the equilibrium?
[Hint: The force on a charge Q by the electric field E is F = QE in the direction of E.]
Answers
For the equilibrium in vertical direction,
N + QE = mg
N = mg - QE
Friction force on upper block,
f =µN
= µ(mg - QE)
For the equilibrium in the horizontal direction
F=T + fric.
Consider the equilibrium of lower block in vertical direction,
N = Mg + mg - QE,
Since the surface is smooth there is no friction on the lower surface of lower block, but the upper surface of the lower block will have friction force P by the upper block in the opposite direction. So for the equilibrium in the horizontal direction.
F = T
= µ(mg-QE),
F = T+ fric.
=2 × fric.
=2µ (mg - QE)
Hope it helps.
Answer:
For the equilibrium in vertical direction,
N + QE = mg
N = mg - QE
Friction force on upper block,
f =µN
= µ(mg - QE)
For the equilibrium in the horizontal direction
F=T + fric.
Consider the equilibrium of lower block in vertical direction,
N = Mg + mg - QE,
Since the surface is smooth there is no friction on the lower surface of lower block, but the upper surface of the lower block will have friction force P by the upper block in the opposite direction. So for the equilibrium in the horizontal direction.
F = T
= µ(mg-QE),
F = T+ fric.
=2 × fric.
=2µ (mg - QE)