**8.1 Newton’s First Law and the concept of force**

Every object will remain at rest or moves with constant velocity unless an external force is applied

The following are forces .

– W = weight in Newtons (mg=mass(kg) x 9.8ms^{-2} )

– R = Reaction (normal reactions is at right angles to the point of contact)

– F = Force of friction (acts opposite to the direction at which the object is moving)

-T = Tension in a string

Assumptions :

– Objects are modelled as masses concentrated at a point , no rotational forces

– Strings are inextensible so any stretch can be disregarded (No need to apply Hooke’s Law)

– Strings and rods are light (no mass ) so, their weight can be neglected

– Pulleys are smooth, so no need to consider any frictional force

**8.2 Newton’s second Law**

The resultant force acting on an object is equal to the acceleration multiplied by its mass

\(\vec{F}\) = \(m\vec{a}\)

**8.3 Weight and motion in a straight line under gravity**

Value of g is not constant, it varies with the altitude and the geographical location

g = 9.8 ms^{-2}

**8.4 Newton’s Third Law **

Action and reaction are equal and opposite.

\(\vec{F_{a/b}}\) = – \(\vec{F_{b/a}}\)

Equilibrium of forces

\(\sum\hat{F}\) = \(\hat{0}\)

There is no net force acting on the object

Pulleys and Connected Particles in 2D

R_{2} = m_{2}g…………………………………………….(1)

P -T – μR = m_{2}a………………………………….(2)

R1 = m_{1}g……………………………………………..(3)

T – μR1 = m_{1}a……………………………………..(4)

Mg – T = Ma……………………(1)

T – mg = ma…………………….(2)

Equilibrium of a particle under coplanar forces

\(\vec{P} + \vec{Q} + \vec{R}\) = \(\vec{0}\)

**8.5 Addition of forces**

**Example4:** F1 and F2 are two forces of magnitudes 9 N and 5 N and the angle between their directions is 100º . Find the resultant force.

**Solution: **Using the cosine rule

R^{2} = 5^{2} + 9^{2} – 2 × 5 × 9 × cos 80

⇒ R = 9.50640

and, using the sine rule ,

\(\frac{5}{sinx}\) = \(\frac{9.50640}{sin 80}\) => x = 31.196º

**Answer:** The resultant force is 9.51 N at an angle of 31.2º with the 9 N force.

Resultant Forces

Resultant Force = \(\vec{F_1} + \vec{F_2}\)

**Example:** Find the resultant of P = 5i – 7j and Q = -2i + 13j.

**Solution: ** R = P + Q = (5i – 7j) + (-2i + 13j) = 3i + 6j.

|R| = \(\sqrt{(3^2 +6^)}\) = \(\sqrt{45N}\)

**Answer:** Resultant is 3 \(\sqrt{5}N\)

Motion in a plane

**8.6 Friction**

Coefficient of friction

μ= \(\frac{F}{R}\) F is the force, R is the Normal reaction and μ is the coefficient of friction.

Motion of a body on a rough surface

If a force is acting on the object and the object does not move, then F < μ_{2}R where μ_{d} is the coefficient of static friction

When the object is moving, F=μ_d R where μ_{d} is the coefficient of dynamic friction.

Limiting Friction

Limiting friction is defined as the friction which is generated between the two static surfaces that come in contact with each other.

**Example:** A box of mass 30 kg is being pulled along the ground by a horizontal force of 60 N. If the acceleration of the trolley is 1.5 ms^{-2} , find the magnitude of the friction force.

**Solution: **No need to resolve as forces are already at 90º to each other.

Resolve horizontally

⇒ 60 – F = 30 × 1.5

=> F = 60 – 45 = 15.

**Answer:** Frictional force is 15 N.

Statics

The branch of mechanics that deals with bodies at rest or forces in equilibrium.