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Thursday, March 19, 2009

APPLICATIONS OF NEWTON’S LAWS – II

1. As a body moves through a body it displaces the fluid. it has to exert a force
on the fluid to push it out of the way. By Newton's third law, the fluid pushes
back on the body with an equal and opposite force. The direction of the fluid
resistance force on a body is always opposite to the direction of the body's velocity
relative to the fluid.
2. The magnitude of the fluid resistance force usually increases with the speed of the
body through the fluid. Typically, v (an empirical law!). Imagine that you
drop a ball bearing into a deep container filled with oil. Af
f =k
final
ter a while the ball bearing
will approach its maximum (terminal) speed when the forces of gravity and friction
balance each other: v from which v / .
3. The above was a simple
mg=k =mg k
example of equilibrium under two forces. In general, while
solving problems you should a)draw a diagram, b)define an origin for a system of
coordinates, c)identify all forces (tension, normal, friction, weight, etc) and their and
components, d)Apply Newton's law separately along the and axes. e) find the
accelerations, then velocities, then displacements. This sounds very coo
x
y xy
k-book, and in
fact it will occur to you naturally how to do this while solving actual problems.
4. Your weight in a lift: suppose you are in a lift that is at rest or moving at constant
velocity. In either case a=0 and the normal force N and the force due to gravity are
exactly equal, N−Mg=0⇒N=Mg.But if the lift is accelerating downwards then
Mg−N=Ma or N=M(g−a) . So now the normal force (i.e. the force with which the
floor of the lift is pushing on you) is decreased. Note that if the lift is accelerating
downwards with acceleration a (which it will if the cable breaks!) then N=0 and you
will experience weightlessness just like astronauts in space do. Finally, if the lift is
accelerating upwards then a is negative and you will feel heavier.
5. Imagine that you are in a railway wagon and want to know how much you are
accelerating. You are not able to look out of the windows. A mass is hung from the
roof. Find the acceleration of the car from the angle made by the mass.
a
F m M
θ
T
mg
We first balance the forces vertically: cos
And then horizontally: sin
From these two equations we find that: tan
Note that the mass doesn't
T mg
T ma
a
g
m
θ
θ
θ
=
=
=
matter - it cancels out!
6. Friction is a funny kind of force. It doe not make up its mind which way to act until
some other force compels it to decide. Imagine a block lying on the floor. If you push
it forward, friction will act backward. And if you push it to the left, friction will act
to the right. In other words, the direction of the frictional force is always in the
opposite direction to the applied force.
7. Let us solve the following problem: a rope of total length L and mass per unit length m
is put on a table with a length l hanging from one edge. What should be l such that the
rope just begins to slip?
To solve this, look at the balance of forces in the diagram below: in the vertical direction,
the normal force balances the weight of that part of the rope that lies on the table:
N=m(L−l)g. In the horizontal direction, the rope exerts a force to the right, which
is counteracted by the friction that acts to the left. Therefore . Substituting
from the first equation we find that
mlg
N mlg N
l L
μ
μ
μ
=
= . Note that if isvery small then even a small
1
piece of string that hangs over the edge will cause the entire string to slip down.
μ
+
9. In this problem, we would like to calculate the minimum force
such that the small block does not slip downwards. Clearly,
since the 2 bodies move together, ( ) . This gives
. W
( )
F
F m Ma
a F
m M
= +
=
+
e want the friction to be at least as large as the downwards force, .
So, we put from which the minimum horizontal force needed to
( )
prevent slippage is ( ) .
N mg
N ma m F
m M
F m M g
μ
μ
⎛ ⎞
= = ⎜⎝ + ⎟⎠
+
=
l
(L−l)
μN
N
m(L−l)g
ml g
0 Δ x x →
1 2 1 2
1
0 1
0
Now add up all the little pieces of work:
To get the exact result let 0 and the number of intervals N : lim
Definition: lim
N
N N n
n
x n n
x n
W W W W F x F x F x F x
x W F x
W F
=

Δ →
=
Δ →
= Δ + Δ + ⋅ ⋅ ⋅ + Δ = Δ + Δ + ⋅ ⋅ ⋅ + Δ ≡ Δ
Δ → → ∞ = Δ
=
Σ
Σ
1
( ) is called the integral of with respect to from
f
i
x
n x
x Fxdx F x

=
Σ Δ ≡ ∫

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