Impulse:
We sometimes encounter examples where a large force acts for a very short duration producing a finite change in momentum of the body.
The product of this force and time is called impulse.
Impulse = Force × time duration = Change in momentum
The force is called an impulsive force.
Example 5.4
A batsman hits back a ball straight in the direction of the bowler without changing its initial speed of 12 m s–1. If the mass of the ball is 0.15 kg, determine the impulse imparted to the ball. (Assume linear motion of the ball)
Answer:
Change in momentum = 0.15 × 12–(–0.15×12) = 3.6 N s,
Impulse = 3.6 N s, in the direction from the batsman to the bowler.
Equilibrium of a particle:
Equilibrium of a particle is the situation when the net external force on the particle is zero.
Equilibrium under three concurrent forces F1, F2 and F3 requires that the vector sum of the three forces is zero. F1 + F2 + F3 = 0
Example 5.6
A mass of 6 kg is suspended by a rope of length 2 m from the ceiling. A force of 50 N in the
horizontal direction is applied at the midpoint P of the rope, as shown. What is the angle the rope makes with the vertical in equilibrium ? (Take g = 10 m s-2). Neglect the mass of the rope.
Friction:
Let us consider a body of mass m at rest on a horizontal table. The force of gravity (mg) is cancelled by the normal reaction force (N) of the table. Now suppose a force F is applied horizontally to the body. We know from experience that a small applied force may not be enough to move the body. The body remains at rest because some other force comes into play in the horizontal direction and opposes the applied force F, resulting in zero net force on the body. This force fs parallel to the surface of the body in contact with the table, is known as frictional force, or simply friction.
Static Friction:
In this case a force is applied but the body does not move.
When there is no applied force, there is no static friction. It comes into play the moment there is an applied force. As the applied force F increases, fs also increases, remaining equal and opposite to the applied force (up to a certain limit), keeping the body at rest. Hence, it is called static friction. Due to this property it is called self balancing.
Static friction opposes impending motion. The term impending motion means motion that would take place (but does not actually take place) under the applied force, if friction were absent.
Laws of static friction:
It is found experimentally that the limiting value of static friction (f)s max is independent of
the area of contact and varies with the normal force(N) approximately as :
(f)s max = μsN
where μs is a constant of proportionality depending only on the nature of the surfaces in contact. The constant μs is called the coefficient of static friction.
The law of static friction may thus be written as
fs ≤ μsN
Laws of Kinetic friction:
If the applied force F exceeds (f)s max then the body begins to slide on the surface. It is found
experimentally that when relative motion has started, the frictional force decreases from the
static maximum value (f)s max.
Kinetic Friction:
Frictional force that opposes relative motion between surfaces in contact is called kinetic or sliding friction and is denoted by fk .
Kinetic friction, like static friction, is found to be independent of the area of contact. Further, it is nearly independent of the velocity. It satisfies a law similar to that for static friction:
fk = μkN
where μk is called the coefficient of kinetic friction, depends only on the surfaces in contact.
As mentioned above, experiments show that μk is less than μs.
Note:
When relative motion has begun, the acceleration of the body according to the Second Law is
a = ( F – fk )/m.
For a body moving with constant velocity, F = fk. If the applied force on the body is removed, its acceleration is a = – fk /m and it eventually comes to a stop.
Example 5.7
Determine the maximum acceleration of the train in which a box lying on its floor will remain stationary, given that the co-efficient of static friction between the box and the train’s floor is 0.15.
Example 5.8
A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle θ = 15° with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?
Example 5.9
What is the acceleration of the block and trolley system shown in a Fig. 5.12, if the coefficient of kinetic friction between the trolley and the surface is 0.04?
What is the tension in the string?
(Take g = 10 m s-2). Neglect the mass of the string.
Rolling friction:
A body like a ring or a sphere rolling without slipping over a horizontal plane will suffer no friction, in principle. At every instant, there is just one point of contact between the body and the plane and this point has no motion relative to the plane. In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity. We know, in practice, this will not happen and some resistance to motion (rolling friction) does occur, i.e. to keep the body rolling, some applied force is needed. For the same weight, rolling friction is much smaller (even by 2 or 3 orders of magnitude) than static or sliding friction.
Question: For the motion of a car on a level road, find maximum speed for turning without slipping.
Answer:
Three forces act on the car.
(i) The weight of the car, mg
(ii) Normal reaction, N
(iii) Frictional force, f
As there is no acceleration in the vertical
direction
N–mg = 0
N = mg
The centripetal force required for circular motion is along the surface of the road, and is provided by the static friction. Static friction opposes the
impending motion of the car moving away from the circle.
mv2/R = fs
But fs ≤ μsN
mv2/R ≤ μsN
But N = mg
mv2/R ≤ μsmg
v2 ≤ μsRg
which is independent of the mass of the car.
Question: What is the need of banking of roads at turns? For the motion of a car on a banked circular road, find an expression for maximum velocity for the safe turn.
Answer:
Since there is no acceleration along the vertical direction, the net force along this direction must be zero. Hence,
N cos θ = mg + f sin θ equation 1
The centripetal force is provided by the horizontal components of N and fs.
N.sinθ + f.cosθ = mv2/R equation 2
But f ≤ μsN
Thus to obtain v(max) we put
f = μsN
Then Equations become
N cos θ = mg + μsN sin θ
Or N.cosθ - μsN.sinθ = mg equation 3
N.sinθ + μsN.cosθ = mv2/R equation 4
Divide equation 4 by equation 3
[N sin θ + μsN cos θ]/[ N cos θ - μsN sin θ] = mv2/Rmg
Dividing numerator and denominator of LHS by cosθ, we obtain
[μs + tanθ]/[ 1 - μs tanθ ]= v2/Rg
v2 = Rg [ μs + tanθ]/[ 1 - μs tanθ ]
the need of banking of roads at turns: We can reduce the contribution of friction to the circular motion of the car if the road is banked. Comparing this we see that maximum possible speed of a car on a banked road is greater than that on a flat road. For μs = 0
vo = ( R g tan θ ) ½
At this speed, frictional force is not needed at all to provide the necessary centripetal force. Driving at this speed on a banked road will cause little wear and tear of the tyres.
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