Rectangular block rests on a horizontal table A horizontal force is applied on the block at a height h above table to move the block. Can the normal on the block depends on height h?
Answers
Answer:
In the absence of any external force in horizontal direction the normal reaction N passes through the centre of mass of the block; when force F is applied, normal reaction shifts in the direction of applied force F.
Since right part of body is having tendency to lift from surface, at the instant of tipping over about the edge the normal reaction passes through edge. From the conditions of equilibrium
In horizontal direction f=F=μ
k
mg
In vertical direction N=mg
Balancing torque about edge Fh−mg
2
b
or h=
2F
mgb
=
2μ
k
mg
mgb
=
2μ
k
b
We can solve the problems by another approach also as the resultant of friction force (μ
k
N) and the normal reaction must pass through the same point through which F passes, since three coplanar forces keeping a body in equilibrium pass through a common point i.e they should be concurrent.
From Lami's theorem:
sin90
o
F
′
=
sin(180
o
+θ)
F
=
sinsin(90
o
+θ)
mg
⇒
sinθ
F
=
cosθ
mg
ortanθ=
mg
F
=
mg
μ
k
mg
=μ
k
From figure it is clear tanθ=
N
μ
k
N
=
h
b/2
or
h
b/2
=μ
k
⇒h=
2μ
k
b
(b) In this case we cannot take normal reaction at the edge
Let normal reaction acts at a distance x from the line of action of mg
Torque of all the forces about C must be zero
(μ
k
N×
2
H
)=Nx
x=
2
μ
k
H
Alternatively we can proceed as in part (a). Resultant of μ
k
,N,N passes through C the interaction of F and mg
Therefore from geometry of the figure.
H/2
x
=tanθ=μ
k
;x=
2
μ
k
H
(c) As the point of appication of force is raised higher, the location of the line of action of the normal reaction N moves to the left. In the limiting case (when the block is about to trip over), x=b/2. The normal reaction passes through edge as shown in figure.
Initial toppling torque about edge:
Fh
max
=mg
2
b
orh
max
=
2F
mgb
...(i)
For initial sliding, in horizontal direction: F=μN
In vertical direction: N=mg
Hence, F=μmg
If sliding and tipping are equally likely to occur, we can eliminate F eqs. (i) and (ii) to get h
max
=
2μ
b
which is independent of weight mg, height H of the body and applied force F.