Because of friction, real machines cannot convert 100% of applied input work into output work. The efficiency of a system measures the fraction of input work converted to output work. Since input and output distances are often fixed by geometry, efficiencies less than 100% tend to reduce output forces. For example, suppose an input force of 1 N is applied to a machine with MA = 10. The ideal output force should be 10 N, but an actual output force might only be 9 N.

(12.3)

$$\eta =\frac{{W}_{o}}{{W}_{i}}$$

η

=

efficiency

W_{o}

=

output work (J)

W_{i}

=

input work (J)

Efficiency of a machine

Ideal mechanical advantage

In an ideal machine all of the input work is converted into output work, so F_{i}d_{i} = F_{o}d_{o}. Since the ratio of the output force to the input force is the mechanical advantage, we can rearrange the terms to determine the MA of an ideal machine: $$MA=\frac{{F}_{o}}{{F}_{i}}=\frac{{d}_{i}}{{d}_{o}}$$ To determine the ideal mechanical advantage of a machine one assumes an efficiency of 100% (i.e., there are no energy losses). The ideal MA is the ratio of the distance moved at the machine’s input to the distance moved by the output.

(12.4)

$$M{A}_{ideal}=\frac{{d}_{i}}{{d}_{o}}$$

MA_{ideal}

=

ideal mechanical advantage

d_{i}

=

input distance (m)

d_{o}

=

output distance (m)

Ideal mechanical advantage

Measuring forces or distances

The ideal mechanical advantage of a system is determined from the input and output distances moved. For the (actual) mechanical advantage, however, you measure the input and output forces. To determine efficiency, you need both force and distance!

Efficiency of a pulley system

A block and tackle lifts a mass of 0.25 kg by 15 cm. The input string is pulled by 45 cm with a force of 0.95 N. What is the efficiency of the system?

Asked:

efficiency η of the pulley system

Given:

mass m = 0.25 kg, input distance d_{i} = 0.45 m, input force F_{i} = 0.95 N, output distance d_{o} = 0.15 cm

Relationships:

efficiency η = W_{o}/W_{i}; work W = Fd

Solution:

The output force is the weight of the mass: $${F}_{o}={F}_{w}=mg=(0.25\text{kg})(9.8{\text{m/s}}^{2})=2.45\text{N}$$ The efficiency is the ratio of the output work to the input work: $$\eta =\frac{{W}_{o}}{{W}_{i}}=\frac{{F}_{o}{d}_{o}}{{F}_{i}{d}_{i}}=\frac{(2.45\text{N})(0.15\text{m})}{(0.95\text{N})(0.45\text{m})}=0.86=86\%$$

Answer:

The efficiency is η = 86%.

Test your knowledge

For the solved problem above, what is the mechanical advantage of the system?

Answer:MA = 2.6 Asked: mechanical advantage MA Given: input force F_{i} = 0.95 N; output force F_{o} = 2.45 N Relationships:MA = F_{o}/F_{i} Solution:$$MA=\frac{{F}_{o}}{{F}_{i}}=\frac{2.45\text{N}}{0.95\text{N}}=2.6$$ Note that the actual mechanical advantage is smaller than the ideal mechanical advantage.