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Physics Notes
CHAPTER - 1
Units and Dimensions
Physical Quantity – If any object can be measured, then it is called a physical quantity. It is a process of comparison of two quantities:
Fundamental Physical Quantities
Derived Physical Quantities
Derived physical quantities are those which are obtained from fundamental physical quantities. They depend on the fundamental physical quantities.
Fundamental Units
Derived Units
S.I. Unit (System International Unit)
7 Basic (Fundamental) Units
Physical Quantity
SI Unit
Symbol
Time
Second
s
Mass
Kilogram
kg
Length
Metre
m
Electric current
Ampere
A
Temperature
Kelvin
K
Luminous Intensity
Candela
cd
Amount of substance
Mole
mol
2 Supplementary Units
Plane Angle
Radian
rad
Solid Angle
Steradian
sr
Physical Quantities Based On:
Plane angle is subtended at the centre of a circle. The angle θ is set to the radian when the length of the arc is equal to the radius.
If arc length AB = radius OA, then: θ = 1 radian
Unit of plane angle = radian.
Length of Arc = length of radius
A solid angle is subtended at the centre of a sphere. It is a 3-dimensional angle.
Ω = 1 steradian
When the surface area of OAB = R², then the surface area should be equal to the square of the radius.
Temperature – Degree of hotness and coldness. When heat is measurement of then this measurement of heat is called temperature.
Chapter – 1 Units and Dimension
Physical Quantity: Which quantities can measured as known as physical quantities. For example, mass, length, velocity, time, force, acceleration etc. are all physical quantities.
Measurement: Measurement is the process which consists of the comparison of an unknown quantity with a known fixed unit quantity. Every measurement whether it be a distance or weight, time or velocity or any other physical quantity required two things: first a number and second a unit or standard.
Unit: It is a standard by which we can measure physical quantities.
Ex: We say 5 metres. Here the numerical fact is 5 and the unit of length is metre. Thus the numerical value and the unit both are needed to completely express a physical quantity.
q = n u q = physical quantity n = numerical value u = unit of the quantity
Properties of a Unit: A unit selected to measure a physical quantity should possess the following properties:
(i) It should be of suitable size – neither too large nor too small. (ii) It should be well defined. (iii) It should not be variable (i.e., its value should remain same under all the circumstances & that external factors like temperature, pressure etc. cannot affect its magnitude.) (iv) It should be accepted internationally. (v) It should be easily reproducible. (vi) It should not vary from place to place.
Ex: (i) M.K.S. (metre, kg, sec.) (ii) C.G.S. (cm, gm, sec.) (iii) F.P.S. (foot, gm, sec.) is mode unit of Britain. (iv) S.I. unit (1960 world congregation, Britain)
Note: S.I. units is based on seven fundamental units and derived units.
Review Questions
1. Define unit. Ans: The standard used to measure a quantity is called its unit, and it is standard by which we can measure physical quantity. Ex = metre, second, kg, etc.
2. What are the various characteristics of units of physical quantity? Ans: Physical quantity is these quantity which can measure these are physical quantity.
3. Define physical quantity. What is difference between physical quantity and unit? Ans: Physical quantity which quantity can measured as known a physical quantity. To example; mass, length, velocity, time, force, acceleration etc. are all physical quantity.
(i) These quantity is these quantity which can measure any quantity is known as physical quantity. (ii) These quantity was physically separate. (iii) These quantity are formed in M.K.S system.
Unit
(i) It is the standard by which we can measure physical quantities.
(ii) These quantities were commonly expressed.
(iii) These are formed in C.G.S. system.
4. Define unit and give its properties.
Unit: The standard used to measure a quantity is called its unit. For example: metre, kilogram, second etc.
Its properties:
(i) It should be of a suitable size — neither too large nor too small. (ii) It should be well-defined. (iii) It should not be variable (i.e., its value should remain same under all circumstances so that external factors like temperature, pressure etc. cannot affect its magnitude). (iv) It should be accepted internationally. (v) It should be easily reproducible. (vi) It should not vary from place to place.
Classification of Units
Units can be classified in two types:
1. Fundamental Unit
(i) The units which cannot be expressed in terms of any other physical quantity are called fundamental units. (ii) They are independent of each other. (iii) Examples:
2. Derived Unit
(i) All other quantities which can be expressed in terms of other quantities are called derived units. (ii) They can be obtained by combining fundamental units.
Examples:
Example 1: Find the derived unit of force.
Force = mass × acceleration = mass × (velocity / time) = mass × (length / time × 1 / time) = (kg × m) / (sec × sec) = kg m/s² = 1 Newton
Example 2: Find the derived unit of density.
Density = mass / volume Volume = length × breadth × height
Density = mass / (m × m × m) = kg / m³ = kg/m³
1. Distinguish between fundamental and derived units. Give some examples of each.
Fundamental Unit
Derived Unit
Velocity = distance / time = m / sec
Force = mass × acceleration = mass × (length / time²) = kg m / s²
Ans: Unit: It is a standard by which we can measure physical quantities. Examples: metre, second, kilogram etc.
(i) Fundamental units are those units which cannot be resolved (expressed) from one unit to another unit. (ii) They are independent of each other. (iii) Fundamental units = (a) Mass = kilogram (kg) (b) Length = metre (m) (c) Time = second (s) (d) Amount of substance = mole (mol) (e) Temperature = kelvin (K) (f) Luminous intensity = candela (cd)
(i) Derived units are those units which can be expressed by multiplying or dividing fundamental units without any numerical value. These are known as derived units.
(ii) They can be obtained by combining fundamental units.
(iii)
Force = mass × acceleration = mass × velocity / time = mass × (length / time) / time = mass × length / time² = kg × m / s² Its derived unit is kg m/s²
Velocity = mass / volume = mass / (length × breadth × height) = kg / m × m × m = kg / m³ Its derived unit is kg/m³
Potential Energy = mgh = (mass × velocity × height) / time = mass × displacement × height / time = kg × m × m / s = kg m² / s Its derived unit is kg m²/s
Momentum = mass × velocity = mass × displacement / time = kg × m / s Its derived unit is kg m/s
Power = Work / Time = Force × Distance / Time = Mass × Acceleration × Distance / Time = Mass × Distance / Time × Time × Time = kg × m × m / s × s × s = kg·m²/s³
Moment of inertia = HOSS × (radius of gyration)² = Mass × k² = kg·m² Its derived unit is kg·m²
Impulse = Force × Time = Mass × Acceleration × Time = Mass × Velocity × Time / Time = kg × m/s = kg·m/s Its derived unit is kg·m/s
= Force × Distance = Mass × Acceleration × Distance = Mass × Distance × Velocity / Time = Mass × Distance × Displacement / Time² = kg × m × m / s² = kg·m²/s²
(i) Area = l × b = m × m = m²
(ii) Force = Mass × Acceleration = Mass × Velocity / Time = Mass × Displacement / Time × Time = kg × m / s² = kg·m/s²
(i) Force (ii) Volume (iii) Velocity (iv) Length Ans: Length is a fundamental quantity.
(i) Force (ii) Length (iii) Work (iv) Time Ans: Force is a derived unit.
There are the following systems of measurements commonly in use:
It is a system of measurement in which the fundamental units of measurement of length, mass and time are centimeter, gram and second respectively.
It is a system of measurement in which the fundamental units of measurement of length, mass and time are foot, pound and second respectively.
It is a system of measurement in which the fundamental units of measurement of length, mass and time are meter, kilogram and second respectively.
We are free to express any physical quantity in the above three systems, but intermixing is not allowed. In M.K.S unit, force will be kg·m/s². In C.G.S, force will be g·cm/s². In F.P.S, unit of force will be 1 ft·1 lb·1 s². But we cannot express force as kg·cm/s² or g·m/s² since mixing of the three systems is not allowed in physics.
It is a standard by which we can measure physical quantities. It the general conference of weight and measures held in 1960, was introduced a new and logical system of units known as System International units (S.I Units) whose abbreviated form is S. I. It consists of total nine units in which seven are basic or fundamental units and two Supplementary units along with their symbols. The list of the units is given below.
S.No
Name of Quantity
Name of units
Symbols
1.
meter
2.
kilogram
3.
S
4.
5.
kelvin
k
6.
Condela
7.
Quantity of matter
mole
| 1 | Plane Angle | Radian | rad | | 2 | Solid Angle | steradian | Sr |
Review questions
1 cm (centimeter) = 10⁻² m 1 mm (millimeter) = 10⁻³ m 1 µm (micrometer) = 10⁻⁶ m 1 nm (nanometer) = 10⁻⁹ m 1 Å (Angström) = 10⁻¹⁰ m 1 f (femtometer) = 10⁻¹⁵ m
★ Astronomical unit (A.U):– The mean distance between the centre of Sun to the centre of earth is known as A.U. Ex: 1 A.U = 1.5 × 10¹¹ m
★ Light year (L.Y):– The distance travel by light in one year through the vacuum is known as L.Y. Ex: 1 L.Y = 9.46 × 10¹⁵ m
★ Parallactic seconds (Parsec):– The distance at which arc of one A.U makes an angle of 1sec is known as par sec.
Ex: 1par.sec = 9.08X1016 m
Unit of Angle
(i) Degree 1∘ = π180 md
(ii) Minute 1′=160∘
(iii) Second 1′′=(160×60)∘
DIMENSIONAL FORMULA
If mass is represented by M, length by L and time by T, any derived unit that is known as dimensional formula.
(i) Area = length × breadth L x L = [M0 L2 T0]
(ii) Volume = length × breadth × height L x L x L = [M0 L3 T0]
(iii) Density = mass / volume M/L3 = [M1 L−3 T0]
(iv) Velocity = displacement / time L/T = [M0 L1 T−1]
(v) Acceleration = velocity / time LT−2 /T = [M0 L1 T−2]
(vi) Force = mass × acceleration M x [M0 L1 T−2], [F] = [M^1 L^1 T^{-2}][F]=[M1L1T−2]
(vii) Work = force × distance [M1 L2 T−2] x L1 = [M^1 L^2 T^{-2}], = [M1 L2 T−2]
(viii) Energy [E] = [M1 L2 T−2], [E] = [M^1 L^2 T^{-2}], [E] = [M1 L2 T−2]
(ix) Pressure = force / area [P]=[M1 L−1 T−2]/L2, [P] = [M^1 L^{-1} T^{-2}], [P] = [M1 L−1 T−2]
(x) Momentum = mass × velocity M x [M0 L1 T−2]
(xi) Impulse = force × time [I] = [M1 L1 T−1]
(xii) Power = work / time [P] = [M1 L2 T−2]/T = [M1 L2 T−3]
(xiii) Torque (couple) = force × distance [τ] = [M1 L1 T−2]x L1 = [M1 L2 T−2]
(xiv) Stress = force / area [σ] = [M1 L−1 T−2]/L2 = [M1 L−1 T−2]
(xv) Strain = change in length / original length Strain is dimensionless [M0L0T0] X [M^0 L^0 T^0] = [M0L0T0]
(xvi) Coefficient of elasticity / Modulus of elasticity = stress / strain = [M1 L−1 T−2]
(xvii) Surface tension = force / length [Ts] = [M1 L1T−2] / L1 = [M1 T−2]
(xviii) Surface energy = energy / area [Es] = [M0 L0 T−2]
(xix) Radius of gyration [k]=[L][k] = [L][k] = [L]
(xx) Moment of inertia = mass × L2 = ML2
Principle of Homogeneity of Dimensions
To check the correctness of a given equation, the physical quantities on the two sides of the equation are expressed in same dimensional terms.
Example: v = u+at
Dimensional formula of v = [LT−1] Dimensional formula of u = [LT−1] Dimensional formula of at = [LT−2]×[T1]
= [LT−1] = [LT−1] + [LT−2 x T1]
= [LT−1] = [LT−1] + [LT−1]
[LT−1] = [LT−1]
So, LHS = RHS in dimensional terms, hence the formula is correct.
2. Method: Dimensional Formula of V = [LT−1]
U = [LT−1]
At = [LT−2] x T
All terms have same diamentional formula hence the given formula in correct.
Some worked checks
[L1] = [LT−1] x T1 + ½ [LT−1]
[L1] = [L1] + ½ [LT−1]
[L1] = L1 + L2
L.H.S = R.H.S
Hence prove: this formula is correct.
[M1 L2 T−2] = [M1 L2 T−2] – [M1 L2 T−2] → correct.
[T−1] = 2 π L1/LT−2 , [T1] = [T−2] → not correct.
[T1] = under root L1/ LT−2]
[T1] = [T1] → dimensions of LHS and RHS same → correct.
[M L−1 T−2] = [L1] [ML−3] [LT−1]
[M L−1 T−2] = [M L−1 T−2] LHS = RHS → formula correct.
Angle
Arc / Radius = L1 / L1 = 1 → angle in radians is dimensionless.
Angular momentum = linear momentum × radius L = mv x r [L] = [M1 L2 T−1] x [L] = [M^1 L^2 T^{-1}] = [M1 L2 T−1]
Planck’s Constant (h)
Energy E = hν,
h = E / V
[E] = [M1 L2 T−2] / T−1 [h] = [M L2 T−1], v = frequency
Universal Gravitational Constant (G)
From Newton’s law of gravity:
F x m1 m2 -------1 and F x 1/r2 ------- 1
From equation 1 and 2
F x m1 m2 / r2
F = Gm1m2 / r2
When G is a constant known as universal gravitational constant.
G m1 m2 = F x r2
[G] = F x r2 / [m1m2] = M 2T−2 x L2 / M x M [G] = [M−1 L3 T−2]
Centripetal Force
Centripetal force = mass x (velocity)2 / Radius
F = M [LT−1] / L [F] = [M1 L1 T−2]
Coefficient of Viscosity (η)
η=force × distance / area×velocity
[η] = [M1 LT−2 L1] / L x LT−1 = [M1 L−1T−1]
Application: Derive the relation among various physics quantities.
Q2. Obtain an expression for centripetal force required to move a body of mass m with velocity v in a circle of radius r.
Assume F ∝ ma bvb rc
F x = Kma vb rc
[M L T−2] = K [LM]a [LT−1]b [L]c
[M L T−2] = K [Ma Lb T−b x Lc]
[M L T−2] = K [Ma Lb+c T−b]
Through dimensional comparison (equating exponents of M, L, T) you get: a=1, b=2, c=−1a = 1,\; b = 2,\; c = -1a=1,b=2,c=−1
So F=kmv2/rF
Q3. Velocity of sound through a medium depends on density ρ\rhoρ of the medium and modulus of elasticity E.
Assume v ∝ ρa Eb
v = K Pa Eb ------- 1
[M0 LT−1] = K [ML−3]a [ML−1 T−2]b
[M0 LT−1] = K [Ma L−3a] x [Mb L−b T−2b]
[M0 LT−1] = K [Ma+b L−3a+b T−2b]
Equation both side
A+b = 0, a = -b, a = -1/2
-3a + b = 1, -3a = -b, a = b/3, a = ½ x 3, a = 1/6
-2b = -1, b = ½
P from equation 1
V = K [P-1/2 E1/2]
V = K [E1/2] / P1/2
V = K under E/P
Review Question
Check the accuracy of F = GMM / r2
When F is the fore, G is gravitational constant, Mandm and masses and r is radius.
[M1 L−1 L−2] = [M−1 L3 T−2] [M1] [M1] / [L] 2
[M1 L−1 L−2] = [M1 L3 L−2] / [L2]
[M1 L1 L−2] = [M1 L1 L−2]
Dimensional check shows LHS and RHS both have → formula is correct.
1. LHS = RHS This is the correct formula.
2. Show S = 1/2 gt is incorrect. [L1] = [LT−2] = [LT1] [L1] = [LT−1] RHS ≠ LHS So, this equation is incorrect.
3. Show t = 2π√(t/g) is incorrect. [T1] = [T1] / [LT−2] [T1] = [LT−3] LHS = RHS This formula is correct.
4. Check the correlation of the following relation by dimensional analysis. (a) s = ut + ½ at2 [L1] = [L1 T1] [T1] + [LT−2] [T2] [L1] = [L1] [L1] RHS = LHS This is correct formula.
(b) T = h / mv [L1] = [L1]/ [M1] [LT−2] [L1] = [M−2 T1] RHS = LHS This is correct formula.
(c) t = 2π√(t/g)
[T1] = √L1 / LT−2
[T1] = √L2
[T1] = [T1]
L.H.S = R.H.S this is correct formula
Application: To convert 1 unit into another, m2 = m1 [m1 / m2]a [L1 / L2]b [T1 / T2]c
Q.1 Convert one newton (kg m/s²) into dyne (g cm/s²). A newton is a unit of force, hence dimensional formula of force is [M L T]. Hence a = 1, b = 1, c = -2.
N2 = 1[Kg/g]1 [m/cm]1 [s/s]2
[103 g /g] [102cm/cm] Result: N = 10⁵ dyne
Review Questions:
Q.2 Convert a work of 1 Joule into energy. Joule is a unit of energy, hence the dimensional formula of energy is [M L² T−2].
N = 1, a = l, b = 2, c = -2 n₂ = n₁ [m1/m2]a [L1/L2]b [T1/T2]c
1 = [m1/m2]1 [L1/L2]2 [T1/T2]−2
[Kg/g] [m/cm]2 [1] −2 Result: 10³ x 10⁴ = 10⁷ erg.
Q.1 Convert a work of 1 Joule into energy. Joule is a unit of energy, hence the dimensional formula of energy is [M L² T−2].
[Kg/g] [m/cm]2 [sec/s] −2 Result: 10³ x 102 = 105 erg.
Q.2 Express a pressure of 448 pound per sq. inch into dyne per sq. cm by Dimensional analysis. Pressure = 448 pounds per square inch into dyne per sq.cm.
= 4482/ inch² --------- dyne/cm2 Dimensional formula of Planck’s constant = Force/Distance = MLT−2/l = [MT−2] A² = 1, b = 0, c = -2 n² = n1, [m1/m2]a [L1/L2]b [T1/T2]C
448 [pound/gm]1 [1foot/1cm]0 [s/s]2 448 x 4536 = 203212.8 g = 9.8 m/s² g = 9.8 m/s² = km/m² g = 9.8 x 1 x 60 x 60 x 60 /100 352.8 = 35.28
V = 332 m/s = -----------------------km/hr. a = 0, b = 1, c = -1, n=332 n2 = n1, [m1/m2]a [L1/L2]b [T1/T2]c 332 [m1/m2]0 [m/km]1 [s/hr-2] 332 x 1 x [m/1000m] [s/3600s] -1
332 x 1 / 1000 x 3600
= 5976 / s = 11.95 02
Chapter 10
Force and Motion Force = A push or pull on a body, due to which it changes or tends to change the state of the body and force is a vector quantity.
Basic Forces in Nature The following four basic forces are operating in nature:
F ∝ 1 / r2 ------- 2
From 1 and 2
F ∝ m1 m2 / r2 F = G m₁ m₂ / r² Where G is a constant known as the universal gravitational constant. G = 6.67 x 10⁻¹¹ N m² kg⁻²
Its Properties
Note:
Resolution of Force and Rectangular Components: The resolution of a vector (force) into two mutually perpendicular components is called the rectangular resolution of a vector and the forces obtained are known as components.
Using Pythagoras theorem:
X2 + y2
= (Cosθ)2 + (Fsinθ)2
F2sin2Q + cos2Q = F2
X2 + y2 = F2
In ∆AOAB sinα=y/F y=Fsin y=10sin30∘ y=10×0.5 y=5
x=Fcos x=10× under root 3/2 =5under root 3 5×1.732=8.66N
x2+y2 =(5)2+(5 root3)2 =25+75 = 100
F = 100N
X = 50
Using Pythagoras’ theorem: F2 = y2 + x2 100 = y2 + 2500
1000 - 2500 = y2
7500 = y2
Root 7500 = y y = 86.60N
X = Vcosθ 250 = Vcos30∘
250 / root 3/2 = V 250 x 2/root3 = v
V = 500/root 3
500/1.73 = 288.67
For the vertical component: y=Vsin30∘
y=288.67×0.5=144.335km/h
Parallelogram Law: It state that if a particle is subjected to two forces represented in magnetic and direction by the two adjustment sides of a parallel gram drawn from a point, the resulting is presented in magnitude and direction was the diagonal of the parallel gram passing through that same point.
Let two forces P and Q acting at a point Q on a body are represented by OB and OA in the figure. The resultant by these forces will be represented by the diagonal OC(R), and the angle ∠AOB=α.
To find the magnitude of resultant: BC=BA=Q ∠CBD=∠LAOB=o In △BDC, cosα=BD\Q
BD cosQ = Bx,
Also, sinQ = CD/Q
CD = Q sinQ
Now, int ODC = (OC)2 = (OD)2 + (CD)2
DC = R cosQ
CD = R sinQ
Now using the law of cosines: R2=p2+Q2 (cos2α + sin2Q + 2 pcosQ) R= root p2+ Q2 + 2p cosQ
This is the explanation of magnitude of resultant vectors.
Direction: Let resultant makes an angle 'a' with a force 'P'. In right angle triangle ODC.
tan a = CD/OD = (1)
tan a = BC * sin a / (OB + BC * cos a)
tan a = a * sin a / (P + a * cos a)
Case I: When the two forces P and Q are acting in the same direction.
i.e. a = 0°. R = √(P² + Q² + 2Pa * cos 0°) R = √(P² + Q² + 2PQ cosQ)
√(P² + Q² + 2PQ) = √ (P + Q)2 = PQ tan a = Q * sin 0° / (P + Q * cos 0°) tan a = 0 / P+Q a = 0° or 180°
Case 2: When the two forces P and Q are perpendicular to each other. a = 90°.
∴ R = √(P² + Q² + 2P * Q * cos 90°) R = √(P² + Q²) tan a = Q * sin 90° / (P + Q * cos 90°) tan a = Q / T
Case III: When the two forces are in opposite direction.
a = 180°. R = √(P² + Q² + 2Pa * cos 180°) R = √(P² + Q² - 2PQ)
= √(P + Q)² = (P - a)
tan a = Q * sin 180° / (P + Q * cos 180°) = O/P-Q tan a = 0 a = 0° or 180°
Two forces of 10N and 6N at upon a body. The direction of force are unknown. The direction of forces upon the body may lie between which limits.
P = 10N, a = 6N Maximum resultant force: R_max = P + Q = 10N + 6N = 16N
Minimum resultant force: R_min = P - Q = 10N - 6N = 4N
Problem:
Two forces of 50N and 30N are exerted to each other and angle of 60° find the resultant. P = 50N, a = 30N, a = 60° R = √(P² + a² + 2Pa * cos a) R = √(50² + 30² + 25030*cos 60°) = √(2500 + 900 + 1500) = √4900 R = 70N
tan a = sin Q / (P + Q * cos Q) = sin 60° / (50 + 30 * cos 60°) = √3/2 / (50 + 30 * 1/2) = √3 / 80 = 0.3997
Q: Resultant of two equal forces acting at right angles to each other. Find the magnitude of the resultant.
Given: P = X θ = 90°
Solution: R=√P2 + Q2 + 2PCos 1414 = √x2 + x2 + 2x2 cos 90 1414 = √2x2 + 2x2 x 0
1414 = √2x2 = x√2 1414/√2 = 1414/1.414 x 1000 = 1000N
Limitations of the parallelogram method: This method is only applicable of the resultant of two forces from more than two forces. This method is not applicable for more than two forces.
Methods of Components: This method is used to find the resultant of more than two forces.
Rx=F1cosθ1+F2cosθ2+F3cosθ3 Ry=F1sinθ1+F2sinθ2+F3sinθ3 Rz=√efx2 – egy2
Q: Find the resultant force of 15 N and 10 N acting at angles 20° and 60° respectively with the horizontal.
Solution: Rx=15cos30∘+10cos60∘ Ry=15sin30∘+10sin60∘
Rx=15×√3/2+10×1/2 Rx=25×√3/2 Rx=12.97+5=17.97
Ry=15×1/2+10×3/2 Ry=7.5+8.66=16.10
Resultant force: R=√efx2 + efy2 R=√ (17.97)2 + (16.10)2 R=√582.13+259.22R R=24.12 NR
Q: A man using a 70 kg roller on a level surface exerts a force of 19 N at 45° to the ground. Calculate the vertical force on the ground (a) if he pulls, (b) if he pushes the roller.
M = 70kg
Q = 45∘
F = 196N
(a) When he pulls the roller: The vertical force on the ground = Weight of roller's vertical component when pulled.
70×9.8=196 sin 45∘ Vertical component: 70 x 9.8 -196sin45∘
686 – 138.6 = 547.4N
(b) When he pushes the roller: The vertical force on the ground = Weight of roller's vertical component when pushed.
70 x 9.8 + 196sin45∘= 826.6 N
Concurrent: If a number of forces act at the same point, they are called concurrent forces.
Equilibrium of Force: If a number of forces act at the same point, they are called equilibrium forces.
Lami’s Theorem: It states that if three forces acting at a point are in equilibrium, then each force is proportional to the sine of the angle between the other two forces is directly proportional to the sine of the angle between the two P/sinθ = Q/sinβ = R/sinγ
Proof: We know that if these forces, P, Q, and R are in equilibrium, then these forces can be completely represented by the three sides BE, CA, and AD as shown in the figure.
From triangle of forces, we have From trigonometry, we have:
BC/sin(180-x) =CA/sin(180° - β)
BC/sin x CA/sin β = AB/sin r
From equation (i):
ρ/sina =Q/sinβ = R/sin r
Law of Motion
(i) It states that if a body is in rest position, then it will be continuously in rest and if it is in uniform motion, then it will be continuous in its state of motion continuously, and there is no external force change in first state.
Inertia: The tendency of a body to remain in its state of motion or rest due to the fact that a body by itself is unable to change its position is called inertia.
Types of Inertia: (i) Inertia of rest: The property by which the body is unable to change itself from the rest position.
Example: When a bus suddenly starts moving, the passengers bend backward. Initially, the passengers are in rest position with the bus. When the bus suddenly starts moving with the bus due to inertia of rest, the upper position wants to go to its rest position and hence they bend then.
(ii) When a car (bicycle) is battered with a stick, the dust particles fly down.
(iii) When the branches of shaken tree faults for touch.
Inertia of motion: When a moving bus suddenly stops, the passengers bend forwards. Example: When a fast-moving horse suddenly stops, the riders fall forwards.
Inertia of Direction: (i) When a bus suddenly turns, the passengers bend in the opposite direction. (ii) The mud comes out of the wheel of a cycle tangentially.
MASS AND WEIGHT Mass: Mass of a body is the quantity of matter contained in it. Weight: Weight of the body is the gravitational force with which the Earth attracts the body downward. Its unit is the same as that of the body.
Principal of conservation of linear momentum: It states that if no external force acts on a system, the momentum of the system remains constant if P and P2 are intent then is the absence of external force P1 = P2 (constant)
If due to Newtonian interaction, the momentum of the body becomes P1 P2 that according to the law of conservation of linear momentum P1+P2 = P12 + P12
Impulse: The force which acts for a very short time but produces an appreciable change of momentum is called impulse. F = mv – mv / t Ft = mv = total change momentum.
Parabolic Motion – Projectile: A body when projected in air, it covers a parabolic path. It's known as a projectile and the path following is called trajectory.
Circular Motion: Consider a body moving in a circle of radius r, starting from an angle described by a body moving in a circle is called angular displacement.
Chapter 11: Work, Energy, and Power When an object is pushed or pulled in the direction of force, then it is said to be work. It is the product of force and displacement. Work is done when: (i) A force is acting on a body. (ii) The body moves through some distance in the direction of the force.
1. How can we say that work is done? Work is said to be done when a force moves through a distance in the direction of the force and is measured by the product of the force and the distance moved in the direction of the force. Work = F x d
Case I: If a force F is acting along the direction of motion, then the force F displaces a body through displacement d in the direction of force. Hence, the work is done: W = F.S
F & S = magnitude of force vector.
S = displacement vector in the direction of force.
CASE II: Work Done by a Force:
W = Fx⋅s = (Fcosθ)⋅s or w = Fcosθ
Dimension Formula of Work:
If force is expressed in Newtons and distance in meters, the unit of work is Newton meter (Nm), and 1 Joule = 1 Newton × 1 meter.
In CGS units, 1 erg = 1 dyne × 1 cm.
Law of Conservation of Energy:
Conservation of Energy for a Freely Falling Body:
Now, if the body is allowed to fall freely,
Now if body reaches at point B, P.E of the body = mg(h – x) = mgh – mgx
K.E of the body = ½ m v_B² = ½ m (2gx) = mgx
∴ Total energy at B = K.E + P.E = mgx + mgh – mgx = mgh
Let the body now reaches at the L.O.
K.E of the body = ½ m v_C² = ½ m (2gh)
∴ Total Energy = K.E + P.E = mgh + 0 = mgh
In lifting a mass, work is done against the gravity. For example, if a body of mass m is lifted to a height h, the work is done on the body to move it vertically up through a distance h against the gravity attraction force mg i.e.
Consider a body of mass m being pulled up by a force F along an inclined plane AB as shown. In this case the body is moving along the direction of force F.
Forces acting on the body are: (i) Weight mg of the body acting vertically downward. (ii) Normal reaction R. (a) mg cosθ which balances the normal reaction R. (b) mg sinθ downward along the plane.
As is clear, from the minimum forces required to move the body up the plane, F = mg sinθ
Hence the work done in moving the body from A to B: W = F × AB W = mg sinθ × AB
Now h = sinθ × AB or sinθ × AB = h
∴ W = mgh
The body is under the action of the following forces: (a) Weight mg of the body acting downwards. (b) Normal reaction R acting vertically. (c) Applied force F.
Now frictional force = F = μmg
∴ Force required F = μmg If body covers a distance S, then work done W = F × S = (μmg) S.
Various forces acting on the body are: (i) Weight mg of the body acting vertically downward. (ii) Normal reaction R. (iii) Force of friction acting along the plane downward. (iv) Force applied F.
(a) mgcosθ which balances normal reaction
R = mgcosθ
(b) mgsinθ acting downwards along the plane
Net force acting downward:
F = mgsinθ
= Umgcost + mgsinθ
Work done:
W = FS = (cosθ+sinθ)mgs.
The energy possessed by a body by virtue of its motion is called kinetic energy.
Examples from daily life of bodies possessing kinetic energy: (i) The kinetic energy of air is used to run windmills. (ii) A bullet fired from a gun can pierce a target due to its kinetic energy. (iii) The kinetic energy of a hammer is made use of in driving a nail into a piece of wood.