## MACHINE (WORK 1)

(A) M.A = F.R, M.A = load/effort or output force/input force

Where F.R = force ratio

(B) Velocity ratio = distance moved by effort/ distance moved by load = e/L

(C) Efficiency = M.A/V.R × 100%

Or

Efficiency = work output/work input × 100%

EXAMPLE 1

A machine has an efficiency of 80%, if the machine is required to overcome a load of 60N with a force of 40N, calculate its velocity ratio. Neco 2003

Solution

Efficiency = 80%, Load = 60N, Effort = 40N, V.R =?

To get the velocity ratio we’ve to first find mechanical advantage (M.A)

Mechanic advantage(M.A) = Load/ effort

60/40 = 1.5

Efficiency = M.A/V.R × 100%, 80 = 1.5/V.R × 100

V.R = (1.5 × 100)/80, = 1.875

EXAMPLE 2

A machine has an efficiency of 60%, if the machine is required to overcome a load of 30N with a force of 20N;

(I) calculate the mechanical advantage

(II) velocity ratio

Solution

Efficiency = 60%, load= 30N, effort = 20N

Mechanical advantage (M.A) = Load/effort; = 30/20

M.A = 1.5

Efficiency = M.A/V.R × 100%, 60 = 1.5/V.R × 100

V.R = (1.5 × 100)/80, = 2.5

EXAMPLE 3

A machine has a velocity ratio of 6 and an efficiency of 75%. Calculate the effort needed to raise a load of 90N.

Solution

Velocity ratio (V.R) = 6, Efficiency = 75%, load = 90N, effort = ?

First find mechanical advantage (M.A),

M.A = Load/Effort

Efficiency = M.A/V.R × 100%, 75 = M.A/6 × 100

M.A = (6×75)÷100; = 4.5

Calculate effort

M.A = Load/Effort; 4.5 = 90/Effort

Effort = 20N

Another formula for solving efficiency, work 2

Efficiency = (Load × distance moved by load) ÷ (effort × distance moved by effort) × 100%

Or effort = work output/work input × 100%

EXAMPLE 4

A block and tackle system is used to lift a load of 20N through a height of 10m. If the efficiency of the system is 40% how much work is done against friction.

Work done = work input = effort × distance moved by effort

Now using,

Efficiency = work output/ work input × 100%

Work output = load × distance moved by load

Where

L = 20, distance (h) = 10m, Efficiency = 40%, work output = ?

Efficiency = (Load × distance moved by load) ÷ (work input) × 100%

40 = (20×10)÷(work input) × 100

Work input = (20×10×100)÷40

20000/40 = 500N

How much work is done against friction?

Work output = Load × distance moved by load; 20×10 = 200N

Work input – work output = 500 – 200; = 300N

## Another formula, work 3

Velocity ratio (V.R) = numbers of poles supporting the pulley

Example 5

A block and tackle system has six(6) pulley. A force of 50N applied to it lifts a load of weight (W). If the efficiency is 40% calculate the value of w.

Solution

Velocity ratio (V.R) = Number of pulley = 6, effort (force) = 50N, Efficiency = 40%, weight,w(load) = ?

Efficiency = M.A/V.R × 100%

40 = M.A/6 × 100%

M.A = (40×6)/100

M.A = 2.4

Now using,

M.A = Load/Effort; 2.4 = Load/50

Weight (load) = 50×2.4; 120N

Another formula, inclined plane (work 4)

V.R = distance moved by load / distance moved by effort

Or Length of inclined plane/height of inclined plane

Sinø = opposite/hypotenuse = H/L

Rearranging, 1/sin∅ = L/H

Therefore V.R; = L/H = 1/sinø

Example 6

Example 6

An inclined plane of angle 15⁰ is used to raise a load of 4500N, through a height of 2m if the plane is 75% efficient, calculate;

(I) velocity ratio of the plane

(II) work done on the load

Solution

Angle of inclination (Ø) = 15⁰, load = 4500N, H = 2m, efficiency = 75%

V.R = 1/sinø; 1/sin15

= 3.86

Work done = work input

Using,

Efficiency = (work output/work input) × 100%

Work input = (work output/Efficiency) × 100%

Work input = (Load × height of plane/ efficiency) × 100%

(4500×2×100)/75 = 12000J

## Another formula, work 4

The wheel and axle

Radius of the wheel (R) , radius of the axle (r)

Work input(work done by effort) = work done by load

Effort × distance moved by effort = load × distance moved by load

Or effort × circumference of wheel = load × circumference of axle

E × 2πR = L ×2πr

Therefore, velocity ratio (V.R) = distance moved by effort/distance moved by load

V.R = 2πR/2πr

Efficiency = (load/effort × 2πr/2πR) × 100%

(L/E × r/R) × 100%

Example 7

A wheel and axle is used to raise a load of 500N by the application of an effort 250N, if the radii of the wheel and axle are 0.4cm and 0.1cm respectively. The efficiency of the machine will be?

Solution

Load = 500N, effort = 250N, radius of the wheel (R) = 0.4cm, radius of the axle (r) = 0.1cm

Using

Efficiency = (L/E × r/R) × 100

E = 500/250 × 0.1/0.4 × 100

E = (500×0.1×100) / (250×0.4)

5000/100 = 50%