If a force acting on a body of mass 1 kg produces an acceleration of 1 ms^-2, then the force is said to be 1 newton. 1 N = 1 kg ms^-2.

(i) It is measured by the product of anyone of the forces (F) and the perpendicular distance between two forces (d). ፒ = F x d.
(ii) The turning effect of couple is measured by the magnitude of its moment.
(iii) Moment of the couple = Force x perpendicular distance between the force = F x S.
(iv) The unit of moment of the couple is newton meter (Nm).

(i) Torque is a vector quantity. It is acting along the direction which is perpendicular to the plane of action of force and distance.
(ii) Its SI unit is N m.
(iii) Examples : Turning a tap, winding or S unwinding a screw, spinning of a top etc.

(i) When a number of like or unlike parallel forces act on a rigid body and the body is in equilibrium, then the algebraic sum of the moments in the clockwise direction is equal to the algebraic sum of the moments in the anti-clockwise direction.
(ii) Moment in clockwise direction = Moment in anticlockwise direction.
(iii) F₁ x d₁ = F₂ x d₂.

Let, F₁ = 5 N
F₂ = 15 N
F_net = F₂ - F₁            [∵ F₂ > F₁]
= 15 - 5 = 10 N
F_net i.e, the resultant force acts along the direction of the greater force 15 N.

(i) Let M be the mass of the Earth and m be the mass of the object.
(ii) The entire mass of the earth is assumed to be concentrated at its centre. The radius of the earth is R (= 6378 km = 6400 km approximately).
(iii) By Newton's law of gravitation, the force acting on the object is given by
F = G M m / R²  ............. (A)
(iv) According to Newton's second law, the force acting on the object is given by the product of its mass and acceleration. Here acceleration of the is under action of gravity hence a =g.
F=ma =mg
F = weight = mg
Comparing equations (A) and (B), we get
g= GM/ R²
Acceleration due to gravity
g=\(\frac { GM }{ { R }^{ 2 } } \).

Given: The ratio of masses of two planets \(M_{1}: M_{2} \) is 2: 3 The ratio of their radii \(R_{1}: R_{2}\) is 4: 7
\(g_{1}: g_{2}= ? \)
\(\mathrm{g}=\frac{\mathrm{G \times M}}{\mathrm{R}^{2}} ; \quad \mathrm{g}_{1}=\frac{\mathrm{G\times M}_{1}}{\mathrm{R}_{1}^{2}} ; \quad \mathrm{g}_{2}=\frac{\mathrm{G \times M}_{2}}{\mathrm{R}_{2}^{2}} \)
\(\frac{g_{1}}{g_{2}}=\frac{\frac{G \times M_{1}}{R_{1}^{2}}}{\frac{G \times M_{2}}{R_{2}^{2}}} ; \quad \frac{g_{1}}{g_{2}}=\frac{M_{1}}{R_{1}^{2}} \times \frac{R_{2}^{2}}{M_{2}}=\frac{2 \times 7 \times 7}{4 \times 4 \times 3}=\frac{49}{24}\)
∴ \(g_{1}: g_{2}\) = 49: 24

Application of Newton's law of gravitation
(i) Dimensions of the heavenly objects can be measured using gravitation law. Mass of the earth, radius of the earth, acceleration due to gravity etc. can be calculated with a higher accuracy.
(ii) Helps in discovering new stars and planets. Mass of the double stars can be calculated.
(iii) One of the irregularities in the motion of stars is called "Wobble" which leads to the disturbance in the motion of planet nearby. In this condition mass of the star can be calculated using law of gravitation.
(iv) Helps to explain germination of roots due to the property of geotropism, which is the property of root responding to the gravity.
(v) Helps to predict the path of the astronomical bodies.

Given: K₁ = K₂ = K, m₁ = 4m₂
The kinetic energy of the truck \(=\frac{1}{2} m_{1} v_{1}^{2} \Rightarrow v_1= \sqrt\frac{2 k}{m_{1}}\)
The kinetic energy of the bike \(=\frac{1}{2} \mathrm{~m}_{2} \mathrm{v}_{2}^{2}\Rightarrow v_2= \sqrt\frac{2 k}{m_{2}}\)
∴ Momentum p = mv
∴ Momentum of the two bodies are given by,
\(P_1=\sqrt{2m_1K,}\)
\(P_2=\sqrt{2m_2K,}\)
\(\therefore \frac{P_1}{P_2} =\sqrt{\frac{2 m_1K}{2m_{2}K}}=\sqrt{\frac{m_1}{m_{2}}}=\sqrt{\frac{4m_2}{m_{2}}}=\frac{\sqrt { 4}}{\sqrt 1}= \frac{2}{1}\)
Ratio of momenta = 2: 1

Given:
Weight, W = 100 dyne
Acceleration due to gravity, g = 10 m/s²
g = 1000 cm/ s²
To find: Mass of the body, m = ?
W= mg
m=\(\frac { w }{ g } \)
m=\(\frac { 100\quad dyne }{ 100\quad cm/{ s }^{ 2 } } =\frac { 1 }{ 10 } \)g
Mass, m=10^-1 g