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Week 7, Lesson 1 Simple Machines 1 
Simple Machines 
Week 7, Lesson 1 
References/Reading Preparation: 
Schaum’s Outline Ch. 7 
Principles of Physics by Beuche – Ch.5 
Definition of a Machine 
The Principle of Work 
Mechanical Advantage 
Efficiency of a machine
Week 7, Lesson 1 Simple Machines 2 
Simple Machines 
Machines: 
Are devices we use to help us do work. 
A Simple Machine: 
Is a mechanical device that can exert a force on an object at one point when an external force is applied to the device at another point. 
Stated another way, it: 
Is any device by which the magnitude, direction, or 
Method of application of a force is changed so as to 
achieve some advantage.
Week 7, Lesson 1 Simple Machines 3 
Some examples are: 
F 
The Lever 
F 
Wheel and Axle 
F 
Pulley 
F 
Inclined Plane 
Jackscrew
Week 7, Lesson 1 Simple Machines 4 
Discussion - Machines 
Simple machines cannot create energy! 
According to the principle of conservation of energy: 
- 
a machine can provide no more work output than the 
amount of work supplied to it. 
In reality, machines are subject to some friction, and so work 
output is less than work input by an amount equal to the 
thermal energy produced (by the friction).
Week 7, Lesson 1 Simple Machines 5 
The Principle of Work That Applies to a Machine 
The Principle of Work that applies to a continuously operating 
machine is as follows: 
Work input = useful work output + work to overcome friction 
In machines that operate for only a short time, some of the 
input work may be used to store energy within the machine – 
for example, an internal spring may be stretched.
Week 7, Lesson 1 Simple Machines 6 
Efficiency of a Machine 
The efficiency of a machine measures the degree to which it 
converts work input to work output. 
% efficiency = 
work output 
work input 
x 100 
If a machine could operate at 100% efficiency, it would be 
called an ideal machine.
Week 7, Lesson 1 Simple Machines 7 
Mechanical Advantage 
Although machines cannot create energy, they can magnify the input force. 
Consider the lever: 
m 
F 
H 
h 
x 
X 
A force F is applied to raise 
the mass m a distance h. 
The work done by F to do this = work input = FH 
The weight mg is lifted through a displacement h. 
Thus, Work output = mgh 
We assume that there is no friction at the fulcrum, so that we are dealing with an 
ideal machine. 
fulcrum 
Thus, work input = work output 
or 
FH = mgh 
or 
F = mgh/H 
From similar triangles, h/H = x/X 
therefore 
F = mgx/X
Week 7, Lesson 1 Simple Machines 8 
The Mechanical Advantage (cont’d) 
F = mgx/X 
This equation tells us that, to lift a load, we need to exert a force F that is 
less than mg by the ratio x/X.. 
For example: if x = ½ X, then F would have to be only ½ mg. 
The lever has thus multiplied the input force by 2!! 
We call the ability of a simple machine to multiply forces the 
mechanical advantage of the machine.
Week 7, Lesson 1 Simple Machines 9 
Actual Mechanical Advantage 
If Fo is the force output of the machine, 
and 
Fi is the force applied (the input force), 
Then we define 
Actual Mechanical Advantage (AMA) = 
Fo 
Fi 
In the case of our lever, with x = ½ X, then 
Fo = mg and Fi = ½ mg 
Therefore, AMA = 2 
The price you pay for magnifying a force with a simple machine is that 
the distance through which the load is moved is shorter than the distance 
through which you exert the applied force!!
Week 7, Lesson 1 Simple Machines 10 
Ideal Mechanical Advantage 
As we saw, the distance moved by Fi is greater than the output force Fo . 
This difference in distance is simply a consequence of the conservation 
of energy. 
Thus, for an ideal machine, 
Fi si = Fo so 
where si is the distance over which the applied force is exerted, and 
so is the distance the load is moved 
The mechanical advantage of an ideal machine can be expressed as the 
ratio of the input and output displacements: 
Ideal Mechanical Advantage (IMA) = si / so
Week 7, Lesson 1 Simple Machines 11 
Example 
A motor furnishes 120 hp to a device that lifts a 5000 kg load to a height 
of 13.0 m in 20 s. Compute a) the work output, b) the power output and 
power input, c) the efficiency of the system.
Week 7, Lesson 1 Simple Machines 12 
Examples 
In the figure, the 300N load is balanced by a force F in both systems. 
Assuming efficiencies of 100%, how large is F in each system? Assume 
all ropes are vertical. 
F 
300N 
300N 
F

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Physics 7

  • 1. Week 7, Lesson 1 Simple Machines 1 Simple Machines Week 7, Lesson 1 References/Reading Preparation: Schaum’s Outline Ch. 7 Principles of Physics by Beuche – Ch.5 Definition of a Machine The Principle of Work Mechanical Advantage Efficiency of a machine
  • 2. Week 7, Lesson 1 Simple Machines 2 Simple Machines Machines: Are devices we use to help us do work. A Simple Machine: Is a mechanical device that can exert a force on an object at one point when an external force is applied to the device at another point. Stated another way, it: Is any device by which the magnitude, direction, or Method of application of a force is changed so as to achieve some advantage.
  • 3. Week 7, Lesson 1 Simple Machines 3 Some examples are: F The Lever F Wheel and Axle F Pulley F Inclined Plane Jackscrew
  • 4. Week 7, Lesson 1 Simple Machines 4 Discussion - Machines Simple machines cannot create energy! According to the principle of conservation of energy: - a machine can provide no more work output than the amount of work supplied to it. In reality, machines are subject to some friction, and so work output is less than work input by an amount equal to the thermal energy produced (by the friction).
  • 5. Week 7, Lesson 1 Simple Machines 5 The Principle of Work That Applies to a Machine The Principle of Work that applies to a continuously operating machine is as follows: Work input = useful work output + work to overcome friction In machines that operate for only a short time, some of the input work may be used to store energy within the machine – for example, an internal spring may be stretched.
  • 6. Week 7, Lesson 1 Simple Machines 6 Efficiency of a Machine The efficiency of a machine measures the degree to which it converts work input to work output. % efficiency = work output work input x 100 If a machine could operate at 100% efficiency, it would be called an ideal machine.
  • 7. Week 7, Lesson 1 Simple Machines 7 Mechanical Advantage Although machines cannot create energy, they can magnify the input force. Consider the lever: m F H h x X A force F is applied to raise the mass m a distance h. The work done by F to do this = work input = FH The weight mg is lifted through a displacement h. Thus, Work output = mgh We assume that there is no friction at the fulcrum, so that we are dealing with an ideal machine. fulcrum Thus, work input = work output or FH = mgh or F = mgh/H From similar triangles, h/H = x/X therefore F = mgx/X
  • 8. Week 7, Lesson 1 Simple Machines 8 The Mechanical Advantage (cont’d) F = mgx/X This equation tells us that, to lift a load, we need to exert a force F that is less than mg by the ratio x/X.. For example: if x = ½ X, then F would have to be only ½ mg. The lever has thus multiplied the input force by 2!! We call the ability of a simple machine to multiply forces the mechanical advantage of the machine.
  • 9. Week 7, Lesson 1 Simple Machines 9 Actual Mechanical Advantage If Fo is the force output of the machine, and Fi is the force applied (the input force), Then we define Actual Mechanical Advantage (AMA) = Fo Fi In the case of our lever, with x = ½ X, then Fo = mg and Fi = ½ mg Therefore, AMA = 2 The price you pay for magnifying a force with a simple machine is that the distance through which the load is moved is shorter than the distance through which you exert the applied force!!
  • 10. Week 7, Lesson 1 Simple Machines 10 Ideal Mechanical Advantage As we saw, the distance moved by Fi is greater than the output force Fo . This difference in distance is simply a consequence of the conservation of energy. Thus, for an ideal machine, Fi si = Fo so where si is the distance over which the applied force is exerted, and so is the distance the load is moved The mechanical advantage of an ideal machine can be expressed as the ratio of the input and output displacements: Ideal Mechanical Advantage (IMA) = si / so
  • 11. Week 7, Lesson 1 Simple Machines 11 Example A motor furnishes 120 hp to a device that lifts a 5000 kg load to a height of 13.0 m in 20 s. Compute a) the work output, b) the power output and power input, c) the efficiency of the system.
  • 12. Week 7, Lesson 1 Simple Machines 12 Examples In the figure, the 300N load is balanced by a force F in both systems. Assuming efficiencies of 100%, how large is F in each system? Assume all ropes are vertical. F 300N 300N F