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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

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


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%

Oluwamuyide Peter

My name is seyi, the main aim of creating this platform is to help users get information like school updates, electrical engineering topics and many more for free

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