__What is called Mechanical Advantage?__

*Mechanical advantage of machine is defined as the ratio of load lifted at output end to the effort applied at the input end of the machine.*

Therefore, \quad \text {Mechanical Advantage} = \left ( \frac{\text {Output force}}{\text {Input force}} \right )

= \left ( \frac {\text {Load}}{\text {Effort}} \right )

Mechanical Advantage is simply written as MA.

### Ideal Mechanical Advantage of machine

In ideal conditions, friction is neglected. But, all machines have some friction and a part of applied effort is lost in overcoming this friction.

Therefore, \quad \text {Actual effort} = \text {Ideal effort} + \text {Effort lost in friction}

*Ideal mechanical advantage of a machine is defined as the ratio of load to the ideal effort for the machine.*

Hence, \quad \text {Ideal Mechanical A} = \left ( \frac {\text {Load}}{\text {Ideal Effort}} \right )

- Ideal effort is less than actual effort.
- Therefore, Ideal MA of a machine is greater than actual MA.

## Mechanical Advantage of Simple Machines

In this article, detail process is explained to find the MA of some *simple machines*.

__MA of Wedge__

Woodcutter’s Axes is a simple example of a *wedge*.

Consider about an axes as shown in figure. When an *effort* is applied on the head of axes, it digs into the wood.

If we apply an effort ( P ) on axes head, it penetrates producing forces perpendicular to its inclined surfaces as shown in figure. Let, the resistance offered by the wood is ( W ) .

Let, the width of the head of axes is ( w ) and its length is ( h ) . Then, for complete digging of the axes, distance moved by the effort will be ( h ) and distance moved by the load will be ( w )

We know that, \quad \text {work done} = \text {Force} \ \times \ \text {displacement}

Then, *work done* by the effort is ( P \times h ) and work output in overcoming the resistance of wood is ( W \times w )

Therefore, by the law of *conservation of energy* we have –

\text {Work Input} = \text {Work Output}

So, \quad ( P \times h ) = ( W \times w )

Or, \quad \left ( \frac {W}{P} \right ) = \left ( \frac {h}{w} \right )

Hence, \quad MA = \left ( \frac {h}{w} \right )

*Therefore, Mechanical advantage of a wedge is determined by dividing its length by its width.*

MA = \left ( \frac{\text {Length of wedge}}{\text {Width of wedge}} \right )

Therefore, MA of axes shown in figure will be –

MA = \left ( \frac {30}{10} \right ) = 3

If angle of wedge is more acute or narrow, then ratio of its length and width will be greater and so, the mechanical advantage will be more.

__MA of Inclined Plane__

Consider about an inclined plane as shown in figure. A body of *mass* ( M ) is being lifted up to height ( H ) . For this purpose, the body is pushed along the slant surface of the inclined plane with a force ( F ) .

In ideal conditions if there is no *friction force*, the required push will be equal to the component ( Mg \sin \theta ) of weight of body which is parallel to slant surface.

So, \quad F = W \sin \theta = M g \sin \theta

Therefore, ideal MA of inclined plane will be –

MA = \left ( \frac {\text {Weight lifted}}{\text {Applied \ force}} \right )

= \left ( \frac { Mg }{ Mg \sin \theta } \right )

= \left ( \frac { 1 }{ \sin \theta } \right )

But, \quad \sin \theta = \left ( \frac{ H }{ L } \right )

Therefore, \quad MA = \left [ \frac { 1 }{ \left ( \frac { H }{ L } \right ) } \right ]

= \left ( \frac { L }{ H } \right )

*So, ideal mechanical advantage of an inclined plane is also given by –*

= \left ( \frac{\text {Slant length}}{\text {Height of plane}} \right )

__MA of Wheel & Axle__

Consider about a wheel and axle arrangement as shown in figure.

Let, a force ( F_{Input} ) is applied tangentially to the surface of the wheel disc of radius ( R ) .

Then, turning effect or *torque* developed in disc will be –

T_{Wheel} = F_{Input} \times R

If, there is no friction, then same torque will be transmitted to the axle.

Therefore, \quad T_{Axle} = T_{Wheel}

Let, ( F_{Output} ) is the force developed at surface of axle.

Then, turning effect or torque developed in axle will be –

T_{Axle} = F_{Output} \times r

So, \quad F_{Output} \times r = F_{Input} \times R

Or, \quad \left ( \frac {F_{Output}}{F_{Input}} \right ) = \left ( \frac {R}{r} \right )

Therefore, MA of Wheel & Axle will be –

MA = \left ( \frac {F_{Output}}{F_{Input}} \right )

= \left ( \frac {R}{r} \right )

*Mechanical advantage of wheel and axle is the ratio of radius of wheel over radius of axle.*

__MA of Screw__

Consider about a *screw* as shown in figure. Diameter of the screw cylinder is ( d ) . On its body surface, an inclined plane is wrapped which forms thread on the cylinder surface.

Let, radius of screw head is ( R ) . The * lead *or

*of screw is ( L ) . ( Lead or pitch is the measure of axial distance of advancement of nut for one complete rotation of screw. )*

**pitch**In general, \quad \text {Lead of screw} = n \times \text {pitch} .

Where ( n ) is the number of starts of thread.

Since, lead or pitch of screw is similar to Rise or Height ( H ) of an *inclined plane.* Therefore, when screw is rotated for one complete turn, it advances or retracted about a distance ( L ) .

Let, a force ( F_{Input} ) is applied on the head of the screw. Then for one complete rotation of the screw –

Input work = F_{Input} \times 2 \pi R

If, ( F_{Output} ) is the generated force by screw then –

Output work = F_{Output} \times L

If, friction is negligible, then –

**Output Work = Input Work.**

Therefore, \quad F_{Output} \times L = F_{Input} \times 2 \pi R

Or, \quad \left ( \frac {F_{Output}}{F_{Input}} \right ) = 2 \pi \left ( \frac {R}{L} \right )

Therefore, MA of screw is given by –

MA = \left ( \frac {F_{Output}}{F_{Input}} \right )

= 2 \pi \left ( \frac {R}{L} \right )

__MA of Compound Machines__

A *compound machine* or a complex machine is made up of series of components. Each component is basically a *simple machine*. It may be a lever, a screw, a wedge, an inclined plane or a wheel & axle etc.

Thus, if a machine is a combination of 2 components, then its mechanical advantage will be –

MA_{compound} = MA of component 1 × MA of component 2

Similarly, if a machine is a combination of ( n ) components, then –

Mechanical advantage = MA of component 1 × MA of component 2 × ….. × MA of component ‘n’

*Therefore, Mechanical Advantage of a compound machine = Product of individual Mechanical Advantages of each components.*