We shall require a substantially new manner of thinking if mankind is to survive.
– Albert Einstein
Even though most of you will be remembering the important points in connection with Newton’s Laws of Motion, it will be better to have a glance at the following:
(1) Inertia is a basic property of any material body, by virtue of which it resists any change in its state of rest or of uniform motion.
(2) A force is required to change the state of rest or of uniform motion of a body. The resultant force acting on a body at rest or in uniform motion is zero. [Note that a body in uniform motion has uniform velocity].
(3) Newton’s second law is mathematically expressed as
Fnet = ma where Fnet is the net (resultant) force and a is the acceleration.
This can be written in terms of momentum p as
Fnet = dp/dt
Often we write this as F= dp/dt, understanding that F is indeed the net force.
[Remember that the mass m can be treated as constant only at speeds negligible compared to the speed of light].
(4) Impulse given to an object (by a force) = F∆t where F is the force and ∆t is the time for which the force acts.
If the force is not constant and it acts from the instant t1 to the instant t2, we have
This gives the area under the force-time graph between the ordinates corresponding to the times t1 and t2 (Shaded area in fig.)
Since force F = ∆p/∆t where ∆p is the change in momentum during the time ∆t, we can write
Impulse = F∆t = (∆p/∆t) ∆t = ∆p
Thus impulse = change of momentum
(5) Motion in a lift
The weight of a body of mass ‘m’ in a lift can be remembered as m(g-a) in all situations if you apply the proper sign to the acceleration ‘a’ of the lift. The acceleration due to gravity ‘g’ always acts vertically downwards and its sign may be taken as positive. The following cases can arise in this context:
(i) Lift moving down with acceleration of magnitude ‘a’:
In this case ‘a’ also is positive and the weight is m(g-a) which is less than the real weight of the body (when it is at rest).
(ii) Lift moving up with acceleration:
In this case ‘a’ is negative and the weight is m[g-(-a)] = m(g+a).
(iii) Lift moving down with retardation (going to stop while moving down):
In this case also ‘a’ is negative and the weight is m[g-(-a)] =m(g+a) which is greater than the actual weight.
(iv) Lift moving up with retardation (going to stop while moving up):
In this case ‘a’ is positive and the weight is m(g-a)
(v) Lift moving up or down with uniform velocity:
In this case ‘a’ is zero and the weight is mg.
(vi) Lift moving down with acceleration of magnitude ‘g’ (falling freely under gravity as is the case when the rope carrying the lift breaks):
In this case ‘a’ is positive and the weight is m(g-g) which is zero.
If you have a clear idea of the weight of a body in a lift, you will be able to use it in other similar situations as well (for instance, the motion of bodies connected by a string passing over a pulley).
[We will discuss conservation of momentum separately in due course].
The force of friction, Ffric ≤ μN where μ is the coefficient of friction and N is the normal reaction (normal force).
In the adjoining figure, a body being pulled along a horizontal surface by a horizontal force F is shown. The frictional force Ffric is maximum when the body just begins to move and is called limiting force of static friction (Fs)max so that we have
(Fs)max = μs N. This gives the value of the coefficient of static friction μs as
μs = (Fs)max /N
When the body slides along the surface, the friction called into play is called kinetic friction. The force of kinetic friction Fk is less than the above limiting value (Fs)max and the corresponding coefficient of kinetic friction μk is less than μs. We have
μk = Fk/N
If the body rolls along the surface, The friction called into play is called rolling friction which is much less than kinetic friction.
Angle of friction λ is the angle between the the normal force N and the resultant reaction S. As shown in the figure, the resultant reaction is the resultant of the normal force N and the frictional force Ffric. Since tan λ = Ffric/N, it follows that
μ = tan λ
A body of mass m placed on a ramp (inclined plane) is shown in the adjoining figure. The component mg sinθ of the weight mg of the body is the force trying to move the body down the plane. The normal reaction is the reaction (force) opposing the component mg cosθ of the component of the weight of the body normal to the inclined plane. The frictional force Ffric is opposite to the component mg sinθ (of the weight of the body) parallel to the plane. Note that friction is a self adjusting force up to its maximum value (Fs)max and if the body shown in the figure is at rest, Ffric is just sufficient to balance the component mg sinθ (of the weight of the body).
If the inclination of the plane is gradually increased from a small value, the body placed on it will begin to slide down when the angle is equal to the angle of friction, λ. The angle of repose is therefore equal to the angle of friction.
Often you may be asked to draw a free body diagram (FBD), indicating the forces acting on the body. In the case of the body placed on the inclined plane, the free body diagram is as shown. The body is represented by a dot. The forces to be shown are the weight mg of the body, the normal force N (equal to mg cosθ) exerted by the inclined surface on the body and the frictional force Ffric since they are the actual forces acting on the body. Don’t worry about the components mg sinθ and mg cosθ of the weight. The real force is the weight mg which we have shown already. We consider the components just for the convenience of explanation. The normal reaction (force) offered by the surface and the frictional force between the body and the surface are to be accommodated in addition to the weight of the body.
[Note that if you want, you can draw the FBD showing the components mg sinθ and mg cosθ of the weight of the body. But in that case you will not show the weight mg in the FBD].
If the inclined plane is smooth, the frictional force Ffric will be absent in the free body diagram.
If the body is held on a smooth incline by a spring fixed to the incline, the spring force Kx has to be shown in place of the frictional force Ffric. Here K is the spring constant and x is the elongation (or compression as the case may be) of the spring.
If the body moves down the incline and the viscous drag force (air resistance) is significant, that too is to be shown up the incline.
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