Applications of Laws of Motion

Equilibrium Of A Particle

In physics, a particle is an object whose size and shape are irrelevant for the situation being analysed. Often, real-world objects are treated as particles when their rotational motion or deformation is not significant for the problem. Understanding the conditions under which a particle remains in equilibrium is a fundamental application of Newton's Laws.

A particle is said to be in equilibrium if its velocity is constant. This means that the particle is either at rest ($ \vec{v} = 0 $) or moving with a constant speed in a straight line ($ \vec{v} = \text{constant} \ne 0 $ and $ \vec{a} = 0 $).

Condition for Equilibrium

According to Newton's First Law, an object maintains a constant velocity if and only if the net external force acting on it is zero. Newton's Second Law confirms this: $ \vec{F}_{net} = m\vec{a} $. If the velocity is constant, the acceleration $ \vec{a} $ is zero ($ \vec{a} = \frac{d\vec{v}}{dt} = 0 $). Therefore, the net force must be zero.

The condition for a particle to be in equilibrium is that the vector sum of all the external forces acting on it must be zero.

$ \sum \vec{F}_{ext} = 0 $

If the forces are $ \vec{F}_1, \vec{F}_2, \vec{F}_3, \ldots, \vec{F}_n $, then for equilibrium:

$ \vec{F}_1 + \vec{F}_2 + \ldots + \vec{F}_n = 0 $

Resolution of Forces

Since force is a vector, the vector sum being zero means that the sum of the components of the forces along any chosen set of perpendicular axes must also be zero. If we choose a Cartesian coordinate system (x, y, z), the condition $ \sum \vec{F}_{ext} = 0 $ is equivalent to:

$ \sum F_x = 0 $ $ \sum F_y = 0 $ $ \sum F_z = 0 $

where $ F_x, F_y, F_z $ are the components of the individual forces along the x, y, and z axes, respectively.

Types of Equilibrium

Equilibrium can be categorised into two types:

1. Static Equilibrium: The particle is at rest ($ \vec{v} = 0 $). Since the velocity is constant (zero), the acceleration is zero, and the net force is zero. An example is a book lying on a table.

2. Dynamic Equilibrium: The particle is moving with a constant, non-zero velocity ($ \vec{v} = \text{constant} \ne 0 $). The acceleration is zero, and thus the net force is zero. An example is a satellite moving at a constant speed in a circular orbit under the influence of gravity (gravity provides the necessary force for the *change in direction*, resulting in circular motion, but if considered in an inertial frame, the forces balancing the centripetal force from other reference frames would lead to zero net force in that context, or more correctly, if velocity is constant, net force is zero. Note: uniform circular motion *is* accelerated motion. A particle in uniform circular motion is NOT in equilibrium because its velocity direction changes. Dynamic equilibrium refers to constant velocity *in a straight line*). Let's refine this - in common usage based on $\sum \vec{F} = 0 \implies \vec{a} = 0$, equilibrium means $\vec{v}$ is constant in magnitude and direction. A particle in uniform circular motion has constant speed but changing velocity, hence it is accelerating, and the net force (centripetal force) is NOT zero. So, dynamic equilibrium should strictly mean constant velocity linear motion.

Let's re-state the types based on the strict definition derived from $ \sum \vec{F}_{ext} = 0 \implies \vec{a} = 0 \implies \vec{v} = \text{constant}$.

1. Static Equilibrium: The particle is at rest ($ \vec{v} = 0 $). Since $ \vec{v} $ is constant (zero), $ \vec{a} = 0 $, and thus $ \sum \vec{F}_{ext} = 0 $.

2. Dynamic Equilibrium: The particle is moving with a constant, non-zero velocity in a straight line ($ \vec{v} = \text{constant} \ne 0 $). Since $ \vec{v} $ is constant in both magnitude and direction, $ \vec{a} = 0 $, and thus $ \sum \vec{F}_{ext} = 0 $.

Examples of Equilibrium:

Answer:

The traffic light is in static equilibrium, meaning its velocity is zero and remains zero. Therefore, the net force acting on it must be zero.

Forces acting on the traffic light (considered as a particle):

1. Weight ($ \vec{W} $): 200 N downwards.

2. Tension in cable 1 ($ \vec{T}_1 $): Upwards and to the left, making $ 30^\circ $ with the horizontal.

3. Tension in cable 2 ($ \vec{T}_2 $): Upwards and to the right, making $ 60^\circ $ with the horizontal.

We choose a coordinate system with the x-axis horizontal and the y-axis vertical. The origin is at the traffic light.

Resolving the forces into components:

Weight: $ \vec{W} = (0 \, \hat{i} - 200 \, \hat{j}) $ N.

Tension $ \vec{T}_1 $: The angle with the positive x-axis is $ 180^\circ - 30^\circ = 150^\circ $ or $ 90^\circ + 60^\circ $. Using the angle $ 30^\circ $ with the horizontal, the angle with the vertical is $ 90^\circ - 30^\circ = 60^\circ $. Horizontal component: $ -T_1 \cos(30^\circ) $ Vertical component: $ +T_1 \sin(30^\circ) $

Tension $ \vec{T}_2 $: The angle with the positive x-axis is $ 60^\circ $. The angle with the vertical is $ 90^\circ - 60^\circ = 30^\circ $. Horizontal component: $ +T_2 \cos(60^\circ) $ Vertical component: $ +T_2 \sin(60^\circ) $

Condition for equilibrium ($ \sum \vec{F} = 0 $), resolving into components ($ \sum F_x = 0 $ and $ \sum F_y = 0 $):

Sum of x-components = 0:

$ -T_1 \cos(30^\circ) + T_2 \cos(60^\circ) = 0 $

$ -T_1 \left(\frac{\sqrt{3}}{2}\right) + T_2 \left(\frac{1}{2}\right) = 0 $

$ \sqrt{3} T_1 = T_2 \quad \text{(Equation 1)} $

Sum of y-components = 0:

$ T_1 \sin(30^\circ) + T_2 \sin(60^\circ) - 200 = 0 $

$ T_1 \left(\frac{1}{2}\right) + T_2 \left(\frac{\sqrt{3}}{2}\right) - 200 = 0 $

$ \frac{1}{2} T_1 + \frac{\sqrt{3}}{2} T_2 = 200 $

$ T_1 + \sqrt{3} T_2 = 400 \quad \text{(Equation 2)} $

Now we solve the system of equations. Substitute $ T_2 = \sqrt{3} T_1 $ from Equation 1 into Equation 2:

$ T_1 + \sqrt{3} (\sqrt{3} T_1) = 400 $

$ T_1 + 3 T_1 = 400 $

$ 4 T_1 = 400 $

$ T_1 = 100 \, \text{N} $

Now find $ T_2 $ using Equation 1:

$ T_2 = \sqrt{3} T_1 = \sqrt{3} \times 100 \approx 1.732 \times 100 = 173.2 \, \text{N} $

The tension in the cable making $ 30^\circ $ with the horizontal is 100 N, and the tension in the cable making $ 60^\circ $ with the horizontal is approximately 173.2 N.

Problems involving equilibrium often require drawing a Free Body Diagram, resolving forces into components, and applying the conditions $ \sum F_x = 0 $, $ \sum F_y = 0 $, and $ \sum F_z = 0 $.

Common Forces In Mechanics

To apply Newton's Laws effectively, it is important to be able to identify and represent the various types of forces that commonly occur in mechanics problems. These forces can be broadly categorised as contact forces (arising from objects touching each other) and non-contact forces (action-at-a-distance forces).

Common Force Types:

• Gravitational Force (Weight, $ \vec{W} $ or $ \vec{F}_g $): The attractive force between any two objects with mass. Near the Earth's surface, the gravitational force on an object is its weight, $ \vec{W} = m\vec{g} $, directed towards the centre of the Earth (downwards).
• Normal Force ($ \vec{N} $): A contact force exerted by a surface on an object. It is always perpendicular (normal) to the surface and pushes away from the surface, preventing the object from passing through it. Its magnitude adjusts as needed to prevent penetration, up to a limit.
• Tension Force ($ \vec{T} $): A force exerted by a stretched string, rope, cable, or wire. It always pulls along the direction of the string. Assuming the string is massless and inextensible, the tension is uniform throughout the string.
• Spring Force ($ \vec{F}_s $): A force exerted by a compressed or stretched spring. According to Hooke's Law (for ideal springs within elastic limits), the force exerted by the spring is proportional to the displacement from its equilibrium length and acts in the direction opposing the displacement ($ \vec{F}_s = -k\vec{x} $), where $ k $ is the spring constant and $ \vec{x} $ is the displacement vector.
• Friction Force ($ \vec{f} $): A contact force that opposes relative motion or the tendency of relative motion between two surfaces in contact. It acts parallel to the surfaces.
• Air Resistance (Drag Force, $ \vec{F}_D $): A force exerted by air (or any fluid) on an object moving through it. It opposes the motion and generally increases with the speed of the object.
• Applied Force ($ \vec{F}_{app} $): A general term for any force applied to an object by another object through physical contact or a mechanism, like pushing or pulling.

Friction

Friction is a fundamental force that plays a crucial role in everyday life, from walking to driving to operating machinery. It arises from the interaction between the surfaces of objects in contact.

Nature of Friction:

Even seemingly smooth surfaces have microscopic irregularities (peaks and valleys). When two surfaces are in contact, these irregularities interlock, and also, attractive intermolecular forces (electromagnetic in origin) act between the molecules of the surfaces at the points of contact. These interactions give rise to the force of friction, which opposes relative motion between the surfaces.

Types of Friction:

Friction is broadly classified into two types:

1. Static Friction ($ \vec{f}_s $): This force acts when there is no relative motion between the surfaces in contact, but there is a tendency for motion. Static friction opposes the impending motion. Its magnitude is not constant; it varies from zero up to a maximum value. As the applied force trying to initiate motion increases, the static friction force also increases, remaining equal in magnitude and opposite in direction to the applied force, thus keeping the object at rest. The maximum possible static friction force is given by:

$ f_{s, max} = \mu_s N $

where $ \mu_s $ is the coefficient of static friction (a dimensionless constant depending on the nature of the two surfaces in contact) and $ N $ is the magnitude of the normal force between the surfaces.

So, the magnitude of static friction $ f_s $ is $ 0 \le f_s \le \mu_s N $.

2. Kinetic Friction ($ \vec{f}_k $): This force acts when there is relative motion between the surfaces in contact. Kinetic friction opposes the actual motion. Unlike static friction, the magnitude of kinetic friction is generally considered to be constant, independent of the relative speed (for typical speeds) and the area of contact (if the normal force is constant). It is given by:

$ f_k = \mu_k N $

where $ \mu_k $ is the coefficient of kinetic friction (also a dimensionless constant depending on the surfaces) and $ N $ is the magnitude of the normal force.

Generally, $ \mu_k < \mu_s $. This means that once an object starts moving, the frictional force opposing its motion is often slightly less than the maximum static friction that had to be overcome to start the motion.

Properties of Friction:

• Friction force is always parallel to the surfaces in contact.
• Friction force always opposes the relative motion or the tendency of relative motion.
• The magnitude of kinetic friction is proportional to the normal force.
• The magnitude of maximum static friction is proportional to the normal force.
• The coefficient of friction depends only on the nature of the surfaces in contact, not on their apparent area of contact.

Examples involving Friction:

Answer:

Friction is a dissipative force, meaning it converts mechanical energy into heat, often felt as surfaces getting warmer when rubbed together.

Circular Motion

Circular motion is a type of motion where an object moves along a circular path. While the speed of the object might be constant (uniform circular motion), its velocity is constantly changing because the direction of motion is changing. According to Newton's First Law, a change in velocity (acceleration) requires an unbalanced force (Newton's Second Law). Therefore, an object moving in a circle must be acted upon by a net force.

Uniform Circular Motion

In uniform circular motion, the object moves around a circular path at a constant speed ($ v $). The radius of the circle is $ r $. Although the speed is constant, the velocity vector is continuously changing direction. This change in velocity means there is an acceleration.

The acceleration in uniform circular motion is always directed towards the centre of the circle. This is called centripetal acceleration ($ a_c $).

Its magnitude is given by:

$ a_c = \frac{v^2}{r} $

where $ v $ is the speed and $ r $ is the radius of the circular path.

We can also express this in terms of angular speed ($ \omega $). The relationship between linear speed $ v $ and angular speed $ \omega $ is $ v = r\omega $. Substituting this into the formula for centripetal acceleration:

$ a_c = \frac{(r\omega)^2}{r} = \frac{r^2\omega^2}{r} = r\omega^2 $

So, $ a_c = \frac{v^2}{r} = r\omega^2 $.

The direction of $ \vec{a}_c $ is always towards the centre of the circle, perpendicular to the instantaneous velocity vector.

Centripetal Force

According to Newton's Second Law ($ \vec{F}_{net} = m\vec{a} $), if an object is accelerating, there must be a net force acting on it in the direction of the acceleration. For uniform circular motion, the acceleration is centripetal (towards the centre), so the net force must also be directed towards the centre of the circle. This net force is called the centripetal force ($ F_c $).

The magnitude of the centripetal force is given by:

$ F_c = m a_c = m \frac{v^2}{r} $

or using angular speed:

$ F_c = m r\omega^2 $

The centripetal force is not a new fundamental type of force. It is the net force required to keep an object moving in a circle. This net force can be provided by various physical forces or combinations of forces:

• For a stone whirled on a string: Tension in the string provides the centripetal force.
• For a planet orbiting the sun: Gravitational force provides the centripetal force.
• For a car turning on a level road: Static friction between the tyres and the road provides the centripetal force.
• For a vehicle on a banked curve: A component of the normal force and/or friction provides the centripetal force.
• For an electron orbiting a nucleus (in a simple model): Electrostatic attraction provides the centripetal force.

Non-Uniform Circular Motion

If the speed of the object in a circular path is not constant, the motion is called non-uniform circular motion. In this case, there is not only a centripetal acceleration (changing direction of velocity) but also a tangential acceleration ($ a_t $) (changing magnitude of velocity).

The total acceleration ($ \vec{a} $) is the vector sum of the centripetal acceleration ($ \vec{a}_c $) and the tangential acceleration ($ \vec{a}_t $).

$ \vec{a} = \vec{a}_c + \vec{a}_t $

Since $ \vec{a}_c $ is perpendicular to $ \vec{a}_t $, the magnitude of the total acceleration is:

$ a = \sqrt{a_c^2 + a_t^2} $

According to Newton's Second Law, the net force ($ \vec{F}_{net} $) is in the direction of the total acceleration. The net force can be resolved into two components:

$ \vec{F}_{net} = \vec{F}_c + \vec{F}_t $

where $ \vec{F}_c = m\vec{a}_c $ is the centripetal component of the net force (directed towards the centre) responsible for changing the direction of velocity, and $ \vec{F}_t = m\vec{a}_t $ is the tangential component of the net force (parallel or anti-parallel to the velocity) responsible for changing the speed.

Examples of Circular Motion Applications:

Answer:

Circular motion problems often involve identifying which real force (or combination of forces) provides the necessary centripetal force to maintain the circular path.

Solving Problems In Mechanics

Solving problems in mechanics that involve forces and motion requires a systematic approach based on Newton's Laws. A structured method helps in correctly identifying the forces, applying the laws, and solving the resulting equations.

Steps for Solving Mechanics Problems

Here is a general procedure for solving problems involving forces:

1. Understand the Problem:

• Read the problem carefully.
• Identify what is given and what needs to be found.
• Visualise the physical situation.

2. Draw a Diagram:

• Draw a clear diagram of the physical situation described in the problem.
• Include all relevant objects and their positions.
• Show velocities and accelerations if known or assumed.

3. Identify the System and Object(s) of Interest:

• Clearly define the system or the specific object(s) you will apply Newton's Laws to. Sometimes you might need to analyse multiple objects within a system separately.

4. Draw a Free Body Diagram (FBD):

• This is a crucial step. For each object of interest, draw a separate diagram showing only that object and all the external forces acting on it.
• Represent the object as a point mass or a simplified shape.
• Draw vectors representing each force acting on the object. Label each force clearly (e.g., $ \vec{W} $ for weight, $ \vec{N} $ for normal force, $ \vec{T} $ for tension, $ \vec{f} $ for friction). Indicate the direction of each force.
• Do NOT include forces exerted *by* the object on other objects.
• Do NOT include internal forces if analysing a system of multiple objects as a whole (unless you're considering the forces between parts of the system to calculate internal effects).

5. Choose a Coordinate System:

• Select a suitable coordinate system (e.g., Cartesian x-y axes).
• Align the axes conveniently, often aligning one axis with the direction of acceleration (if any) or along inclined surfaces.

6. Resolve Forces into Components:

• Break down each force vector in the FBD into its components along the chosen coordinate axes.
• Pay attention to angles and signs.

7. Apply Newton's Second Law (or First Law):

• Apply Newton's Second Law ($ \sum \vec{F}_{ext} = m\vec{a} $) to the object or system.
• Write the equation in vector form or, more commonly, as separate scalar equations for each axis: $ \sum F_x = ma_x $, $ \sum F_y = ma_y $, $ \sum F_z = ma_z $.
• If the object is in equilibrium ($ \vec{a} = 0 $), use Newton's First Law, which is $ \sum \vec{F}_{ext} = 0 $, resulting in $ \sum F_x = 0, \sum F_y = 0, \sum F_z = 0 $.

8. Apply Other Relevant Principles:

• If there are interactions between multiple bodies, use Newton's Third Law ($ \vec{F}_{AB} = -\vec{F}_{BA} $) to relate forces acting on different bodies.
• Use definitions or formulas for specific forces (e.g., $ W = mg $, $ f_k = \mu_k N $, $ f_s \le \mu_s N $, Hooke's Law $ F_s = kx $).
• If motion is involved, you might need kinematic equations ($ v = u + at $, $ s = ut + \frac{1}{2}at^2 $, $ v^2 = u^2 + 2as $) to relate acceleration, velocity, displacement, and time.

9. Solve the Equations:

• Solve the system of equations derived from Newton's Laws and other principles to find the unknown quantities.

10. Check the Answer:

• Does the answer make sense in the context of the problem? (e.g., is the direction of acceleration reasonable? Is the magnitude of forces within expected limits?)
• Check units.

Example Problem Solution Walkthrough:

Answer:

1. Understand the Problem: We have two connected masses, one on an incline and one hanging. We need to find how fast they accelerate and the force in the connecting string.

2. Draw a Diagram:

3. Identify Objects of Interest: The two objects are the 2 kg block ($ m_1 $) on the incline and the 3 kg hanging mass ($ m_2 $).

4. Draw Free Body Diagrams (FBDs):

FBD for the 2 kg block ($ m_1 $):

Forces acting on $ m_1 $:

• Weight ($ \vec{W}_1 $): $ W_1 = m_1 g = 2 \times 9.8 = 19.6 $ N, vertically downwards.
• Normal Force ($ \vec{N} $): Perpendicular to the incline, upwards.
• Tension ($ \vec{T} $): Along the string, upwards parallel to the incline.

FBD for the 3 kg mass ($ m_2 $):

Forces acting on $ m_2 $:

• Weight ($ \vec{W}_2 $): $ W_2 = m_2 g = 3 \times 9.8 = 29.4 $ N, vertically downwards.
• Tension ($ \vec{T} $): Along the string, vertically upwards. (Note: The tension is the same throughout the string because the pulley is frictionless and massless).

5. Choose Coordinate System:

• For $ m_1 $ on the incline: Choose x-axis parallel to the incline (positive direction upwards, along the string's pull) and y-axis perpendicular to the incline (positive direction upwards, away from the plane).
• For $ m_2 $ (hanging): Choose y-axis vertically (positive direction upwards, or downwards if we assume $ m_2 $ goes down). Let's assume the 3 kg mass accelerates downwards, so take positive y downwards for this mass for simplicity in signs, or consistently take upward as positive and let the sign of acceleration tell us the direction. Let's stick to upward as positive for both, and $ a_1 $ will be $ -a $ (down incline) and $ a_2 $ will be $ -a $ (downwards) if the system moves such that $m_2$ descends. Let's assume $m_2$ descends, so $m_1$ goes up the incline with acceleration $a$. Thus, positive x is up the incline for $m_1$, and positive y is downwards for $m_2$. The magnitudes of acceleration are the same for both masses, let this magnitude be $a$. So $a_1 = a$ up the incline, $a_2 = a$ downwards.

6. Resolve Forces into Components:

For $ m_1 $:

Weight $ \vec{W}_1 $ needs to be resolved along the incline axes. The angle between $ \vec{W}_1 $ (vertical) and the negative y-axis (perpendicular to incline, downwards) is $ 30^\circ $.

Component parallel to incline (x-component): $ -W_1 \sin(30^\circ) $ (down the incline)

Component perpendicular to incline (y-component): $ -W_1 \cos(30^\circ) $ (into the incline)

$ W_{1x} = -19.6 \sin(30^\circ) = -19.6 \times 0.5 = -9.8 \, \text{N} $ $ W_{1y} = -19.6 \cos(30^\circ) = -19.6 \times \frac{\sqrt{3}}{2} \approx -17.0 \, \text{N} $

Normal force $ \vec{N} $: $ N_x = 0 $, $ N_y = +N $.

Tension $ \vec{T} $: $ T_x = +T $, $ T_y = 0 $.

For $ m_2 $:

Weight $ \vec{W}_2 $: $ W_{2y} = +W_2 = +29.4 \, \text{N} $ (downwards, if positive y is down).

Tension $ \vec{T} $: $ T_{2y} = -T $ (upwards, opposite to positive y). There are no x-components for $ m_2 $.

7. Apply Newton's Second Law ($ \sum F = ma $):

For $ m_1 $ (up the incline is positive x, perpendicular up is positive y):

Sum of x-components: $ \sum F_x = m_1 a_x $

$ T + W_{1x} = m_1 a $

$ T - 9.8 = 2a \quad \text{(Equation A)} $

Sum of y-components: $ \sum F_y = m_1 a_y $

Since the block does not move perpendicular to the incline, $ a_y = 0 $.

$ N + W_{1y} = 0 $

$ N - 17.0 = 0 \implies N = 17.0 \, \text{N} $

(We found the normal force, but it is not needed to find the acceleration or tension in this problem as there is no friction, which depends on N).

For $ m_2 $ (downwards is positive y):

Sum of y-components: $ \sum F_y = m_2 a_y $

$ W_2 + T_{2y} = m_2 a $

$ 29.4 - T = 3a \quad \text{(Equation B)} $

8. Solve the Equations:

We have two equations (A and B) and two unknowns ($ T $ and $ a $):

Equation A: $ T - 9.8 = 2a $

Equation B: $ 29.4 - T = 3a $

Add Equation A and Equation B:

$ (T - 9.8) + (29.4 - T) = 2a + 3a $

$ 29.4 - 9.8 = 5a $

$ 19.6 = 5a $

$ a = \frac{19.6}{5} = 3.92 \, \text{m/s}^2 $

The acceleration of the system is $ 3.92 \, \text{m/s}^2 $ (the 2 kg block moves up the incline, and the 3 kg mass moves downwards).

Now substitute the value of $ a $ back into either Equation A or Equation B to find $ T $. Using Equation A:

$ T - 9.8 = 2a $

$ T - 9.8 = 2 \times (3.92) $

$ T - 9.8 = 7.84 $

$ T = 7.84 + 9.8 = 17.64 \, \text{N} $

The tension in the string is $ 17.64 \, \text{N} $.

10. Check the Answer:

The acceleration $ a \approx 3.92 \, \text{m/s}^2 $ is less than $ g = 9.8 \, \text{m/s}^2 $, which is reasonable (the incline and the other mass resist the acceleration). The tension $ T \approx 17.64 \, \text{N} $ is less than the weight of the 3 kg mass (29.4 N), which is also reasonable because the 3 kg mass is accelerating downwards, meaning the net force on it is downwards ($ W_2 - T > 0 $). The tension is greater than the component of the 2 kg weight down the incline (9.8 N), which is necessary for the 2 kg mass to accelerate up the incline.

Mastering these steps and practising with various problems (involving different forces, multiple objects, inclines, pulleys, etc.) is key to successfully applying Newton's laws to solve mechanics problems.