Consider the situation shown in the figure (6-E9). 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?Concept of Physics - 1 , HC VERMA , Chapter 6:Friction "
Answers
Answered by
3
Solution :
****************************************************
The force on a charge Q by the electric field E is F=QE in the direction of E
From Free Body Diagram :
R1+QE-mg=0
R1=mg-QE-----------1
F-T=μR1
From equation I
F-T=μ(mg-QE)
F-T=μmg+μQE=0
T-μR1=0-------------3
T=μR1=μ[mg-QE]
From equation 2:
F-μmg+μQE-μmg+μQE=0
F-2μmg+2μQE=0
F=2μmg-2μQE
F=2μ(mg-QE)
∴Maximum horizontal force that can be applied is 2μ(mg-QE)
****************************************************
The force on a charge Q by the electric field E is F=QE in the direction of E
From Free Body Diagram :
R1+QE-mg=0
R1=mg-QE-----------1
F-T=μR1
From equation I
F-T=μ(mg-QE)
F-T=μmg+μQE=0
T-μR1=0-------------3
T=μR1=μ[mg-QE]
From equation 2:
F-μmg+μQE-μmg+μQE=0
F-2μmg+2μQE=0
F=2μmg-2μQE
F=2μ(mg-QE)
∴Maximum horizontal force that can be applied is 2μ(mg-QE)
Attachments:
raminder1:
hii
mam
Similar questions