Force Notes

Force, Work, Power and Energy

Force

• When a force is applied to a rigid body, it can cause motion.
• When applied to a non-rigid body, it can change its size, shape, and cause motion.
• Mathematically, force is the rate of change of linear momentum: $\vec{F} = \frac{d(m\vec{v})}{dt} = \frac{d\vec{p}}{dt}$ or $\vec{F} = m\vec{a}$ (if mass m is constant).
• Force is a vector quantity.
• The S.I. unit of force is the newton (N) or kilogram-force (kgf).
• $1 \text{ kgf} = g \text{ N}$, where $g$ is the acceleration due to gravity (average $9.8 \text{ m/s}^2$).

Moment of a Force and Equilibrium

Translational and Rotational Motions

• A rigid body acted upon by a force can undergo:
  1. Linear or translational motion.
  2. Rotational motion.

Linear or Translational Motion:

• When a force acts on a stationary rigid body free to move, it moves in a straight path in the direction of the force.
• Example: Pushing a ball on the floor.

Rotational Motion:

• If a body is pivoted and a force is applied suitably, it rotates about the axis passing through the pivot.
• This turning effect of the force is called rotational motion.
• Example: A wheel pivoted at its center with a tangential force applied on its rim rotates about its center.
• Example: Applying force normally on a door handle causes the door to rotate about the hinges.

Moment (Turning Effect) of a Force or Torque

• Consider a body pivoted at point O.
• If a force $F$ is applied horizontally, the body rotates about the vertical axis through point O.
• Moment of force about the axis through point O is given by: $\text{Moment of force} = \text{Force} \times \text{Perpendicular distance from point O}$
• $\text{Moment of force} = F \times OP$
• For maximum turning effect, apply the force at a point where the perpendicular distance from the axis of rotation is maximum.

Factors Affecting the Turning of a Body

1. Magnitude of the force applied.
2. Distance of the line of action of the force from the axis of rotation (or pivoted point).

• The turning effect depends on the product of these factors, called the moment of force (or torque).
• The turning effect on the body about an axis is due to the moment of force (or torque) applied on the body.

Measurement of Moment of Force (or Torque)

• The moment of a force (or torque) equals the product of the magnitude of the force and the perpendicular distance of the line of action of the force from the axis of rotation.

Units of Moment of Force

• Unit of moment of force = unit of force × unit of distance
• The S.I. unit of force is newton (N), and that of distance is meter (m), so the S.I. unit of moment of force is newton-meter (Nm).
• The C.G.S. unit of moment of force is dyne × cm.
• If force is measured in gravitational units:
  • S.I. system: kgf × m
  • C.G.S. system: gf × cm
• Relations:
  • $1 \text{ Nm} = 10^5 \text{ dyne} \times 10^2 \text{ cm} = 10^7 \text{ dyne cm}$
  • $1 \text{ kgf} \times \text{m} = 9.8 \text{ Nm}$
  • $1 \text{ gf} \times \text{cm} = 980 \text{ dyne cm}$

Clockwise and Anticlockwise Moments

• Anticlockwise moment: The effect on the body is to turn it anticlockwise and is taken as positive.
• Clockwise moment: The effect on the body is to turn it clockwise and is taken as negative.
• The moment of force is a vector quantity.
• The direction of the anticlockwise moment is along the axis of rotation outwards.
• The direction of the clockwise moment is along the axis of rotation inwards.
• The direction of rotation depends on:
  • The point of application of the force.
  • The direction of the force.

Common Examples of Moment of Force

1. Opening or shutting a door: Applying force normal to the door at the handle, which is at the maximum distance from the hinges.
2. Hand flour grinder: The upper circular stone has a handle near its rim for easy rotation with a small force.
3. Steering wheel: Applying a tangential force on the rim to turn the wheel.
4. Bicycle: Applying a small force on the foot pedal of a toothed wheel larger than the rear wheel.
5. Spanner: A long handle produces a large moment of force by a small force applied normally at the end.

• Conclusion: Turning a body depends not only on the magnitude of the force but also on the perpendicular distance of the line of action of the applied force from the axis of rotation. The larger the perpendicular distance, the less force is needed.

Couple

• A single force on a pivoted body alone does not cause rotation.
• Rotation results from a pair of forces.
• The force of reaction at the pivot is equal in magnitude but opposite in direction to the applied force.
• Two equal and opposite parallel forces not acting along the same line form a couple.
• A couple is always needed to produce rotation.
• Examples: opening a door, turning a water tap, tightening a cap, turning a key, winding a clock, steering a car, driving a bicycle pedal.

Moment of Couple

• Consider a bar AB pivoted at point O. Equal and opposite forces F are applied at ends A and B.
• The perpendicular distance between the forces is AB = d (couple arm).
• The forces cannot produce translational motion, but each force has a turning effect on the bar in the same direction.
• Moment of force F at end A = $F \times OA$ (anticlockwise)
• Moment of force F at end B = $F \times OB$ (anticlockwise)
• Total moment of couple = $F \times OA + F \times OB = F \times (OA + OB) = F \times AB = F \times d$
• $\text{Moment of couple} = \text{Either force} \times \text{perpendicular distance between the two forces (couple arm)}$

Equilibrium of Bodies

• When a force acts on a body, it produces translational or rotational motion.
• It's possible to apply multiple forces such that:
  • Resultant of all forces is zero (no change in rest or motion).
  • Algebraic sum of moments of all forces about the fixed point is zero (no change in rotational state).
• Then the body is in equilibrium.
• When a number of forces acting on a body produce no change in its state of rest or of linear or rotational motion, the body is said to be in equilibrium.

Kinds of Equilibrium

1. Static equilibrium.
2. Dynamic equilibrium.

Static Equilibrium

• When a body remains in the state of rest under the influence of several forces, it is in static equilibrium.
• Examples:
  • A body on a table pulled by equal and opposite forces remains at rest.
  • A book lying on a table with its weight balanced by the reaction force from the table.
  • A beam balance when balanced in a horizontal position.

Dynamic Equilibrium

• When a body remains in the same state of motion (translational or rotational) under the influence of several forces, it is in dynamic equilibrium.
• Examples:
  • A raindrop reaching the Earth's surface with a constant velocity.
  • An airplane moving at a constant height.
  • A stone tied at the end of a string whirled in a circular path with uniform speed.
  • Motion of a planet around the sun or a satellite around a planet.

Conditions for Equilibrium

1. The resultant of all the forces acting on the body should be equal to zero.
2. The algebraic sum of moments of all the forces acting on the body about the point of rotation should be zero (sum of anticlockwise moments = sum of clockwise moments).

Principle of Moments

• When several forces act on a pivoted body, they tend to rotate it about an axis through the pivot.
• The resultant moment of all forces about the pivoted point is obtained by taking the algebraic sum of the moment of each force about that point (anticlockwise is positive, clockwise is negative).
• Principle of moments: If the algebraic sum of moments of all the forces acting on the body about the axis of rotation is zero, the body is in equilibrium.
• In equilibrium, sum of the anticlockwise moments = sum of the clockwise moments.
• A physical balance (or beam balance) works on this principle.

Verification of the Principle of Moments

1. Suspend a meter rule horizontally from a fixed support by a thread at point O.
2. Suspend two spring balances A and B with slotted weights $W<em>1$ and $W</em>2$ on either side of the thread.
3. Adjust either the slotted weights or the position of the spring balances until the meter rule is horizontal.
4. Let $W<em>1$ be the weight suspended on the right at distance $l</em>1$, and $W<em>2$ be the weight on the left at distance $l</em>2$.
5. Clockwise moment of weight $W<em>1$ about point O = $W</em>1 \times l_1$
6. Anticlockwise moment of weight $W<em>2$ about point O = $W</em>2 \times l_2$
7. In equilibrium, $W<em>1 l</em>1 = W<em>2 l</em>2$, which verifies the principle of moments.

Examples

1. A body is pivoted at a point. A force of 10 N is applied at a distance of 30 cm from the pivot. Calculate the moment of force about the pivot.
   • Given: $F = 10 \text{ N}$, $r = 30 \text{ cm} = 0.3 \text{ m}$
   • Moment of force = $F \times r = 10 \times 0.3 = 3 \text{ Nm}$
2. The moment of a force of 5 N about a point P is 2 Nm. Calculate the distance of the point of application of the force from the point P.
   • Given: Moment of force = 2 Nm, $F = 5 \text{ N}$
   • $2 = 5 \times r$ => $r = \frac{2}{5} = 0.4 \text{ m}$
3. A mechanic can open a nut by applying a force of 150 N while using a lever handle of length 40 cm. How long handle is required if he wants to open it by applying a force of only 50 N?
   • First case: $F = 150 \text{ N}$, $r = 40 \text{ cm} = 0.4 \text{ m}$
   • Moment of force needed = $150 \times 0.4 = 60 \text{ Nm}$
   • Second case: $F = 50 \text{ N}$
   • $50 \times L = 60$ => $L = \frac{60}{50} = 1.2 \text{ m}$
4. The iron door of a building is 3 m broad. It can be opened by applying a force of 100 N normally at the middle of the door. Calculate: (a) the torque needed to open the door, (b) the least force and its point of application to open the door.
   • (a) $F = 100 \text{ N}$, $r = \frac{1}{2} \times 3 \text{ m} = 1.5 \text{ m}$
   • Torque needed = $100 \times 1.5 = 150 \text{ Nm}$
   • (b) Least force at the farthest point (3 m):
   • $F' \times 3 = 150$ => $F' = \frac{150}{3} = 50 \text{ N}$
5. The wheel has a fixed axle through O. It's stationary under a horizontal force $F<em>1$ at A and a vertical force $F</em>2$ at B.
   • (a) Direction of $F<em>2$ is downward to balance the clockwise moment of $F</em>1$.
   • (b) $F_2$ is greater because its perpendicular distance from O is smaller.
   • (c) $F<em>1 \times OA = F</em>2 \times OO'$ => $\frac{F<em>2}{F</em>1} = \frac{OA}{OO'} = \frac{2.5}{2.0} = \frac{5}{4}$
6. Two parallel and opposite forces $F<em>1$ and $F</em>2$, each of magnitude 5 N, are separated by 2 m. Point X is midway, and point Y lies on $F_2$.
   • (a) (i) Moment about X = $5 \times 1 + 5 \times 1 = 10 \text{ Nm}$ (clockwise)
   • (ii) Moment about Y = $5 \times 2 + 5 \times 0 = 10 \text{ Nm}$ (clockwise)
   • (b) Both X and Y experience clockwise rotation.
7. Two forces of 2 N act vertically upwards and downwards at the ends of a 1 m rod pivoted at its center.
   • OA = OB = 0.5 m
   • Moment of force at A = $2 \times 0.5 = 1 \text{ Nm}$ (clockwise)
   • Moment of force at B = $2 \times 0.5 = 1 \text{ Nm}$ (clockwise)
   • Total moment = 2.0 Nm (clockwise).
8. A meter rule balances horizontally on a knife edge at the 60 cm mark when a mass of 10 g is suspended from one end.
   • (a) Mass must be suspended at the 100 cm mark to balance the anticlockwise moment.
   • (b) Mg × (60 - 50) = 10g × (100 - 60)
   • M × 10 = 10 × 40 => M = 40 g.
9. On a seesaw, two children of masses 30 kg and 50 kg are sitting on one side at distances 2 m and 2.5 m from the middle. Where should a man of mass 74 kg sit to balance it?
   • Total anticlockwise moment = $30 \times 2 + 50 \times 2.5 = 60 + 125 = 185 \text{ kgf m}$
   • Clockwise moment = 74 × x
   • $74x = 185$ => $x = \frac{185}{74} = 2.5 \text{ m}$
10. A meter rule AB pivoted at end A (0 mark) is supported at end B by a spring balance when a weight of 40 kgf is suspended at the 40 cm mark. Find the reading of the spring balance when the rule is of (i) negligible mass, (ii) mass 20 kg.
    • (i) $40 \times 40 = F \times 100$ => $F = \frac{40 \times 40}{100} = 16 \text{ kgf}$
    • (ii) $40 \times 40 + 20 \times 50 = F \times 100$ => $F = \frac{40 \times 40 + 20 \times 50}{100} = 26 \text{ kgf}$

Centre of Gravity

• The gravitational force attracts every particle towards the Earth's center (weight w).
• A body consists of particles with weights $w<em>1, w</em>2, w_3, \dots$
• These parallel forces can be replaced by a single resultant force W (total weight).
• The weight W acts at point G (center of gravity), such that the algebraic sum of moments due to weights $w<em>1 , w</em>2 \dots$ about G is zero.
• The center of gravity (C.G.) of a body is the point about which the algebraic sum of moments of weights of all the particles constituting the body is zero. The entire weight can be considered to act at this point.

Notes on Center of Gravity

• The position depends on the shape, i.e., mass distribution.
• It changes if the body is deformed.
• The center of gravity is not always within the material of the body.
• A body of weight W can be considered as a point particle of weight W at its center of gravity.

Center of Gravity of Some Regular Objects

• Rod: Mid-point.
• Circular disc: Geometric center.
• Solid or hollow sphere: Geometric center.
• Solid or hollow cylinder: Mid-point on the axis.
• Solid cone: Height h/4 from the base on its axis.
• Hollow cone: Height h/3 from the base on its axis.
• Circular ring: Center of ring.
• Triangular lamina: Intersection of medians.
• Parallelogram, rectangular lamina, square, rhombus: Intersection of diagonals.

Center of Gravity and the Balance Point

• A solid body can be balanced by supporting it at its center of gravity.
• Example: A uniform meter rule balanced at the 50 cm mark.
• The algebraic sum of moments of the weights of all particles about the knife edge (or finger tip) is zero.
• If a body is freely suspended, it comes to rest with its center of gravity vertically below the point of suspension.

Determination of Center of Gravity of an Irregular Lamina

1. Make three fine holes (a, b, c) near the edge of lamina A.
2. Suspend the lamina from hole a with a plumb line.
3. Draw a straight line ad along the plumb line.
4. Repeat by suspending from holes b and c, drawing lines be and cf.
5. The intersection of lines ad, be, and cf is the center of gravity G.

Uniform Circular Motion

Uniform Circular Motion

• When a particle moves with a constant speed in a circular path, its motion is said to be the uniform circular motion.
• The particle travels equal distance along the circular path in equal intervals of time, so the speed of the particle is uniform.
• The direction of motion of the particle changes at each point of the circular path.
• The continuous change in direction of motion implies that the velocity of the particle is non-uniform (or variable), i.e., the motion is accelerated.

Direction of Velocity at Any Instant in Circular Path

• The direction of motion is along the tangent drawn at that point of the circular path.
• At any point, the direction of motion is along the tangent drawn at that point of the circular path.
• The velocity of particle in circular motion is variable or the circular motion is accelerated even though the speed of particle is uniform.

Difference Between Uniform Circular Motion and Uniform Linear Motion

• In uniform linear motion, the speed and velocity are constant and acceleration is zero, i.e., the uniform linear motion is an unaccelerated motion.
• In a uniform circular motion the velocity is variable (although speed is uniform), so it is an accelerated motion.

Centripetal and Centrifugal Force

Centripetal Force

• For circular motion, a force is needed, which is termed as the centripetal force.
• At each point of circular path, this force is directed towards the centre of the circle.
• The direction of acceleration also changes at each point of the circular path, but its magnitude remains the same i.e.,the acceleration is variable (or non-uniform).

Examples:

1. An electron moving around the nucleus in an atom.
2. A planet moving around the sun.
3. The moon moving around the earth.
4. A stone tied at the end of a string whirled in a circular path.

Centrifugal Force

• A force acting on a body away from the centre of circular path is called the centrifugal force.
• The centrifugal force is in a direction opposite to the direction of centripetal force. Its magnitude is the same as that of the centripetal force.
• Centrifugal force is not the force of reaction of the centripetal force because action and reaction do not act on the same body.
• It is not the real force, but is a fictitious force assumed by an observer moving in circular path along with the body.
• Centrifugal Force != Real Force.
• Force which really does not exist != Fictitious Force (or Virtual Force).