Contents
Many modern technologies, such as computers and smartphones, are built on the principles of Physics Topics such as quantum mechanics and information theory.
What is the Equation of Motion Solution?
Analysis of Different Motions
Walking on a horizontal plane: AB is a horizontal plane [Fig.]. A man wants to walk on the plane towards B. He applies an oblique force F on the plane. The plane too, exerts a reaction force R on the man. Vertical component of R acts opposite to the weight of the man and the horizontal component H provides the force for the forward movement. So, the motion of the man is not directly due to the force F that he exerts, but due to the opposite reaction.
In fact, this force decides the type of the motion. As the frictional force is low on a smooth surface, it is not possible to exert a large force. If the horizontal component of F becomes much larger than the frictional force, the man may slip.
Hence one cannot walk fast on a very smooth surface as there is a high chance of slipping.
Flight of birds: If a bird intends to fly along OC, it flaps its wings along OA and OB, thus exerting some force on the air [Fig.]. At the same time, the reaction forces on the bird are OE and OD.
If both the wings of the bird exert equal force on air, the resultant reaction force will be along OC. The bird moves forward due to this reaction force. By adjusting the force applied by the wings, the resultant may be shifted to the left or right of the direction of OC, and the bird can thus change its course.
In the absence of air, i.e., in vacuum, the reaction forces OD and OE cannot be generated, and so birds cannot fly.
Motion of a hand-pulled rickshaw: A man is trying to pull a rickshaw on a horizontal plane, towards the point B. He exerts an oblique force P on the plane. The reaction R of the plane has a vertical component N and a horizontal component F. Force N is balanced by the weight of the man. Horizontal component F sets up the motion of the man [Fig.].
Let the force exerted by the man on the handles of the rickshaw be T. The rickshaw, also pulls the man backward with the same force T. Due to the motion of the wheel on the plane, frictional force F’ acts at the point of contact of the wheel and the plane in a direction opposing the motion.
Hence, resultant of forces acting on the man = F – T in the forward direction and that on the rickshaw in the same direction = T – F’
Let the masses of the man and the rickshaw be m1 and m2 respectively and common acceleration in the forward direction be a.
For the man,
F – T = m1a or, T = F – m1a
For the rickshaw,
T – F’ = m2a or, T F’ + m2a
∴ F – m1a = F’ + m2a or, (m1 + m2)a = F – F’
or, a = \(\frac{F-F^{\prime}}{\left(m_1+m_2\right)}\)
The expression for the acceleration a shows that,
- individual masses m1 or m2 are not significant but the sum of the masses (m1 + m2) matters,
- the rickshaw will start moving, i.e., accelerating when F > T > F’,
- during the early stages of motion, F > T > F’. It attains a uniform speed when F = T = F’ and then of course a = 0, and
- if F < F’, the rickshaw cannot be set into motion at all.
There are many similar examples in nature, where a force (action) is applied backwards, and the reaction force generated is actually responsible for the intended forward motion.
Equation of motion and its solution
i) The body, whose motion is to be analysed, should be identified at first. This ‘body’ may consist of a single body or a combination of bodies (often called ‘a system of bodies’) or even a part of a single body.
ii) Then we have to consider the surroundings (or environ-ment). Every other body having a direct influence on our chosen body is a part of the surroundings. In general, action-reaction-type forces act between the body and its surroundings.
iii) Now we search for every force acting on our body. Each of the forces is either a force of action on the body or a force of reaction on it due to some other body in the surroundings. Once the forces are identified, we regard them on the same footing, i.e., we no longer differentiate between actions and reactions. In this sense, the body under consideration acts as a free body, subject to a system of forces acting on it.
iv) If the active forces are \(\vec{F}_1\), \(\vec{F}_2\), \(\vec{F}_3\), ……., then the resultant or net force is F = \(\vec{F}_1\) + \(\vec{F}_2\) + \(\vec{F}_3\) + ………
v) For a body of constant mass m, we get from Newton’s 2nd law,
m\(\vec{a}\) = \(\vec{F}\) = \(\vec{F}_1\) + \(\vec{F}_2\) + \(\vec{F}_3\) + ………
This equation (1) is to be solved to find out the acceler-ation \(\vec{a}\) of the body.
As \(\vec{a}\) is related to the velocity \(\vec{v}\) as \(\vec{a}\) = \(\frac{d \vec{v}}{d t}\), equation (1) may be written as
m\(\frac{d \vec{v}}{d t}\) = \(\vec{F}\) …… (2)
Clearly, this is the differential equation to be solved if the velocity of the body is to be found out. Similarly, as \(\vec{v}\) = \(\frac{d \vec{r}}{d t}\)(\(\vec{r}\) = position vector of the body), we may write
m\(\frac{d^2 \vec{r}}{d t^2}\) = \(\vec{F}\) …….(3)
This 2nd order differential equation is to be solved to obtain \(\vec{r}\) at any instant.
Any of the equations (1) to (3) is then called the equation of motion of a body.
vi) It is to be noted that the equation of motion is a vector equation. Each vector has three independent components. So the solution of an equation of motion involves, in general, a solution of three independent equations corresponding to three mutually orthogonal coordinate axes. However, this is not always the case- we come across plenty of examples of one or two- dimensional motions where one or two equations, respectively, are relevant.
Horizontal Motion with the Help of a Smooth Pulley
A body A of mass m is resting on a frictionless horizontal plane. To set the body A into motion along the plane, it is tied, using a massless, inextensible string passing over a massless, frictionless pulley, with another body B of mass M as shown in Fig.
Both the bodies, A and B, get the same constant acceleration and the tension in the string remains constant (as the pulley is frictionless)
Let the tension in the string = T; acceleration produced = a
∴ For the motion of A, T = ma ……. (1)
For the downward motion of B, Mg – T = Ma …. (2)
From (1) and (2),
(M + m)a = Mg
or, a = \(\frac{M}{M+m} g\) ….. (3)
As \(\frac{M}{M+m}\) < 1 > a < g, though the surface and the pulley are frictionless.
Also, inserting the value of a in equation (1) we get,
T = \(\frac{M m}{M+m} g\) = \(\frac{M}{1+\frac{M}{m}^M} g\)
string is less than the weight of the body B.
Vertical Motion With the Help of a Smooth Pulley
To raise a body A of mass m vertically, a heavy body B of mass M is connected to A using a massless inextensible string passing over a massless smooth frictionless pulley as shown in Fig.
Let the tension developed in the string be T and the common acceleration of A and B be a. Mass of B is more than that of A. Hence B will move downwards, A will move upwards.
Considering the motion of B,
Mg – T = Ma …… (1)
and the motion of A,
T – mg = ma …… (2)
Adding equations (1) and (2),
(M – m)g = (M + m)a
or, a = \(\frac{M-m}{M+m} g\) ….. (3)
From (3), a < g though the pulley is smooth. Again, inserting the value of a in (1),
T = Mg – \(\frac{M(M-m)}{M+m} g\) = Mg(1 – \(\frac{M-m}{M+m}\)) = \(\frac{2 M m}{M+m} g\) ….. (4)
The value of a in equation (3) is positive (i.e., acceleration of A is upwards), if M > m, i.e., if the mass of B is more than that of A. As M > m, we get from equation (4), T > mg and T < Mg, i.e., the tension in the string is more than the weight of A (mg), but less than the weight of B. The net force on the pulley,
2T = \(\frac{4 M m}{M+m} g\) …… (5)
Numerical Examples
Example 1.
A block of mass m2 = 100 g is suspended from one end of an inextensible the string is tied to ml = 200 g, kept on a frictionless table surface. The passes over a [Fig.]. Find tension in the string and the acceleration of the blocks.
Solution:
Let the tension in the string = T, and acceleration of each block be a.
So, the equations of motion of the two blocks are,
T = m1a and m2g – T = m2a
On adding the two equations, we get,
m2g = m1a + m2a
or, a = \(\frac{m_2}{m_1+m_2} \cdot g\) = \(\frac{100}{200+100}\) × 980 = 326.7 cm ᐧ s-2
and T = m1a = 200 × 326.7 = 65340 dyn.
Example 2.
Two blocks of masses 2.9 kg and 1.9 kg are suspended from a rigid support, using two inextensible wires, each of length 1 m, as shown in Fig. The mass of the upper string is negligible and that of the lower string is 0.2 kg ᐧ m-1. If the whole system is moving up with an acceleration of 0.2 m ᐧ s-2, find
(i) the tension at the midpoint of the lower string and
(ii) the tension at the midpoint of the upper string. Acceleration due to gravity = 9.8 m ᐧ s-2.
Solution:
i) Mass of the lower string = 0.2 × 1 = 0.2 kg. Its mid point is B, and so the total mass hanging from B,
mB = (1.9 + \(\frac{0.2}{2}\)) = 2 kg.
Hence, downward force at B = mBg
Upward force at B = TB
So applying Newton’s 2nd law of motion for the mass below B,
mBa = TB – mBg
or, TB = mBa + mBg = mB(a + g)
= 2(0.2+ 9.8) = 20 N
ii) Midpoint of the upper string is A; total mass hanging below A,
mA = 2.9 + 0.2 + 1.9 = 5.0 kg.
Hence, downward force at A = mAg
Upward force at A = TA
Hence applying Newton’s 2nd law of motion for the whole system,
mAa = TA – mAg
or, TA = mA(g + a) = 5(9.8 + 0.2) = 50 N.
Example 3.
Two bodies of masses 7 kg and 5 kg are joined with a rope of mass 4 kg [Fig.]. An upward force of 200 N is applied on the upper body.
Find (i) the acceleration of the sys-tem,
(ii) tension at the top end of the rope and
(iii) tension at the midpoint of the rope.
Solution:
Let the acceleration be a.
i) From Newton’s 2nd law,
16a = 200 – 16g or, 16a = 200 – 16 × 9.8
∴ a = 2.7 m ᐧ s-2.
ii) Upper end of the rope carries the masses of the second body and the rope. If the tension is T, then
T – (4 + 5)g = (4 + 5)a
or, T = 9(g + a) = 9(9.8 + 2.7) = 112.5 N.
iii) Mass carried by the midpoint of the rope is half the mass of the rope and the whole mass of the lower body. Hence, if the tension is T’ at the midpoint, the equation of motion is
r’ – (2 + 5)g = (2 + 5)a
or, T = 7(g + a) = 7(9.8 + 2.7) = 87.5 N.
Example 4.
A double inclined plane A is placed on a horizontal table and two blocks of masses m1 and m2 are placed on the two frictionless inclined planes of A [Fig.].
Two ends of a rope wound over a pulley are tied to the two blocks. Find out the horizontal acceleration to be imparted to the system so that the blocks are at rest relative to A. In this condition, what will be the tension in the rope?
Solution:
Let the tension in the rope be T and the horizontal acceleration be a.
In Fig. the forces acting on the two blocks have been shown. They are at rest relative to A, if
T – m2gsinβ – m2acosβ = 0 ……. (1)
T – m1gsinα + m1cosα = 0 …….. (2)
Subtracting (2) from (1), we get,