Physics Newton's 3rd Law, Conservation of Momentum, EQUILIBRIUM OF A PARTICLE

Topics Covered

• Newton's Third Law of Motion
• Conservation of Momentum
• Equilibrium of Particle
• Common Forces in Mechanics
• Friction
• Rolling Friction

Newton's Third Law of Motion

`color{red}✍️ ` This idea was expressed by Newton in the form of the third law of motion.
`color{blue}"To every action, there is always an equal and opposite reaction."`

`color{red}✍️ ` A simple and clear way of stating the third law is as : `color(green)("Forces always occur in pairs.")` Force on a body `A` by `B` is equal and opposite to the force on the body `B` by `A.`

`color{red}✍️ ` The terms action and reaction in the third law may give a wrong impression that action comes before reaction i.e action is the cause and reaction the effect. There is no cause-effect relation implied in the third law. The force on `A` by `B` and the force on `B` by `A` act at the same instant.

`color{red}✍️ ` Action and reaction forces act on different bodies, not on the same body. Consider a pair of bodies `A` and `B.`
`color{blue}" According to the third law,"`

`color{red}{vecF_(AB) = – vecF_(BA)}`

`color{green}{"(force on A by B) = – (force on B by A)"}`
Q 1712080839

Two identical balls strike a rigid wall with the same
speed but at different angles and get reflected
without any change in speed as in figure. What is
the ratio of the magnitudes of impulses imparted to
the balls by the wall?


An instinctive answer to (i) might be that the force on the wall in case (a) is normal to the wall, while that in case (b) is inclined at 30° to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases.

How to find the force on the wall? The trick is to consider the force (or impulse) on the ball due to the wall using the second law, and then use the third law to answer (i). Let u be the speed of each ball before and after collision with the wall, and m the mass of each ball. Choose the x and y axes as shown in the figure, and consider the change in momentum of the ball in each case :

`ul("Case a")`

`color{orange}{(P_x)_(Initial) = m u, (P_y)_(Initial) = 0}`

`color{green} {(P_x)_text(final) = - m u, (P_y)_text(final) = 0}`

x component of impulse= - mu - mu = - 2mu

y component of impulse = 0 - 0 = 0

Impulse and force are in the same direction. Clearly, from above, the force on the ball due to the wall is normal to the wall, along the negative x-direction. Using Newton’s third law of motion, the force on the wall due to the ball is normal to the wall along the positive x-direction. The magnitude of force cannot be ascertained since the small time taken for the collision has not been specified in the problem.
`ul("Case (b)")`

`color{green} {(P_x)_(Initial) = m u cos 30^o , (P_y)_("initial") = - m u sin 30^o}`

`color{orange} {(P_x)_text(final) = - m u cos 30^o, (P_y)_(final) = - mu sin 30^0}`

Note, while px changes sign after collision, `p_y` does not. Therefore,

x-component of impulse `= –2 m u cos 30°`
y-component of impulse = 0

The direction of impulse (and force) is the same as in (a) and is normal to the wall along the negative x direction. As before, using Newton’s third law, the force on the wall due to the ball is normal to the wall along the positive x direction.

The ratio of the magnitudes of the impulses imparted to the balls in (a) and (b) is

`(2mu//(2 mu cos 30^0) = 2 /sqrt 3 ~~ 1.2`

Conservation of Momentum

`color{red}✍️ ` The total momentum of an isolated system of interacting particles is conserved.

`\color{fuchsia} {★ \mathbf\ul"A bullet is fired from a gun"}`

Force on the bullet by the gun `=color{blue}{vecF}`

Force on the gun by the bullet `=color{blue}{ –vecF}`

According to the third law. The two forces act for a common interval of time Δt.

According to the second law,

Change in momentum of the bullet `=color{blue}{vecF Δt}`

Change in momentum of the gun `=color{blue}{– vecF Δt}`

`color{green}("Since initially, both are at rest, the change in momentum equals the final momentum for each.")`

Thus if `color{blue}{vecp_b}=` momentum of the bullet after firing

`color{blue}{vecp_g}=` recoil momentum of the gun

`vecp_g = – vecp_b`

i.e. `color{blue}{p_b + p_g = 0}`.

`color{green}{"That is, the total momentum of the (bullet + gun) system is conserved."}`

`color{red}✍️ ` An `color(red)("important example")` of the application of the law of conservation of momentum is the

`\color{fuchsia} {★ \mathbf\ul"Collision of two bodies"}`

Consider two bodies `A` and `B`, with initial momenta `color{blue}{vecp_A}` and `color{blue}{vecp_B}`.

The bodies collide, get apart, with final momenta `color{blue}{vec(p_A^')}` and `color{blue}{vec(p_B^')}` respectively.

`"By the Second Law,"`

`vecF_(AB)Δt = vec(p_A^') − vec(p_A)` and

`vecF_(BA)Δt = vec(p_B^') − vec(p_B)`

Since `color{red}{vecF_(AB) = −vecF_(BA)}` by the third law,

`vec(p_A^') − vec(p_A) = −(vec(p_B^') − vec(p_B) )`

i.e. `color{blue}{vec(p_A^') + vec(p_B^') = vec(p_A) + vec(p_B)}` which shows that

`color{green}{"the total final momentum of the isolated system equals its initial momentum."}`

Equilibrium of a Particle

`color{red}✍️ ` Equilibrium of a particle in mechanics refers to the situation
`color(blue)("when the net external force on the particle is zero.")`

`color{red}✍️ ` Equilibrium of a body requires `color(blue)("not only translational equilibrium")` (zero net external force) but `color(blue)("also rotational equilibrium")` (zero net external torque).

`color{red}✍️ ` According to the first law, this means that, the particle is either at rest or in uniform motion.

`color{red}✍️ ` If two forces `color{blue}{vecF_1}` and `color{blue}{vecF_2}`, act on a particle, `color{blue}{"equilibrium requires,"}`

`color(blue)(vecF_1 = − vecF_2)`

i.e. the two forces on the particle must be equal and opposite.

`color{red}✍️ ` Equilibrium under three concurrent forces `vecF_1, vecF_2` and `vecF_3` requires that the vector sum of the three forces is zero.

`color(blue)(vecF_1 + vecF_2 + vecF_3 = 0)` or

`F_(1x) + F_(2x) + F_(3x) = 0`

`F_(1y) + F_(2y) + F_(3y) = 0`

`F_(1z) + F_(2z) + F_(3z) = 0`

where `F_(1x), F_(1y)` and `F_(1z)` are the components of `F_1` along `x, y` and `z` directions respectively.

In other words, the resultant of any two forces say `vecF_1` and `vecF_2`, obtained by the parallelogram law of forces must be equal and opposite to the third force, `vecF_3`.

`color{red}✍️ ` A particle is in equilibrium under the action of forces `vecF_1, vecF_2,... vecF_n` if they can be represented by the sides of a closed n-sided polygon with arrows directed in the same sense.
Q 1773412346

Figure given below, shows a mass of 6 kg
suspended by a rope of length 2 m from the ceiling.
A force of 50 N in the horizontal direction is
applied at the mid-point P of the rope. What is the
angle the rope makes with the vertical in
equilibrium `(g = 10 m//s^2 )`? Neglect the mass of
the rope.


Figures 5.8(b) and 5.8(c) are known as free-body diagrams. Figure 5.8(b) is the free-body diagram of W and Fig. 5.8(c) is the free-body diagram of point P.

Free-body diagrams are diagrams used to show the relative magnitude and direction of all forces acting upon an object in a given situation.

Consider the equilibrium of the weight W. Clearly, `T_2 = 6 × 10 = 60 N`.

Consider the equilibrium of the point P under the action of three forces - the tensions `T_1` and `T_2`, and the horizontal force 50 N. The horizontal and vertical components of the resultant force must vanish separately :

`T_1 cos theta = T_2 = 60 N`
`T_1 sin theta = 50 N` which gives that

`tan theta =5/6 ` or ` theta = tan^-1 5/6 = 40^0`

Common Forces in Mechanics

`color{red}✍️ ` In mechanics, we encounter several kinds of forces. The gravitational force is, of course, all pervasive.

`color{red}✍️ ` Every object on the earth experiences the force of gravity due to the earth. Gravity also governs the motion of celestial bodies. The gravitational force can act at a distance without the need of any intervening medium.

`\color{fuchsia} {★ \mathbf\ul"Contact Forces"}`

`color{red}✍️ ` A contact force on an object arises due to contact with some other object: solid or fluid.

`color{red}✍️ ` When bodies are in contact (e.g. a book resting on a table, a system of rigid bodies connected by rods, hinges and other types of supports), there are mutual contact forces (for each pair of bodies) satisfying the third law.

`color{red}✍️ ` The component of contact force normal to the surfaces in contact is called normal reaction.

`color{red}✍️ ` The component parallel to the surfaces in contact is called friction.

`color{red}✍️ ` Contact forces arise also when solids are in contact with fluids. For example, for a solid immersed in a fluid, there is an upward buoyant force equal to the weight of the fluid displaced. The viscous force, air resistance, etc are also examples of contact forces (Fig. 5.9).

`\color{fuchsia} {★ \mathbf\ul"Tension Force"}`

`color{red}✍️ ` The tension force is the force that is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends.

`color{red}✍️ ` The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.

`color{red}✍️ ` For an inextensible string, the force constant is very high.

`\color{fuchsia} {★ \mathbf\ul"Spring Force"}`

`color{red}✍️ ` When a spring is compressed or extended by an external force, a restoring force is generated.

`color{red}✍️ ` This force is usually proportional to the compression or elongation (for small displacements). The spring force `F` is written as `F = – k x` where `x` is the displacement and `k` is the force constant.

`color{red}✍️ ` The negative sign denotes that the force is opposite to the displacement from the unstretched state.

Static Friction

`color{red}✍️ ` When there is no relative motion between the surfaces in contact, friction is `color(blue)("known as static friction.")`

`color{red}✍️ ` Static friction is an adjustable force which may have any value from zero to certain maximum value, under a given case. The maximum possible value of this static friction is known as limiting static friction.

• `color(red)(0<= f_s <= "Limiting static friction")`

• `color(red)(f_s = mu_s N)`

where `color(blue)(mu_s)` is a constant known as `color(red)("coefficient of static friction")` and `color(blue)(N)` is `color(red)("normal reaction.")`
Q 3220467311

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.


Since the acceleration of the box is due to the static friction,

`ma = fs <= μ_s N = μ_s m g`

i.e. `a <= μ_s g`

`:. a_(max) = μ_s g = 0.15 x 10 m s^-2`

`= 1.5 m s^-2`
Q 3230467312

A mass of 4 kg rests on a horizontal plane. The plane is gradually inclined until at an angle `theta = 15°` with the horizontal, the mass just begins to slide. What is the coefficient of static friction between the block and the surface ?


The forces acting on a block of mass m at rest on an inclined plane are
(i) the weight mg acting vertically downwards
(ii) the normal force N of the plane on the block, and
(iii) the static frictional force fs opposing the impending motion. In equilibrium, the resultant of these forces must be zero. Resolving the weight mg along the two directions shown, we have

`color{pink}{m g sin theta = f_s , m g cos theta = N}`

As q increases, the self-adjusting frictional force `f_s` increases until at `theta = theta_(max)`, achieves its maximum value, `(f_s ) _(max) f = μ_s N.`

`tan theta_(max) = μ_s` or ` theta_(max) = tan^-1 μ_s`

When q becomes just a little more than qmax , there is a small net force on the block and it begins to slide. Note that `theta_(max)` depends only on `μ_s` and is independent of the mass of the block.

For `theta_(max) = 15°`,

`μ_s = tan 15°`

= 0.27
Q 3240467313

What is the acceleration of the block and trolley system shown in a Fig. 5.12(a), 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.


As the string is inextensible, and the pully is smooth, the 3 kg block and the 20 kg trolley both have same magnitude of acceleration. Applying second law to motion of the block (Fig. 5.12(b)),

`30 – T = 3a`

Apply the second law to motion of the trolley (Fig. 5.12(c)),

`T – f_k = 20 a`.

Now `f_k = μ_k N`,

Here `μ_k = 0.04`,

`N = 20 x 10`

= 200 N.

Thus the equation for the motion of the trolley is

`T – 0.04 x 200 = 20 a` Or `T – 8 = 20a`.

These equations give `a = 22/23` `m s^-2 = 0.96 m s^-2`
and `T = 27.1 N`.

Kinetic Friction

`color{red}✍️ ` Whenever there is relative motion between the surfaces in contact, friction is known as kinetic friction (`f_k`) which is a constant force such that

`color(blue)(f_k =mu_k N)`

where `color(blue)(mu_k)` is known as `color(green)("the coefficient of kinetic friction between the surfaces")`

Rolling Friction

`color{red}✍️ ` 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.

`color{red}✍️ ` In this ideal situation, `color(green)("kinetic or static friction is zero")` and the body should continue to roll with constant velocity.

`color{red}✍️ ` 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 than static or sliding friction.