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What you will learn:
• Simple machines• Mechanical advantage / force ratio• Movement ratio / velocity ratio• Machine efficiency• Gears: gear box ratio, gear efficiency• Directions of rotations
Machines• Is a device that can change magnitude or line of action, or
both magnitude and line of action of a force.
• A simple machine usually amplifies an input force, call the effort , to give a larger output force, called the load.
• Examples; pulley systems, screw jacks, gear systems and lever systems.
Machines
Machines• A machine is a device that receives energy in
some available form and converts it to a useable form.
• For example, a person may wish to lift a weight of 2 tonnes; this is not possible unaided but the use of a jack or a hoist permits the person to achieve their objective. A lever such as a pry bar is an example of a simple machine. A relatively small manual force can be converted into a large force in order to lift or move an object.
Levers• Leverage and the use of levers occurs in the
use of tools such as spanners, pry bars, pliers etc., and in many vehicle mechanisms such as clutch and brake pedals, throttle linkages and suspension units.
• The mechanical advantage of a machine is the ratio of the load to the effort:
• Lifting machines such as jacks and cranes have a large mechanical advantage so that relatively small manual forces can be used to raise heavy weights.
Mechanical advantage
mechanical advantage, MA = load effort
Velocity ratio (movement ratio)
• The velocity ratio, VR, of a machine, or movement ratio, MR, is the ratio of the distance moved by the effort to the distance moved by the load:
movement ratio, MR = distance moved by effort distance moved by load
Efficiency of a machine
• The efficiency of a machine = (energy output/energy input) in the same time and this also
Efficiency of machine = MA ×100% VR
Example problem• In the trolley jack example shown before an
effort of 250 newtons is lifting a load of 2 tonnes. In lifting the load through a distance of 15 cm the operator performs 40 pumping strokes of the handle each of which is 50 cm long. Calculate the mechanical advantage, velocity ratio and efficiency of the jack.
• MA = 785• MR =1333• The efficiency = 59%
Gears• Leverage and gears
• The gear ratio in this case is 4:1
The gear ratio = revolutions of input gear revolutions of output gear.
Gearbox• Gear trains
• Ratio of gear set
= driven × driven driver driver = 35×28 = 1225 25×32
Problem 1
A simple machine raises a load of 160 kg through a distance of 1.6m. The effort applied to the machine is 200N and moves through a distance of 16m. Taking g as 9.8m/s2,
determine the force ratio, movement ratio and efficiency of the machine.
Solution 1• Force ratio = Load/effort
= 160kg/200N = (160 x 9.8)N/200N = 7.84
• Movement ratio = distance moved by effort/distance moved by the load
= 16m/1.6m= 10
• Efficiency = (force ratio/movement ratio) x 100%= (7.84/10) x 100= 78.4%
Problem 2
For the simple machine of Problem 1, determine: (a) the distance moved by the effort to move the load through a distance of 0.9m, (b) the effort which would be required to raise a load of 200kg, assuming the same efficiency, (c) the efficiency if, due to lubrication, the effort to raise the 160kg load is reduced to 180N.
Solution 2• Distance moved by the effort
= 10 x distance moved by the load= 10 x 0.9 = 9m
• Since the force ratio is 7.84effort = load/7.84
= (200 x 9.8) / 7.84= 250N
Solution 2
• The new force ratioload/effort = (160 x 9.8)/180
= 8.711• The new efficiency after lubrication
= (8.711/10) x 100= 87.11%
Problem 3
•A driver gear on a shaft of a motor has 35 teeth and meshes with a follower having 98 teeth. If the speed of the motor is 1400 revolutions per minute, find the speed of rotation of the follower.
Solution 3
Speed of driver teeth on follower
speed of follower teeth on driver
1400/speed of follower = 98/35
Speed of follower = (1400 x 35)/98
= 500 rev/min
=
Problem 4
•A compound gear train similar to that shown in last figure consists of a driver gear A, having 40 teeth, engaging with gear B, having 160 teeth. Attached to the same shaft as B, gear C has 48 teeth and meshes with gear D on the output shaft, having 96 teeth. Determine;
(a) the movement ratio of this gear system
(b) the efficiency ratio when the force ratio is 6
Solution 4
(a) The speed of D = speed of A x TA/TB x TC/TD
movement ratio = NA/ND = TB/TA x TD/TC
= 160/40 x 96/48
= 8
(b) The efficiency = (force ratio/movement ratio)100%
= 6/8 x 100
= 75%
THANK YOU
