Chapter 1

Newton’s Laws of Motion

Theory, worked examples and 5 previous-year questions with full solutions — IIT JAM, GATE, JEST and CSIR NET-JRF, 2020 onwards.

Theory

1.1 Newton's three laws

Classical mechanics rests on three laws, stated by Newton in 1687, which connect the motion of a body with the forces acting on it. The laws are simple to state, but applying them correctly requires two skills that this chapter develops in detail: isolating each body and identifying every force on it (the free-body diagram), and expressing the geometrical restrictions on the motion (the equations of constraint). Bodies whose mass changes, and motion under velocity-dependent forces, are treated in Chapter 2; rotating frames of reference are treated in Chapter 3.

(a) The first law: inertia and inertial frames

Matter resists any change in its state of motion. This property is called inertia, and the mass of a body is its quantitative measure: the larger the mass, the harder it is to change the velocity of the body. The first law, also called the law of inertia, states that a body on which no net external force acts either remains at rest or continues to move with constant velocity.

The law does not hold in every frame of reference. An observer in an accelerating car sees a ball on the floor start to roll although no horizontal force acts on it. The first law therefore serves to define the frames in which mechanics takes its simplest form: an inertial frame is a frame of reference, with its clock, in which a free body moves with constant velocity. Any frame that moves with constant velocity relative to an inertial frame is itself inertial, whereas a frame that accelerates or rotates relative to it is not. For most laboratory purposes a frame fixed to the Earth is inertial to a good approximation.

In terms of the linear momentum \(\mathbf p=m\mathbf v\), the first law states that

\[\sum\mathbf F=0\quad\Longrightarrow\quad\frac{d\mathbf p}{dt}=0,\]

so that the momentum of the body is constant in both magnitude and direction.

(b) The second law

The second law states that, in an inertial frame, the net external force on a body equals the rate of change of its linear momentum:

\[\boxed{\mathbf F=\frac{d\mathbf p}{dt}=\frac{d(m\mathbf v)}{dt}}.\]

When the mass is constant this reduces to the familiar form

\[\boxed{\mathbf F=m\frac{d\mathbf v}{dt}=m\mathbf a},\]

in which \(\mathbf F\) is the vector sum of all the external forces. The force is the cause and the acceleration is its effect; the same force produces a smaller acceleration in a more massive body. The form \(\mathbf F=d\mathbf p/dt\) is the more general one, and it is the form required for systems whose mass changes (Chapter 2). At speeds comparable with the speed of light the law must be modified according to the special theory of relativity (Chapter 15).

(c) The third law

To every action there is an equal and opposite reaction. If body 1 exerts a force \(\mathbf F_{12}\) on body 2, then body 2 simultaneously exerts on body 1 a force \(\mathbf F_{21}\) that is equal in magnitude and opposite in direction:

\[\boxed{\mathbf F_{12}=-\mathbf F_{21}}.\]

The two forces of an action–reaction pair always act on different bodies. They therefore never appear together in the free-body diagram of a single body, and they never cancel one another in the equation of motion of that body.

(d) Impulse

When a force \(\mathbf F\) acts during a time interval \(\Delta t\), its impulse is

\[\mathbf I=\int\mathbf F\,dt.\]

The concept is most useful when a large force acts for a very short time (\(\Delta t\to0\)), as in a collision, where the details of \(\mathbf F(t)\) are unknown but its integral is not. Since \(\mathbf F=d\mathbf p/dt\), the impulse equals the change of momentum,

\[\boxed{\mathbf I=\Delta\mathbf p},\]

which, for a body of constant mass, is \(\mathbf I=m\,\Delta\mathbf v\). Collisions are treated in Chapter 4.

(e) The equations of motion in components

In practice the second law is used in component form. In two dimensions, for each body,

\[\sum F_x=ma_x,\qquad \sum F_y=ma_y.\]

The axes may be chosen freely, and a good choice, for example along and perpendicular to an inclined plane, simplifies the algebra considerably. The equations of motion of all the bodies, together with the equations of constraint (Section 1.5), must provide as many independent equations as there are unknowns, so that a complete and unique solution can be found.

1.2 Free-body diagrams

A free-body diagram (FBD) shows a single body, isolated from its surroundings, with every external force that acts on it. To draw one we treat the body as a point mass (or as a simple outline), mark each force as an arrow at the body, and choose a convenient pair of axes along which the forces are resolved. A separate diagram is drawn for each body of the system. The forces most often met are described below.

(a) Weight

The weight of a body of mass \(m\) near the Earth’s surface is \(w=mg\), and it always acts vertically downwards, whatever axes are chosen. On an inclined plane it is usually convenient to take the axes along and perpendicular to the plane, and then the weight has components along both axes (Figure 1.1).

Figure 1.1 The weight \(mg\) is vertically downwards, both for a block on a horizontal surface and for a block on an inclined plane with axes along and perpendicular to the plane.

(b) The normal force

When a body is in contact with a surface, the surface exerts on it a force perpendicular to the surface, called the normal force or normal reaction \(N\). If a block rests on a surface, the surface pushes the block perpendicular to itself with a force of magnitude \(N\), and by the third law the block pushes the surface with a force of the same magnitude in the opposite direction. The normal force can only push, never pull, so \(N\ge0\); a body leaves a surface at the instant at which the normal force required to keep it in contact falls to zero. Figure 1.2 shows the normal force exerted by a vertical wall, by an inclined plane and by a horizontal floor.

Figure 1.2 The normal force \(N\) is perpendicular to the surface of contact: a vertical wall, an inclined plane at angle \(\theta\), and a horizontal floor.

Example 1.1 A force \(F=20\;\text{N}\) acts on a block of mass \(m_1=5\;\text{kg}\), which pushes a block \(m_2=3\;\text{kg}\), which in turn pushes a block \(m_3=2\;\text{kg}\), on a smooth horizontal floor, as shown in the figure. Find (i) the acceleration of each block and (ii) the contact force between each pair of blocks.

Solution. (i) The blocks remain in contact, so all three move with the same acceleration \(a\). Treating them as one body of mass \(m_1+m_2+m_3=10\;\text{kg}\), on which the only horizontal external force is \(F\),

\[F=(m_1+m_2+m_3)a\quad\Longrightarrow\quad a=\frac{20}{10}=2\;\text{m s}^{-2}.\]

Let \(N_1\) be the contact force between \(m_1\) and \(m_2\), and \(N_2\) that between \(m_2\) and \(m_3\). The free-body diagrams are shown below; the vertical forces \(N_1'\), \(N_2'\), \(N_3'\) from the floor balance the weights.

The horizontal equations of motion are

\[F-N_1=m_1a,\qquad N_1-N_2=m_2a,\qquad N_2=m_3a.\]

The first gives \(N_1=20-10=10\;\text{N}\) and the last gives \(N_2=2\times2=4\;\text{N}\); the second is then satisfied, since \(10-4=6=m_2a\). The contact force decreases along the train, because each contact force has to accelerate only the blocks ahead of it.

(c) Tension

When a force \(F\) is exerted on a body through a string, every section of the string is pulled by two equal and opposite forces exerted by the neighbouring parts of the string. Either of these forces is called the tension \(T\) in the string at that point. Tension always pulls: at each end the string pulls the attached body towards itself (Figure 1.3, top row).

Consider two masses \(m_1\) and \(m_2\) joined by a string, with a force \(F\) applied to \(m_2\) (Figure 1.3, middle row). In the free-body diagram of \(m_1\) the tension acts towards \(m_2\), and in that of \(m_2\) it acts towards \(m_1\); in both cases it pulls the body towards the string. The tension is the same at every point of the string only if the string is massless (or if its acceleration is zero). For a string of mass per unit length \(\lambda\) accelerating with \(a\), the tension changes along it by \(\lambda a\) per unit length, since each element needs a net force to accelerate it. In this chapter all strings are taken to be massless and inextensible, and a massless frictionless pulley merely changes the direction of the tension without changing its magnitude.

Figure 1.3 Tension. Top: a string pulled by a force \(F\) pulls the block \(M\), and any section of the string is pulled by equal and opposite forces \(T\). Middle: two blocks connected by a string. Bottom: the free-body diagrams of \(m_1\) and \(m_2\).

Example 1.2 Blocks \(m_3\) and \(m_2\) rest on a smooth horizontal table and are connected by a massless string. Block \(m_2\) is connected by a second massless string, passing over a light frictionless pulley at the edge of the table, to a block \(m_1\) that hangs vertically under gravity. Find the acceleration of each mass and the tension in each string.

Solution. Let \(T\) be the tension in the string joining \(m_1\) and \(m_2\), and \(T_1\) that in the string joining \(m_2\) and \(m_3\). The weights of \(m_2\) and \(m_3\) are balanced by the normal forces of the table. Since both strings are inextensible, the three blocks move with the same magnitude of acceleration \(a\), with \(m_1\) moving down and the others moving towards the pulley. The free-body diagrams are

and the equations of motion are

\[m_1g-T=m_1a,\qquad T-T_1=m_2a,\qquad T_1=m_3a.\]

Adding the three equations eliminates the tensions and gives \(m_1g=(m_1+m_2+m_3)a\). Hence

\[a=\frac{m_1g}{m_1+m_2+m_3},\qquad T_1=\frac{m_1m_3g}{m_1+m_2+m_3},\qquad T=(m_2+m_3)a=\frac{m_1(m_2+m_3)g}{m_1+m_2+m_3}.\]

The tension \(T\) is less than the weight \(m_1g\), as it must be for \(m_1\) to accelerate downwards.

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Chapter 1 Newton’s Laws of Motion