Friction Force Equation and Techniques

Revision Notes on Forces

1. Resolution of a force into components Fx and Fy.

2. Friction and viscous forces
The frictional force acting between surfaces at rest with respect to each other is called static friction. Frictional force is of electromagnetic origin.

When a solid moves in a fluid (i.e. liquid or gas), it will also experience a resistance. This is called the viscous or drag force. Viscous force is dependent on the speed of the object. For example, the drag due to air resistance against a car, moving at low speeds, is proportional to the speed of the car and at high speeds, the drag is proportional to the square of the speed of the car. That is,

Fv = kv or kv2

3. Hooke’s Law: F = kx
4. Elastic Potential Energy = Strain Energy = ½ Fx = ½ kx2
5. Torque or moment of a single force: τ = r × F where
r = perpendicular distance between the axis of rotation and the line of action of the force F
6. Torque or moment of a couple = d× F where d = the perpendicular distance between the two forces.
7. Conditions for equilibrium of rigid body:
(i) The resultant force on the object must be zero
(ii) The resultant torque on the object about every axis must be zero

Condition (i) is the condition for translational equilibrium. It tells us that

(a) ∑Fx = 0 and ∑Fy = 0 .
(b) the vector addition of the forces yields a CLOSED polygon.
(c) the acceleration of the object is zero

Condition (ii) is the condition for rotational equilibrium. It tells us that

(a) sum of the clockwise moments about any axis = the sum of the anticlockwise moments about the same axis. This is known as the Principle of Moments
(b) the angular acceleration of the object is zero.

8. Pressure due to liquid column: p = h?g
9. Archimedes’ Principle:
Upthrust = Weight of liquid displaced by the object

Techniques for solving forces in static equilibrium

Fig. 1.1. shows a beam AB of length L and weight W supported by a cable under a tension T. Force R is the force exerted by the wall on the beam.

1. There are 3 forces (such as T, W and R) act on the body (beam) which in static equilibrium. The line of action of the 3 forces must meet at a point, e.g.. they meet at point P as shown in Fig. 1.1.

2. Since the resultant force is zero, the 3 forces must form a closed vector triangle, as shown in Fig. 1.2. Any unknown force(T, W or R) can be found by
(a) drawing the closed vector triangle according to scale and measuring the unknown force.
(b) using trigonometry such tan, sine or cosine rule to solve for the unknown force.

3. An alternative method would be to resolve all force vectors along the horizontal and vertical axes as shown in Fig. 1.3.

Make use of
∑Fx = 0
=> T sin θ + R sin Φ – W = 0
and ∑Fy = 0
=> T cos θ – R cos Φ = 0

4. Since the resultant torque is also zero, taking moments about point A will yield
Total Clockwise moment about A = Total anticlockwise moment about A
T sin θ x L = W x ½ L

kinematics Definition & Equations


Displacement is the distance travelled along a specified direction.

Speed is the rate of change of distance.

Velocity is the rate of change of displacement.

Acceleration is the rate of change of velocity.

Graphical representation of motion

Displacement-time graph:
Gradient of s-t graph = Instantaneous velocity v = ds/dt

Velocity-time graph:
Area under v-t graph = displacement at time t
Gradient of v-t graph = Instantaneous acceleration a = dv/dt

Acceleration-time graph:
Area under a-t graph = change in velocity at time t

The graphs representing a uniformly acceleration motion are as follows:

Equations for uniform accelerated motion

v = u + a t
s = <v>t = ½(u + v) t
v2 = u2+ 2a s
s = u t + ½at2

Projectile motion

A ball is thrown in air at an angle to the ground. Neglecting air resistance, its velocity vector v is changing but the acceleration due to gravity g is constant. The horizontal component of v is constant since there is no horizontal acceleration. But the vertical component of v is changing due to g.

In projectile motion, the horizontal and the vertical motion are treated independently.

For the horizontal motion, the particle is undergoing uniform motion, i.e. its horizontal component of velocity remains constant.
Thus sx = uxt
where uxis the horizontal velocity component and sx is the horizontal displacement (range)

For the vertical motion: the particle undergoing uniform accelerated motion due to g.

Thus sy= uy t + ½ a t2

vy= uy + a t

vy 2= uy2 + 2 a sy

where sy is the vertical displacement, uy is the initial vertical velocity component, and vy is the final vertical velocity component.

On solving the equations for horizontal and vertical motion, it can be deduced that the
Maximum Height H = u2sin2θ/2g
Range R = u2sin 2θ/g