Laws of Motion Detailed Explanation - Physics Class 11 - NEET

Force

Force is a push or a pull which changes or tends to change the state of rest or of uniform motion of a body, or changes the direction of its motion. A force is an interaction between the object and the source that produces the push or pull. Force is a vector quantity. In this document vector quantities are denoted by bold letters such as v or by a bar on top

.

Effect of Resultant Force

• It may change only the speed of the body.
• It may change only the direction of motion.
• It may change both the speed and the direction of motion.
• It may change the size and the shape of a deformable body.

Unit of Force

• SI unit: Newton (N) (MKS system). 1 N =
  
  .
• CGS unit: Dyne. 1 dyne =
  
  .
• 1 newton = 10⁵ dyne.

Kilogram Force (kgf)

• The kilogram-force is the force with which Earth attracts a 1 kg body: kgf =
  
  .
• Dimensional formula of force: [M L T^-2].
• To specify a force fully we require:
  → magnitude of the force
  → direction of the force
  → point of application of the force

Types of Forces

Two broad kinds of forces commonly encountered in mechanics are field (non-contact) forces and contact forces.

1. Field Forces or Non-Contact Forces

Non-contact forces act without mechanical contact between the interacting bodies.

Examples: gravitational force between two masses, electrostatic force between charges, magnetic force between magnets. Weight W = mg is the gravitational force of Earth on a body and is a field force.

2. Contact Forces

Contact force is produced when objects press or rub against each other.

• If contact is frictionless, the contact force is perpendicular to the common surface and is called the normal reaction (N).
• If surfaces are rough and there is a tendency to move (or actual relative motion), a frictional force (f) arises, opposing relative motion along the surface.
• The contact force between two bodies is generally made up of two mutually perpendicular components:
  (i) Normal reaction (N) or (R)
  (ii) Force of friction (f)
  Since these two are perpendicular, the resultant contact force is obtained by vector addition (see diagram)
  
  .

To clarify which forces act on each body we often draw separated free-body diagrams (space between bodies but contact directions preserved):

In the figure the normal reactions (each of magnitude N) are perpendicular to the surface of contact and the frictional forces (each of magnitude f_act) act along the surface opposing motion. (Note: forces on the blocks from the ground are not shown.)

Newton's Laws of Motion

Isaac Newton formulated three laws that describe how forces affect motion. These laws are fundamental: the first law introduces inertia and inertial frames; the second gives a quantitative definition of force; the third describes mutual interactions (action and reaction).

The First Law

Every body continues in its state of rest, or of uniform motion in a straight line, unless compelled to change that state by external forces impressed on it.

• Inertia
  The tendency of a body to preserve its state of rest or uniform motion is called inertia. Galileo first developed this concept. Inertia is not a separate measurable scalar; mass is the measure of inertia: the greater the mass, the greater the resistance to change of motion.
• Inertial Frame of Reference
  A frame of reference in which Newton's first law holds (i.e., acceleration of a body is due only to forces acting on it, not due to acceleration of the frame) is called an inertial frame. Frames moving at constant velocity relative to an inertial frame are also inertial.

The Second Law

The rate of change of momentum of a body is equal to, and in the direction of, the net applied force.

For a body of mass m and velocity

, acted upon by a net external force F:

When mass is constant this reduces to:

Thus, F = m a, giving a quantitative definition of force. The second law is valid only in inertial frames.

SI unit of force is newton (N): 1 N = 1 kg·m·s^-2. Dimensions: M L T^-2.

The Third Law

For every action there is an equal and opposite reaction.

Force is always an interaction between two bodies. Newton's third law states that whenever body A exerts a force on body B, B exerts an equal and opposite force on A. Action and reaction act on different bodies and form a pair.

Concept of Free Body Diagram (FBD)

Why a Free Body Diagram?

Since forces always arise due to interaction of two or more bodies, to apply Newton's laws we must consider each body separately and show only the forces acting on that body. This pictorial representation of the chosen body and the forces on it is the free body diagram.

Definition - Free Body Diagram
A free body diagram is a pictorial representation in which the body under study is considered isolated from the rest of the system and all forces acting on it are shown, along with its shape and orientation.

How to draw a Free Body Diagram (FBD)

• Identify the contact forces acting on the body (normal reaction, friction, tension, etc.).
• Replace the body by a dot or the body's shape, depending on convenience, and indicate its orientation.
• Choose and draw a coordinate system; label positive directions.
• Draw all contact forces on the body with arrows; arrows should point in the correct direction and be labelled. Make arrow lengths roughly proportional to magnitudes when useful.
• Draw and label long-range forces (weight, electrostatic, magnetic) acting on the body.
• If the body is accelerating, show and label the acceleration vector.

Various Field Forces

Field forces discussed here are gravitational, electrostatic and magnetic. We consider gravitational force (weight) in detail for mechanics problems.

Weight

The net gravitational pull of Earth on a body is its weight. Weight acts on every particle of the body; the resultant may be considered to act at the body's centre of gravity.

• Weight is assumed to act at the centre of gravity of the body.
• For bodies small compared to Earth, gravitational field is approximately uniform across the body so the centre of gravity and the centre of mass coincide. For uniform bodies the centre of mass lies at the geometric centre.
• In the figure the weight of a uniform block acts at its geometric centre, coinciding with its centre of mass and centre of gravity.

Various Contact Forces

Common contact forces are tension, normal reaction, friction, spring force, etc.

Tension Force of Strings

• A string or flexible connector transmits force along its length; due to flexibility it can only pull (not push) and the force acts along the string.
• By Newton's third law the body attached to the string pulls back with equal and opposite force, which makes the string taut. This internal force is called tension (T).
• In a system consisting of a person pulling a string attached to a block, the string transmits equal tension along its length (if the string is massless and the pulley ideal).

When analysing motion of each component (block, string, hand) we draw the appropriate FBD showing the relevant forces.

String Passing over an Ideal Pulley

An ideal pulley is massless and frictionless; it changes the direction of tension but offers no resistance, so tension magnitude is the same on both sides of the pulley.

Normal Reaction

When two bodies press on each other they exert equal and opposite forces at the contact. If surfaces are frictionless the force acts along the normal to the surface at the point of contact; this is the normal reaction.

For a block of weight W on a floor the floor exerts a resultant normal reaction N equal to the vector sum of pressures at each contact point. The FBD of the block includes W and N.

For a spherical ball of weight W placed on a floor, the normal reaction acts at the point of contact and together with W forms the action-reaction pair between ball and floor.

If a block sits in a trough, the normal reactions at different contact surfaces act along the corresponding normals; the FBD includes those normal components as required.

Spring Force

A spring resists changes in its length. If a spring is compressed or extended by amount x from its natural length, it exerts a restoring force given by F = -k x, where k is the spring constant (stiffness) in N·m^-1. When the spring is at natural length x = 0 and spring force is zero.

Force by the spring on the support and by the support on the spring form an action-reaction pair. Forces by hands and spring on each other also form such pairs.

Conservation of Momentum

The law of conservation of momentum states that for an isolated system (no external force acting) the total linear momentum remains constant. It follows from Newton's laws and from the homogeneity of space.

Conservation of Linear Momentum

For an isolated system of particles or bodies, the vector sum of momenta before an interaction equals the vector sum after the interaction. Example: collision between two balls.

Remember: In ideal elastic interactions we neglect energy loss to friction, sound or heat. Real collisions may be inelastic and kinetic energy need not be conserved though momentum still is (if external forces are negligible).

Example: A heavy bowling ball (mass m₁) thrown at a stationary football (mass m₂) transfers momentum to the football. If initial velocities are u₁, u₂ and final velocities v₁, v₂, then

Formula:
m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂.

Here m₁u₁ ≠ m₁v₁ in general; masses and velocities change subject to momentum conservation.

• Though momentum of each body may change during interaction, the total momentum of the system is conserved if no external force acts.
• In car accidents momentum is transferred between vehicles; structural failure occurs when forces exceed material strengths.
• In elastic collisions kinetic energy is conserved; in inelastic collisions kinetic energy is converted into internal energy (heat, deformation), but momentum remains conserved.

Examples and Applications

Example: Particles of gas inside a balloon frequently collide with the walls. Though individual particle momenta change, the total momentum of the gas remains effectively constant; heating the gas increases particle speeds and hence pressure, causing the balloon to expand if the container allows.

Application: Rockets travel in space where there is no external medium to push against by ejecting mass (exhaust) at high speed. To conserve momentum, the rocket moves in the opposite direction to the ejected mass.

Equilibrium

Forces that have zero resultant and zero net turning effect (moment) produce no change in motion. Such a system is in equilibrium. For coplanar or concurrent forces we use resolution of forces and moments to examine equilibrium.

Resolution of a Force

To resolve a force means to replace it by components whose vector sum equals the original force. Often forces are resolved into two perpendicular components using trigonometry.

For the force F making angle θ with a chosen axis:

F₁ = F cos θ (component along axis), F₂ = F sin θ (component perpendicular to axis).

Note: The component of a force perpendicular to itself is zero. The component parallel to the force is equal to the magnitude of the force. In the opposite direction the component is negative.

Example 8.4 Resolve a weight of 10 N into components parallel and perpendicular to a slope inclined at 30° to the horizontal.

Solution: Component perpendicular to the plane:

Component parallel to the plane:

Example 8.5 Resolve horizontally and vertically a force F = 8 N which makes an angle of 45° with the horizontal.

Solution: Horizontal component:

= 4√2 N. Vertical component: F_V = F sin 45° .

Example 8.6 A body is supported on a rough plane inclined at 30° by a string attached to the body and held at angle 30° to the plane. Draw a diagram showing forces acting and resolve each of these forces (a) horizontally and vertically, (b) parallel and perpendicular to the plane.

Solution: The forces are: (i) tension T in the string, (ii) normal reaction N from the plane, (iii) weight w and friction f. Resolving as required gives the diagrams and component expressions shown in the figures.

Moment of a Force

The turning effect of a force about a point or axis is called the moment or torque. The magnitude of the moment of a force F about an axis is F multiplied by the perpendicular distance r from the axis to the line of action of the force:

Moment = F × r (often denoted by C or τ). Direction of torque corresponds to the sense of rotation it would cause.

Direction of Torque

Take a convenient sign convention (anticlockwise positive, clockwise negative). For forces F₁ and F₂ with perpendicular distances r₁ and r₂, the moments are +F₁r₁ and -F₂r₂ respectively if anticlockwise is positive.

Zero Moment

If the line of action of a force passes through the axis of rotation, its perpendicular distance is zero and its moment about that axis is zero.

Note: Torque is a vector (we will treat vector torque in the chapter on rotation).

Example 8.7 ABCD is a square of side 2 m and O is its centre. Forces act along the sides as shown. Calculate the moment of each force about (a) an axis through A and perpendicular to the plane, (b) an axis through O and perpendicular to the plane.

Example 8.8 Forces act as indicated on a rod AB pivoted at A. Find the anticlockwise moment of each force about the pivot.

Coplanar Forces in Equilibrium

For an object in equilibrium under several coplanar forces the following must hold:

(i) No resultant linear force along any two mutually perpendicular directions OX and OY: ΣF_x = 0 and ΣF_y = 0.

(ii) No net rotation about any axis: resultant moment about any chosen axis must be zero: Στ = 0.

Thus for coplanar forces we obtain three scalar equations (ΣF_x=0, ΣF_y=0, Στ=0). Hence the number of unknowns in a typical problem should not exceed three for a determinate solution.

Example 8.9 A uniform rod AB rests with end A on rough horizontal ground and end B against a smooth vertical wall. The rod has weight w and is in equilibrium in the position shown. Find the frictional force at A, the normal reaction at A, and the normal reaction at B.

Solution. Let the rod length be 2l. Using the three equilibrium conditions (ΣF_x=0, ΣF_y=0, Στ_O=0) and taking anticlockwise moments as positive, we obtain the three equations shown and solve to find the required reactions and friction.

Equilibrium of Concurrent Coplanar Forces

If several concurrent coplanar forces act on a point and are in equilibrium, the algebraic sum of their components in two perpendicular directions OX and OY must vanish:

ΣF_x = 0 and ΣF_y = 0.

Moment equation is automatically satisfied if taken about the common point of intersection; hence only two scalar equations arise and at most two unknowns can be solved.

Example 8.10 An object is in equilibrium under four concurrent forces in the directions shown. Find the magnitudes of F₁ and F₂.

Solution: Using ΣF_x=0 and ΣF_y=0 we obtain two equations (given in the text) and solve for F₁ and F₂.

Lami's Theorem

When three concurrent forces F₁, F₂ and F₃ keep a point in equilibrium, Lami's theorem states that each force is proportional to the sine of the angle between the other two forces. This provides a useful method for solving three-force equilibrium problems.

Example 8.11 One end of a string 0.5 m long is fixed at A; the other end is fastened to an object of weight 8 N. The object is pulled aside horizontally until it is 0.3 m from the vertical through A. Find the tension T in the string and the horizontal force F.

Solution. AC = 0.5 m, BC = 0.3 m ⇒ AB = 0.4 m. If ∠BAC = θ then cos θ = 0.4/0.5. Apply Lami's theorem to the three concurrent forces (weight, tension, horizontal pull) to obtain T and F.

Example 8.12 A rod of mass 2 kg and length 4 m is held in equilibrium by a hinge and a string as shown. Find the hinge force components and the tension in the string. Take g = 10 m·s^-2.

Solution. Show only forces on the rod: hinge horizontal component H, hinge vertical component V, tension T, weight 20 N at centre. Set ΣF_x=0, ΣF_y=0 and Στ_O=0 to obtain three equations. From these:

ΣF_x=0 ⇒ H - 0.8 T = 0.8 ... (i)

ΣF_y=0 ⇒ V + 0.6 T - 20 = 0 ... (ii)

Στ_O=0 ⇒ 20 × 2 = 0.6 T × 4 ... (iii) (torques of H, V and 0.8T about O are zero)

Solving gives T = 16.67 N, H = 13.33 N, V = 10 N. Hinge force magnitude F =

⇒ Ans.

Applications

Motion of a Block on a Horizontal Smooth Surface

When a block on a smooth horizontal surface is subjected to a horizontal pull F:

Vertical equilibrium gives R = mg. Horizontal equation gives F = m a ⇒ a = F / m.

When the pull acts at an angle θ above the horizontal, resolve F into F cos θ (horizontal) and F sin θ (vertical upward). Vertical equilibrium: R + F sin θ = mg ⇒ R = mg - F sin θ (so R < mg).="" horizontal="" motion:="" f="" cos="" θ="m" a="" ⇒="" a="(F" cos="" θ)="">

• Thus R ≠ mg in general.

If the push acts at angle θ downward, vertical equilibrium gives R = mg + F sin θ (so R > mg). Horizontal equation remains F cos θ = m a ⇒ a = (F cos θ) / m.

Motion of Bodies in Contact

Two-body system. Consider two blocks in contact on a frictionless surface pushed by a horizontal force F applied to mass m₁ (in contact with m₂):

Free-body diagrams (vertical forces are not shown if they do not cause motion) lead to a common acceleration

and contact force f between the blocks equals

. The acceleration of the system is a = F / (m₁ + m₂) .

If the force F acts on m₂ instead, a remains same but internal contact forces change: f′ =

.

Ques 6: Find the contact force between the 3 kg and 2kg block as shown in figure.

Ans: Considering both blocks as a single system to find common acceleration:

F_net = F₁ - F₂ = 100 - 25 = 75 N

Common acceleration:

To find the contact force between A & B draw FBD of 2 kg block. From (ΣF_net)_x = m a_x ⇒ N - 25 = 2 × 15 ⇒ N = 55 N.

Three-body system.

Free-body diagrams yield a =

, and internal contact forces

f₁ =

, f₂ = .

Here f₁ is contact force between m₁ and m₂, f₂ between m₂ and m₃. Contact forces differ if the external force is applied to different blocks.

Ques 7: Find the contact force between the block and acceleration of the blocks as shown in figure.

Ans: Considering all three blocks as a system to find common acceleration:

F_net = 50 - 30 = 20 N

For contact force between B & C consider FBD of 3 kg block:

⇒ N₁ - 30 = 3 × 2 ⇒ N₁ = 36 N.

For contact force between A & B draw FBD of 5 kg block:

⇒ N₂ - N₁ = 5 a ⇒ N₂ = 5 × 2 + 36 ⇒ N₂ = 46 N.

Motion of Connected Bodies

When bodies are connected by massless strings the tensions through the string segments determine accelerations.

Two bodies:

Free-body diagrams:

From these a =

.

Three bodies:

Free-body diagrams:

From these a =

.

Ques 8: A 5 kg block has a rope of mass 2 kg attached to its underside and a 3 kg block is suspended from the other end of the rope. The whole system is accelerated upward at 2 m/s² by an external force F₀.

(a) What is F₀?

(b) What is the force on rope?

(c) What is the tension at middle point of the rope?

(g = 10 m/s²)

Ans: Consider the whole system (both blocks + rope) as one body.

(a)

F₀ - 100 = 10 × 2 ⇒ F = 120 N. ...(1)

(b) Net force on rope of mass 2 kg: F = m a = 2 × 2 = 4 N. ...(2)

(c) For tension at the middle point, draw FBD of the 3 kg block with half the rope (mass 1 kg) attached:

T - 4g = 4 × 2 ⇒ T = 48 N.

Motion on a Smooth Inclined Plane

Natural acceleration of a block down the plane (if unconstrained) is g sin θ.

To produce acceleration a up the plane the required driving force is F = m(a + g sin θ). For acceleration a down the plane, F = m(a - g sin θ).

Ques 9: Find out the contact force between the 2 kg and 4 kg blocks as shown in figure.

Ans: On an inclined plane the acceleration of a block (if unconstrained) is independent of mass. If both blocks would move with the same acceleration (g sin 37°) they will not press against each other and the contact force is zero.

Ques 10: Find out the contact force between 2 kg & 3 kg block placed on the incline plane as shown in figure.

Ans: Treat both blocks as a single system of mass 5 kg. Represent forces on the 5 kg system and find acceleration down the plane. From Newton's second law:

Equations: N = 5 g cos 37° ...(i), 5 g sin 37° - 20 = 5 a ...(ii). Solving gives a = 2 m·s^-2 down the incline. For contact force N₁ between 2 kg & 3 kg consider FBD of 3 kg block:

3 g sin 37° - N₁ = 3 × 2 ⇒ 18 - N₁ = 6 ⇒ N₁ = 12 N.

Pulley-Block System

Ques 11: One end of a string that passes over a pulley is connected to a 10 kg mass at the other end; the free end of the string is pulled by 100 N force. Find the acceleration of the 10 kg mass (g = 9.8 m·s^-2).

Ans: Tension in the string is 100 N. FBD of 10 kg block: 100 - 10 g = 10 a ⇒ 100 - 10 × 9.8 = 10 a ⇒ a = 0.2 m·s^-2.

Ques 12: In the figure shown, find the acceleration of each block.

Ans: Draw FBD for each block and apply Newton's second law.

(1)

⇒ 10 g - 2 T = 10 a₁ ...(i)

(2)

⇒ T - 2 g = 2 a₂ ...(ii)

(3)

⇒ T - 4 g = 4 a₃ ...(iii)

Constraint: 2 a₁ = a₂ + a₃ ...(iv)

Solving (i)-(iv) gives T =

, a₁ = 70/23 m·s^-2 (down), a₂ = 170/23 m·s^-2 (up), a₃ = 30/23 m·s^-2 (down).

Ques 13: Find acceleration of each block in the figure in terms of masses m₁, m₂ and g. Neglect friction.

Ans: Let T be the tension in the massless string. For mass m₁, FBD gives N₁ = m₁ g and T = m₁ a₁. For mass m₂, m₂ g - 2 T = m₂ a₂. Constraint: a₁ = 2 a₂. Solving yields:

a₁ =

and a₂ = .

Ques 14: Two blocks A and B each of mass 20 kg rest on frictionless surfaces as shown. Pulleys are light and frictionless. (a) Time required for block A to move 2 m down the plane from rest? (b) Tension in the connecting string?

Ans: Draw FBDs. For block A: N_A = m_A g cos θ = 20 × 10 × 0.8 = 160 N. Equation along plane: m_A g sin θ - T = m_A a. For block B: N_B = m_B g = 20 × 10 = 200 N and T = m_B a.

From the two equations a =

= 3 m·s^-2 and T = m_B a = 20 × 3 = 60 N. Time to go distance S = 2 m from rest: S = 1/2 a t² ⇒ t = .

Ques 15: Two blocks m₁ and m₂ are placed on a smooth inclined plane and released from rest. Find (i) accelerations of m₁ and m₂, (ii) tension in the string, (iii) net force on the pulley by the string.

Ans: FBD of m₁: m₁ g sin θ - T = m₁ a ⇒

- T = ...(i). FBD of m₂: T - m₂ g sin θ = m₂ a ⇒ T - 1. = 1.a ...(ii). Adding (i) and (ii) gives a = 0; hence T = . FBD of pulley gives resultant string forces combined: F_R = = .

Ques 16: A block of mass m is on the ground as shown. (i) Draw FBD of block. (ii) Are forces acting on block an action-reaction pair? (iii) If not, draw the action-reaction pairs.

Ans: (i) FBD of block: weight mg downward and normal reaction N upward:

. (ii) N and mg are not an action-reaction pair because they act on different bodies. (iii) The action-reaction pair of mg is (block on Earth) -mg and (Earth on block) mg; the pair for N is (surface on block) N and (block on surface) -N. See figures:

Climbing on the Rope

When a man climbs a rope he exerts downward force on rope; tension responds. If T > mg the man accelerates upward; if T < mg="" he="" accelerates="" downward;="" if="" t="mg" acceleration="" is="" zero="" (constant="" speed="" or="" rest).="" the="" man="" always="" exerts="" a="" downward="" force="" on="" the="" rope="" when="" climbing="" or="">

Ques 17: If the breaking strength of a string is 600 N find the maximum acceleration with which a 50 kg man can climb the rope.

Ans: Maximum rope force on the man is 600 N. FBD of man: 600 - 50 g = 50 a ⇒ a_max = 2 m·s^-2.

Ques 18: A 60 kg painter stands on a 15 kg platform. A rope attached to the platform and passing over an overhead pulley allows the painter to raise himself and the platform.

(i) If he pulls the rope down with force 400 N, find acceleration of the painter and platform.

(ii) What force must he exert to attain upward speed 1 m·s^-1 in 1 s?

(iii) What force should he apply to maintain constant upward speed 1 m·s^-1?

Ans: Treat painter + platform as a system. Tension in the rope equals the applied force by the painter.

(i) Apply Newton's second law to the whole system: 2 T - (M + m) g = (M + m) a. Thus a = [2 T - (M + m) g] / (M + m). With M = 60 kg, m = 15 kg, T = 400 N, g = 10 m·s^-2 we get a = 0.67 m·s^-2.

(ii) To reach 1 m·s^-1 in 1 s, a = 1 m·s^-2. Then required applied force is F = 1/2 (M + m) (g + a) = 412.5 N (text expression simplified in figure) ⇒ approx 412.5 N.

(iii) For constant upward speed net acceleration is zero. Thus 2 F - (M + m) g = 0 ⇒ F = (M + m) g / 2 = 375 N.

Spring Force (revisited)

Every spring resists change of length with a force F = -k x. The spring force vs extension graph is linear for an ideal Hooke's spring.

Ques 19: Two blocks are connected by a spring of natural length 2 m. Spring constant k = 200 N·m^-1. Find spring force in:

(a) Both blocks displaced by 0.5 m in the same direction.

(b) Both blocks displaced by 0.5 m in opposite directions.

Ans: (a) When both are displaced equally in same direction the spring extension is zero ⇒ spring force = 0. (b) When displaced in opposite directions by 0.5 m each, total change in separation = 1.0 m; spring extension x = 1.0 - 2.0 = (depending on initial configuration) use Hooke's law F = k x. (Refer to diagram for exact geometry.)

Pseudo Force and Frames of Reference

A frame of reference is a coordinate system used to describe motion. Frames may be inertial or non-inertial.

Inertial Frame of Reference

A non-accelerating frame is inertial. Any frame moving at constant velocity relative to an inertial frame is also inertial.

Non-inertial Frame of Reference

An accelerating frame is non-inertial (for example, rotating frames are accelerating). Earth rotates and revolves, so strictly it is non-inertial; for most problems we approximate Earth as inertial.

Note: When observing motion from a non-inertial (accelerating) frame, apparent forces called pseudo forces must be introduced to use Newton's laws in that frame.

The pseudo force on a body of mass m observed from a frame accelerating with acceleration a (relative to an inertial frame) is

and acts on the body opposite to the acceleration of the frame. When constructing an FBD in a non-inertial frame, include all real forces plus the pseudo force of magnitude m a in the direction opposite to the frame's acceleration.

Example: A block of mass m on an elevator accelerating upward with acceleration a₀. In the ground (inertial) frame:

N - m g = m a₀ ⇒ N = m(g + a₀). In the elevator (non-inertial) frame include a pseudo force m a₀ downward. Since the block is at rest relative to elevator, ΣF = 0 ⇒ N - m g - m a₀ = 0 ⇒ N = m (g + a₀), consistent with the inertial calculation.

Example 8.24 All surfaces smooth. Find F such that a block remains stationary relative to a wedge accelerating horizontally under force F.

Solution can be done in inertial frame (apply real forces, block has acceleration a) or in wedge frame (include pseudo force). Both methods yield the same result: a = g tan θ and therefore F = (M + m) g tan θ. Derivation shown in the figures.

Example 8.25 A pendulum bob of mass m is suspended from the ceiling of a train accelerating horizontally with acceleration a. Find the equilibrium deflection angle θ.

Solution: Using either the ground (inertial) frame (real forces only) or the train (non-inertial) frame (include pseudo force m a horizontally opposite to train acceleration) we find tan θ = a / g. Figure and algebra give the result.