Newton's Laws of Motion

Newton’s First Law — Inertia (CED Unit 2)

An object at rest stays at rest, and an object in motion stays in motion with the same speed and in The same direction, unless acted upon by a net external force.

Explore the simulation above to develop intuition for this topic.

Inertial Reference Frames

Newton’s laws are valid only in inertial reference frames — frames that are not accelerating. A Frame attached to an accelerating car is non-inertial: a pendulum hanging from the ceiling swings Backward even though no horizontal force acts on it (in the car’s frame). In an inertial frame (the Ground), the pendulum tends to stay at rest while the car accelerates forward.

Mass and Inertia

Mass is a measure of an object’s resistance to changes in its motion (inertia). It is an Intrinsic property of an object, independent of location. The SI unit is the kilogram (kg).

Mass is not weight. Mass is a scalar; weight is a vector. Mass does not change with location; weight Depends on the local gravitational field. An astronaut on the Moon has the same mass as on Earth but One-sixth the weight.

Newton’s Second Law — Force and Acceleration (CED Unit 2)

The net external force on an object equals the product of its mass and acceleration:

\vec{F}_{\mathrm{net} = m\vec{a}

In component form:

$$\sum F_x = ma_x, \qquad \sum F_y = ma_y$$

The Kilogram, Newton, and Slug

• 1 Newton (N) = 1 \mathrm{ kg \cdot \mathrm{m/s^2
• 1 slug = 14.59 \mathrm{ kg (used in the US customary system)

Weight vs Mass

$$\vec{W} = m\vec{g}$$

• Weight is a force (measured in N) that depends on the local gravitational field.
• Mass is an intrinsic property (measured in kg) that does not change with location.

On Earth: g \approx 9.8 \mathrm{ m/s^2. On the Moon: g \approx 1.62 \mathrm{ m/s^2.

Proof That $F = ma$ Is Not a Definition

One might think $F = ma$ merely defines force. But force is defined independently through Interactions (contact forces, gravitational forces, spring forces). $F = ma$ is an empirical law That relates the net force to the resulting acceleration. The fact that the same proportionality Constant ($m$) works for all forces (gravitational, electromagnetic, contact) is a deep and Non-trivial result. If $F = ma$ were merely a definition, it would not have predictive power, but it Does: given the forces, we can predict the motion, and given the motion, we can infer the forces.

Newton’s Third Law — Action-Reaction Pairs (CED Unit 2)

For every action, there is an equal and opposite reaction:

\vec{F}_{A \mathrm{ on B} = -\vec{F}_{B \mathrm{ on A}

Key Points About Action-Reaction Pairs

1. The two forces act on different objects.
2. The forces are always equal in magnitude and opposite in direction.
3. The forces are the same type (both gravitational, both normal, both frictional, etc.).
4. Action-reaction pairs cannot cancel each other because they act on different objects.

Systematic Identification of Third Law Pairs

To correctly identify a third law pair, use this algorithm:

1. Identify the force on object $A$ and what object exerts it: “Object $B$ exerts force $\vec{F}$ on object $A$.”
2. The third law pair is: “Object $A$ exerts force $-\vec{F}$ on object $B$.”
3. Verify: same type, equal magnitude, opposite direction, different objects.

Types of Forces

Gravitational Force (Weight)

$$F_g = mg$$

Normal Force ($\vec{N}$)

The normal force is the contact force perpendicular to the surface. On a flat surface with no Vertical acceleration: $N = mg$. On an inclined plane at angle $\theta$:

$$N = mg\cos\theta$$

Why the Normal Force Is Not Always Equal to $mg$

The normal force adjusts to prevent the object from penetrating the surface. It equals $mg$ only When there is no vertical acceleration and no other vertical forces. In a lift accelerating upward, $N = m(g + a) \gt mg$. On an inclined plane, only the component of gravity perpendicular to the Surface ($mg\cos\theta$) must be balanced by the normal force. The parallel component ($mg\sin\theta$) causes acceleration along the slope.

Tension ($\vec{T}$)

Tension is the pulling force transmitted through a string, rope, or cable. For a massless, Inextensible string, tension is the same throughout.

Friction ($\vec{f}$)

Static friction prevents relative motion:

$$F_s \le \mu_s N$$

Where $\mu_s$ is the coefficient of static friction. Static friction adjusts to match the applied Force up to a maximum of $f_{s,\max} = \mu_s N$.

Kinetic friction opposes relative motion:

$$F_k = \mu_k N$$

Where $\mu_k$ is the coefficient of kinetic friction. Note that $\mu_k \lt \mu_s$ .

Why Kinetic Friction Is Less Than Static Friction

At the microscopic level, static friction holds because the surfaces form temporary bonds (atomic-scale cold welds) at the contact points. These bonds must be broken to initiate sliding. Once sliding begins, the surfaces do not have time to form as many bonds, and the kinetic friction Force is lower. The transition from static to kinetic friction is the reason objects “jerk” when They start to move.

Applied Force (\vec{F}_{\mathrm{app})

Any external push or pull.

Spring Force (Hooke’s Law)

$$F_s = -kx$$

Where $k$ is the spring constant and $x$ is the displacement from equilibrium.

Free Body Diagrams (FBDs)

A free body diagram shows all external forces acting on a single object:

1. Isolate the object.
2. Draw vectors for each force, originating from the object’s center.
3. Do not include internal forces or forces exerted by the object on other objects.

Common FBD Errors

• Including velocity or acceleration vectors on the FBD. The FBD shows only forces.
• Drawing the normal force at an angle. The normal force is always perpendicular to the contact surface.
• Including the “centripetal force” as a separate force. The centripetal force is the net radial force, not a separate interaction.

Applications of Newton’s Laws

Atwood Machine

Two masses $m_1$ and $m_2$ ($m_2 \gt m_1$) connected by a massless string over a frictionless Pulley.

For mass $m_1$ (taking up as positive): $T - m_1 g = m_1 a$

For mass $m_2$ (taking up as positive): $m_2 g - T = m_2 a$

Adding: $(m_2 - m_1)g = (m_1 + m_2)a$

$$A = \frac{(m_2 - m_1)g}{m_1 + m_2}$$ $$T = \frac{2m_1 m_2 g}{m_1 + m_2}$$

Derivation of Tension in the Atwood Machine

The tension is the same throughout a massless string. Solving the first equation for $T$:

$$T = m_1(g + a)$$

Substituting the expression for $a$:

$$T = m_1\left(g + \frac{(m_2 - m_1)g}{m_1 + m_2}\right) = m_1 g \cdot \frac{m_1 + m_2 + m_2 - m_1}{m_1 + m_2} = \frac{2m_1 m_2 g}{m_1 + m_2}$$

Note that $T$ is always between $m_1 g$ and $m_2 g$: it must be greater than $m_1 g$ to accelerate $m_1$ upward, and less than $m_2 g$ to accelerate $m_2$ downward.

Connected Bodies (Systems)

When multiple objects are connected, you can either:

1. Analyze each object separately with its own FBD, or
2. Treat the system as a single object (internal forces cancel).

When to use each approach: Use the system approach when you need the acceleration of the entire System. Use the individual approach when you need the tension in a connecting string or the normal Force between two objects in contact.

Elevator Problems

A person of mass $m$ stands on a scale in an elevator. The scale reads the normal force $N$.

• At rest or constant velocity: $N = mg$
• Accelerating upward: $N = m(g + a)$ (scale reads higher)
• Accelerating downward: $N = m(g - a)$ (scale reads lower)
• Free fall ($a = g$): $N = 0$ (apparent weightlessness)

Friction and Circular Motion

For an object moving in a circle, the net force toward the center provides the centripetal force:

\sum F_{\mathrm{radial} = \frac{mv^2}{r}

Banked Curves

On a banked curve of angle $\theta$ and radius $r$The ideal speed (at which no friction is Required) is:

V_{\mathrm{ideal} = \sqrt{rg\tan\theta}

At this speed, the horizontal component of the normal force provides exactly the required Centripetal force. Below this speed, friction acts up the slope; above it, friction acts down the Slope.

Newton’s Law of Universal Gravitation (CED Unit 3)

$$F = \frac{Gm_1 m_2}{r^2}$$

Where G = 6.674 \times 10^{-11} \mathrm{ N \cdot \mathrm{m^2/\mathrm{kg^2.

Gravitational Field

The gravitational field strength at a distance $r$ from mass $M$ is:

$$G = \frac{GM}{r^2}$$

Gravitational Potential Energy

$$U = -\frac{GMm}{r}$$

The negative sign indicates that work must be done against gravity to move masses apart. The Reference point ($U = 0$) is at infinity.

Orbital Mechanics

For a circular orbit of radius $r$ around mass $M$:

$$\frac{GMm}{r^2} = \frac{mv^2}{r} \implies v = \sqrt{\frac{GM}{r}}$$ $$T = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}}$$

These are Kepler’s Third Law: $T^2 \propto r^3$.

Why All Orbits Are Conic Sections

Newton showed that any orbit under an inverse-square gravitational force is a conic section (ellipse, parabola, or hyperbola). Bound orbits (total energy negative) are ellipses. Escape Trajectories (total energy zero) are parabolas. Unbound trajectories (total energy positive) are Hyperbolas. Circular orbits are a special case of elliptical orbits with eccentricity zero.

Common Pitfalls

1. Including forces in the wrong direction on FBDs. Always draw forces in the direction they actually act, not in the direction of motion.
2. Confusing action-reaction pairs. $\vec{N}$ and $\vec{W}$ on the same object are not an action-reaction pair.
3. Using $\mu_s N$ as the static friction force. $f_s \le \mu_s N$; static friction adjusts to the applied force. Only use $\mu_s N$ for the maximum.
4. Forgetting that tension in a massless string is the same on both sides of a pulley (assuming a frictionless, massless pulley).
5. Assuming $\mu_k = \mu_s$. Kinetic friction is generally less than static friction.
6. Using $g = 9.8$ when a different value is specified or when the problem takes place far from Earth’s surface.
7. Ignoring the direction of the normal force. The normal force is always perpendicular to the contact surface, not necessarily vertical.
8. Adding mass to a pulley system without accounting for the pulley’s inertia. The standard Atwood machine assumes a massless pulley. A massive pulley has its own moment of inertia.
9. Assuming the acceleration of a connected system is the same for all parts. This is true only if the connections are rigid or inextensible.

Practice Questions

1. A 10 \mathrm{ kg box is pushed across a floor with $\mu_k = 0.25$ by a horizontal force of 50 \mathrm{ N. Find the acceleration of the box.

2. Two blocks of masses 4.0 \mathrm{ kg and 6.0 \mathrm{ kg are connected by a string over a frictionless pulley (Atwood machine). Find the acceleration and the tension.

3. A 60 \mathrm{ kg person stands on a scale in an elevator that accelerates upward at 2.0 \mathrm{ m/s^2. What does the scale read?

4. A 2.0 \mathrm{ kg block on a $35^\circ$ incline has $\mu_s = 0.5$ and $\mu_k = 0.3$. Find the acceleration if the block is given an initial push up the incline.

5. A satellite orbits Earth at an altitude of 400 \mathrm{ km. Find the orbital velocity and period. (Earth’s mass = 5.97 \times 10^{24} \mathrm{ kgEarth’s radius = 6371 \mathrm{ km.)

6. A 1500 \mathrm{ kg car rounds a banked curve of radius 100 m at 20 \mathrm{ m/s. The banking angle is $15^\circ$. Find the minimum coefficient of static friction.

7. Three blocks of masses $m_1$, $m_2$And $m_3$ are connected by strings on a frictionless table, with $m_3$ hanging off the edge over a pulley. Derive the acceleration in terms of the masses.

8. A 5.0 \mathrm{ kg block is on a frictionless table connected to a 3.0 \mathrm{ kg block hanging over the edge. Find the acceleration of the system and the tension in the string.

9. A block of mass $m$ is placed on a wedge of mass $M$ with angle $\theta$. All surfaces are frictionless. Find the acceleration of the wedge relative to the ground.

10. A 2000 \mathrm{ kg car towing a 1000 \mathrm{ kg trailer accelerates at 1.5 \mathrm{ m/s^2. If the tension in the tow bar is 1800 \mathrm{ NFind the friction force on the car.

10. Friction: A Deeper Look

Microscopic Origin of Friction

At the microscopic level, even apparently smooth surfaces are rough. When two surfaces are pressed Together, the microscopic peaks (asperities) interlock. To slide the surfaces, these interlocking Features must be broken or deformed. The force needed is friction.

Static friction involves cold welding: the atoms at the contact points form temporary bonds. These bonds must be broken simultaneously to initiate motion, which requires a larger force. Once Sliding begins, the surfaces do not have time to form as many bonds, and the kinetic friction force Is lower.

The Angle of Repose

The steepest angle at which an object on an inclined plane remains stationary is the angle of Repose, given by:

$\theta_r = \arctan(\mu_s)$

This is because the block is on the point of sliding when $mg\sin\theta = \mu_s mg\cos\theta$I.e., $\tan\theta = \mu_s$. This relationship is used in geology to predict landslides and in industry to Design hoppers and chutes.

Worked Example: Friction on a Banked Curve Without Friction

A circular road of radius 80 \mathrm{ m is banked at $15^{\circ}$. Find the speed at which a car Can round the curve without needing any friction.

At the ideal speed, the horizontal component of the normal force provides the centripetal force, and The vertical component balances gravity:

$N\sin\theta = \frac{mv^2}{r}, \qquad N\cos\theta = mg$

Dividing: $\tan\theta = \frac{v^2}{rg}$

v = \sqrt{rg\tan\theta} = \sqrt{80 \times 9.8 \times \tan 15^{\circ}} = \sqrt{80 \times 9.8 \times 0.2679} = \sqrt{209.9} = 14.5 \mathrm{ m/s

At this speed, the car can negotiate the curve even on ice. Below this speed, friction acts up the Slope; above it, friction acts down the slope.

11. Connected Bodies: Systematic Approach

Worked Example: Three-Body System with a Pulley

Three blocks are connected as follows: m_1 = 4 \mathrm{ kg and m_2 = 6 \mathrm{ kg on a Frictionless table, connected by a string over a pulley at the edge, with m_3 = 5 \mathrm{ kg Hanging vertically.

System approach: The net force on the system is $m_3 g$ (the weight of the hanging mass). The Total mass being accelerated is m_1 + m_2 + m_3 = 15 \mathrm{ kg.

a = \frac{m_3 g}{m_1 + m_2 + m_3} = \frac{5 \times 9.8}{15} = \frac{49}{15} = 3.27 \mathrm{ m/s^2

Individual approach for tension: For $m_3$:

m_3 g - T = m_3 a \implies T = m_3(g - a) = 5(9.8 - 3.27) = 5 \times 6.53 = 32.7 \mathrm{ N

For the string between $m_1$ and $m_2$ on the table: T' = m_2 a = 6 \times 3.27 = 19.6 \mathrm{ N.

Note that the tension in the string connecting to the hanging mass (32.7 \mathrm{ N) is greater Than the tension in the string between $m_1$ and $m_2$ (19.6 \mathrm{ N), because the latter string Only needs to accelerate $m_2$While the former must also provide the net force to accelerate $m_3$.

12. Gravitational Field Strength: Derivation of Key Results

Variation of $g$ with Altitude

At the Earth’s surface, g \approx 9.8 \mathrm{ m/s^2. At height $h$ above the surface:

$g(h) = \frac{GM}{(R_E + h)^2}$

At $h = R_E$ (one Earth radius above the surface): g = \frac{GM}{4R_E^2} = \frac{g_0}{4} \approx 2.45 \mathrm{ m/s^2.

This rapid decrease is why astronauts in low Earth orbit (altitude about 400 km) are not truly Weightless because of distance from Earth — they are weightless because they are in free fall. The Gravitational field strength at 400 km altitude is g \approx 8.7 \mathrm{ m/s^2About 89% of the Surface value.

Gravitational Field Strength Inside a Uniform Sphere

For a point at distance $r$ from the centre of a uniform sphere of radius $R$ and mass $M$:

$g(r) = \frac{GMr}{R^3} \quad (r \le R)$

$g(r) = \frac{GM}{r^2} \quad (r \gt R)$

Inside the sphere, $g$ increases linearly with $r$ and reaches its maximum at the surface. This is Because only the mass within radius $r$ contributes to the gravitational force at that point (by Newton’s shell theorem).

13. Orbital Mechanics: Extended Worked Examples

Worked Example: Satellite Speed and Period

A satellite orbits at an altitude of 500 \mathrm{ km above Earth’s surface.

r = R_E + h = 6.371 \times 10^6 + 5 \times 10^5 = 6.871 \times 10^6 \mathrm{ m

v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.871 \times 10^6}} = \sqrt{5.80 \times 10^7} = 7615 \mathrm{ m/s

T = \frac{2\pi r}{v} = \frac{2\pi \times 6.871 \times 10^6}{7615} = 5671 \mathrm{ s \approx 94.5 \mathrm{ min

Worked Example: Escape Velocity from a Planet

The escape velocity from a planet of mass $M$ and radius $R$ is:

$v_e = \sqrt{\frac{2GM}{R}}$

For Earth: v_e = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 5.97 \times 10^{24}}{6.371 \times 10^6}} = 11186 \mathrm{ m/s \approx 11.2 \mathrm{ km/s.

A useful approximation: v_e \approx \sqrt{2} \times v_{\mathrm{orbit} for a surface-skimming orbit. The factor of $\sqrt{2}$ arises because escape requires twice the kinetic energy of a circular Orbit: \frac{1}{2}mv_e^2 = 2 \times \frac{1}{2}mv_{\mathrm{orbit}^2.

14. Summary Table: Forces and Their Characteristics

Force	Law	Direction	Depends On
Gravity	$F = GMm/r^2$	Attractive, along line joining centres	Masses, separation
Normal	Adjusts to prevent penetration	Perpendicular to contact surface	Other forces present
Tension	Along string	Away from object	Applied forces, mass
Static friction	$f_s \le \mu_s N$	Opposes tendency to slide	Normal force, coefficient
Kinetic friction	$f_k = \mu_k N$	Opposes motion	Normal force, coefficient
Spring	$F = -kx$	Toward equilibrium position	Displacement, spring constant
Drag	$F_d \propto v^2$	Opposes motion	Speed, cross-section, fluid density

15. Common Pitfalls: Extended

10. Assuming the normal force equals $mg$ on an inclined plane. The normal force is $mg\cos\theta$Not $mg$. Always resolve forces perpendicular to the surface.

11. Forgetting that tension has a maximum value. A string can only pull, not push. If the calculated tension is negative, the string has gone slack and the tension is zero.

12. Applying $F = ma$ to individual objects when using the system approach. The system approach gives the acceleration of the entire system but does not give tensions or internal forces.

13. Ignoring air resistance in orbital calculations. While this is a valid approximation for satellites above the atmosphere, it fails for objects in the lower atmosphere.

14. Using g = 9.8 \mathrm{ m/s^2 for objects far from Earth’s surface. Use $g = GM/r^2$ with the actual distance from Earth’s centre.

15. Confusing the mass of an orbiting body with the mass of the central body. In $v = \sqrt{GM/r}$, $M$ is the mass of the body being orbited (e.g., Earth), not the satellite.

Practice Questions (Additional)

11. A block of mass $m$ is placed on a wedge of mass $M$ with angle $\theta$. All surfaces are frictionless. Find the acceleration of the wedge. (Hint: use conservation of momentum in the horizontal direction.)

12. A 3 \mathrm{ kg block on a frictionless table is connected by a string over a pulley to a 2 \mathrm{ kg block hanging off the edge. The pulley has mass 0.5 \mathrm{ kg and radius 0.1 \mathrm{ m (moment of inertia $I = \frac{1}{2}mr^2$). Find the acceleration of the system and the tension on each side of the pulley.

13. A rocket of mass 1000 \mathrm{ kg is launched vertically. The engine produces a thrust of 15000 \mathrm{ N for 30 \mathrm{ s. Find the maximum height reached. (Assume $g$ is constant at 9.8 \mathrm{ m/s^2 and neglect air resistance.)

14. Two masses m_1 = 2 \mathrm{ kg and m_2 = 3 \mathrm{ kg are connected by a light string over a smooth pulley. $m_1$ rests on a rough table with $\mu_k = 0.3$. Find the acceleration and the tension in the string.

15. Calculate the gravitational field strength at a point halfway between the Earth and the Moon. (Earth-Moon distance = 3.84 \times 10^8 \mathrm{ m, M_E = 5.97 \times 10^{24} \mathrm{ kg M_M = 7.35 \times 10^{22} \mathrm{ kg.)

Extended Worked Examples

Example 16: Atwood Machine with Massive Pulley

Two masses m_1 = 4 \mathrm{ kg and m_2 = 6 \mathrm{ kg are connected by a light string over a Pulley of mass M = 2 \mathrm{ kg and radius R = 0.1 \mathrm{ m (solid disk, $I = \frac{1}{2}MR^2$). Find the acceleration of the system and the tensions on each side of the Pulley.

Step 1: Write equations for each mass and the pulley

For $m_1$ (lighter, accelerates up):

$T_1 - m_1 g = m_1 a$

For $m_2$ (heavier, accelerates down):

$m_2 g - T_2 = m_2 a$

For the pulley (rotates clockwise):

$T_2 R - T_1 R = I\alpha = \frac{1}{2}MR^2 \times \frac{a}{R} = \frac{1}{2}MaR$

$T_2 - T_1 = \frac{1}{2}Ma$

Step 2: Solve the system

From the first two equations:

$T_1 = m_1(g + a), \quad T_2 = m_2(g - a)$

Substituting into the pulley equation:

$m_2(g - a) - m_1(g + a) = \frac{1}{2}Ma$

$(m_2 - m_1)g - (m_1 + m_2)a = \frac{1}{2}Ma$

a = \frac{(m_2 - m_1)g}{m_1 + m_2 + \frac{1}{2}M} = \frac{(6 - 4) \times 9.8}{4 + 6 + 1} = \frac{19.6}{11} = 1.78 \mathrm{ m/s^2

Step 3: Find the tensions

T_1 = 4(9.8 + 1.78) = 4 \times 11.58 = 46.3 \mathrm{ N

T_2 = 6(9.8 - 1.78) = 6 \times 8.02 = 48.1 \mathrm{ N

:::info With a massless pulley, $T_1 = T_2$ and a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{19.6}{10} = 1.96 \mathrm{ m/s^2. The massive pulley Reduces the acceleration because some of the net force goes into rotating the pulley rather than Accelerating the masses. The difference T_2 - T_1 = 1.8 \mathrm{ N is the net torque on the pulley.

Example 17: Object in a Fluid with Drag

A 0.5 \mathrm{ kg steel ball is dropped into a tank of oil. The drag force is given by $F_D = bv$ Where b = 0.8 \mathrm{ N\cdot\mathrm{s/m. Find (a) the terminal velocity, (b) the time to reach 63% Of terminal velocity, and (c) the velocity after 3 \mathrm{ s.

Step 1: Terminal velocity (net force = 0)

$mg = bv_T$

v_T = \frac{mg}{b} = \frac{0.5 \times 9.8}{0.8} = 6.125 \mathrm{ m/s

Step 2: Time constant

The equation of motion is $ma = mg - bv$Which gives:

$v(t) = v_T(1 - e^{-t/\tau})$

Where \tau = m/b = 0.5/0.8 = 0.625 \mathrm{ s.

The time to reach $63\%$ of $v_T$ is t = \tau = 0.625 \mathrm{ s.

Step 3: Velocity after 3 \mathrm{ s

v(3) = 6.125(1 - e^{-3/0.625}) = 6.125(1 - e^{-4.8}) = 6.125(1 - 0.00823) = 6.125 \times 0.9918 = 6.074 \mathrm{ m/s

The ball is essentially at terminal velocity after 3 \mathrm{ s (about $99.2\%$ of $v_T$).

Example 18: Tension in a Cable at an Angle

A 20 \mathrm{ kg traffic light is suspended by two cables. Cable A makes $30^\circ$ with the horizontal And Cable B makes $45^\circ$ with the horizontal. Find the tension in each cable.

Step 1: Resolve forces

Horizontal: $T_A \cos 30° = T_B \cos 45^\circ$

$T_A \times 0.866 = T_B \times 0.707 \implies T_A = 0.816 T_B$

Vertical: $T_A \sin 30° + T_B \sin 45° = mg$

T_A \times 0.5 + T_B \times 0.707 = 20 \times 9.8 = 196 \mathrm{ N

Step 2: Substitute and solve

$0.816 T_B \times 0.5 + 0.707 T_B = 196$

$0.408 T_B + 0.707 T_B = 196$

1.115 T_B = 196 \implies T_B = 175.8 \mathrm{ N

T_A = 0.816 \times 175.8 = 143.4 \mathrm{ N

Check: 143.4 \times 0.5 + 175.8 \times 0.707 = 71.7 + 124.3 = 196.0 \mathrm{ N. Confirmed.

Common Pitfalls Extended

Pitfall 6: Confusing Mass and Weight in Fluid Problems

When an object is submerged in a fluid, the apparent weight is W_{\mathrm{app} = mg - \rho_{\mathrm{fluid} V g. Do not confuse the actual weight $mg$ with the Apparent weight. Also, the buoyant force depends on the fluid density and the submerged volume, Not the object’s density.

Pitfall 7: Drawing Incorrect Normal Force Directions

The normal force is always perpendicular to the surface of contact, not necessarily vertical. On a Curved surface, the normal force direction changes along the surface. Always draw the normal force Perpendicular to the contact surface at the point of contact.

Pitfall 8: Assuming Tension Is the Same Throughout a Rope with Mass

For a “light” (massless) rope, the tension is the same throughout. For a heavy rope, the tension Varies along its length due to the weight of the rope itself. At the top of a hanging heavy rope, The tension equals the total weight; at the bottom, it is zero.

Additional Practice Problems

16. A 3 \mathrm{ kg block on a $25^\circ$ incline is connected by a string over a pulley to a 5 \mathrm{ kg block hanging vertically. The coefficient of kinetic friction is $0.3$. Calculate the acceleration of the system and the tension in the string.

17. A helicopter of mass 3000 \mathrm{ kg is rising at 2 \mathrm{ m/s^2. Calculate the upward force from the rotors. If the helicopter then moves at constant velocity, what upward force is needed?

18. A block of mass $m$ is placed on a frictionless cone with half-angle $\theta$. The cone rotates about its vertical axis with angular velocity $\omega$. Find the height above the apex at which the block remains stationary relative to the cone.

19. A 10 \mathrm{ kg crate is pushed across a rough floor ($\mu_k = 0.4$) by a force of 80 \mathrm{ N applied at $30^\circ$ below the horizontal. Calculate the normal force, the friction force, and the acceleration.

20. Two blocks (m_1 = 3 \mathrm{ kg on a table, m_2 = 2 \mathrm{ kg hanging) are connected by a string over a pulley. The table is frictionless. Calculate (a) the acceleration, (b) the tension, and (c) the speed after $m_2$ has fallen 0.5 \mathrm{ m from rest.

Practice Problems

Question 1: Inclined plane with friction

A block of mass 5 \mathrm{ kg is placed on a $30^\circ$ incline. The coefficient of static friction is $0.4$ and the coefficient of kinetic friction is $0.3$. Determine (a) whether the block slides down, and (b) if it does, the acceleration down the incline.

Answer

Component of gravity along the incline: mg\sin\theta = 5 \times 9.8 \times \sin(30^\circ) = 24.5 \mathrm{ N.

Maximum static friction: f_s = \mu_s N = 0.4 \times 5 \times 9.8 \times \cos(30^\circ) = 0.4 \times 42.4 = 17.0 \mathrm{ N.

Since 24.5 \mathrm{ N > 17.0 \mathrm{ NThe block slides.

Kinetic friction: f_k = \mu_k N = 0.3 \times 42.4 = 12.7 \mathrm{ N.

Net force along incline: F = 24.5 - 12.7 = 11.8 \mathrm{ N.

Acceleration: a = F/m = 11.8/5 = 2.36 \mathrm{ m/s^2.

Question 2: Elevator apparent weight

A person of mass 70 \mathrm{ kg stands on a scale in an elevator. What does the scale read (a) when the elevator accelerates upward at 2 \mathrm{ m/s^2(b) when it moves at constant velocity, and (c) when it decelerates at 3 \mathrm{ m/s^2 while moving upward?

Answer

The scale reads the normal force $N$.

(a) Accelerating up: $N - mg = ma$, N = m(g + a) = 70(9.8 + 2) = 826 \mathrm{ N. Scale reads 826 \mathrm{ N (about $119\%$ of actual weight).

(b) Constant velocity: $a = 0$, N = mg = 70 \times 9.8 = 686 \mathrm{ N.

(c) Decelerating upward (acceleration downward): N = m(g - a) = 70(9.8 - 3) = 476 \mathrm{ N.

Question 3: Connected masses with friction

Two blocks, m_1 = 4 \mathrm{ kg and m_2 = 6 \mathrm{ kgAre connected by a string over a frictionless pulley. $m_1$ rests on a horizontal table with $\mu_k = 0.2$And $m_2$ hangs vertically. Find the acceleration and the tension in the string.

Answer

For $m_2$: $m_2 g - T = m_2 a$. For $m_1$: $T - \mu_k m_1 g = m_1 a$.

Adding: $m_2 g - \mu_k m_1 g = (m_1 + m_2)a$.

a = \frac{m_2 g - \mu_k m_1 g}{m_1 + m_2} = \frac{6 \times 9.8 - 0.2 \times 4 \times 9.8}{10} = \frac{58.8 - 7.84}{10} = \frac{50.96}{10} = 5.10 \mathrm{ m/s^2.

T = m_1 a + \mu_k m_1 g = 4(5.10) + 0.2(4)(9.8) = 20.4 + 7.84 = 28.2 \mathrm{ N.

Question 4: Circular motion on a banked curve

A curve of radius 50 \mathrm{ m is banked at $15^\circ$. At what speed can a car negotiate this curve without relying on friction? If the car travels at 20 \mathrm{ m/sWhat minimum coefficient of friction is required?

Answer

Without friction: $v^2 = rg\tan\theta = 50 \times 9.8 \times \tan(15^\circ) = 50 \times 9.8 \times 0.268 = 131.3$. v = 11.5 \mathrm{ m/s.

At 20 \mathrm{ m/sFriction must provide additional centripetal force. The horizontal component of normal force provides: $N\sin\theta = mv^2/r - f\cos\theta$. The vertical: $N\cos\theta + f\sin\theta = mg$.

Solving simultaneously: $v^2 = \frac{rg(\tan\theta + \mu)}{1 - \mu\tan\theta}$.

$400 = \frac{50 \times 9.8(0.268 + \mu)}{1 - 0.268\mu}$.

$400(1 - 0.268\mu) = 490(0.268 + \mu) = 131.3 + 490\mu$.

$400 - 107.2\mu = 131.3 + 490\mu$.

$268.7 = 597.2\mu$So $\mu = 0.45$.

Question 5: Newton’s third law pairs

A book sits at rest on a table. Identify all action-reaction pairs involving the book. Explain why the normal force on the book is not the reaction force to gravity.

Answer

Action-reaction pairs (Newton’s third law):

1. Earth pulls book down (gravity) AND book pulls Earth up with equal magnitude.
2. Table pushes book up (normal force) AND book pushes table down with equal magnitude.

The normal force and gravity are NOT an action-reaction pair because they act on the SAME object (the book). Newton’s third law pairs always act on DIFFERENT objects. The normal force on the book is a reaction to the book pressing on the table (pair 2), not to gravity (pair 1). The book is in equilibrium because these two forces (gravity and normal) balance, but they are not a third-law pair.

:::