1) A body of mass M is kept on a rough horizontal surface(friction coefficient=μ). A person is trying to pull the body by applying a horizontal force but the body is not moving. The force by the surface on a body is F where-
(A)F=Mg (B)F=μMg (C)Mg≤F≤Mg√1+μ2 (D)Mg≥F≥Mg√1-μ2
11answer c
Now since body is not moving even on application of force so f = uN
Now since there is no net disp in vertical direction N = Mg
f = uN = uMg
Now when F = 0
Fsurface = Mg since f=0
When F ≠0
Fsurface = √(Mg)2 + f2
or Fsurface = Mg√1 + u2
so Mg ≤ Fsurface ≤ Mg√1 + u2
(P.S. The horizontal force on the body I have taken as F in the diagram whereas F is asked as the force by the surface in the question which I have taken as Fsurface .)
1thanx.
Bt jus have one doubt if u can xplain it 2 me.
how did u arrive at dis expression
When F≠0
FSurface=√(Mg)2+f2
62mg and f are perpendicular to each other
force by the surface is the vector sum (it consists of normal reaction and of friction)