Essential Physics

Efficiency and ideal machines

Friction causes you to work harder to generate the same output work

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. Read the text aloud

Read the text aloud

(12.3)

η = \frac{W_{o}}{W_{i}}

η	=	efficiency
W_o	=	output work (J)
W_i	=	input work (J)

Efficiency
of a machine

In an ideal machine all of the input work is converted into output work, so F_id_i = F_od_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:

M A = \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. Read the text aloud

Read the text aloud

(12.4)

M A_{i d e a l} = \frac{d_{i}}{d_{o}}

MA_ideal	=	ideal mechanical advantage
d_i	=	input distance (m)
d_o	=	output distance (m)

Ideal mechanical
advantage

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! Read the text aloud

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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: $F_{o} = F_{w} = m g = (0.25 kg) (9.8 {m/s}^{2}) = 2.45 N$ The efficiency is the ratio of the output work to the input work: $η = \frac{W_{o}}{W_{i}} = \frac{F_{o} d_{o}}{F_{i} d_{i}} = \frac{(2.45 N) (0.15 m)}{(0.95 N) (0.45 m)} = 0.86 = 86 %$
Answer:	The efficiency is η = 86%.

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

Show

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:

M A = \frac{F_{o}}{F_{i}} = \frac{2.45 N}{0.95 N} = 2.6

Note that the actual mechanical advantage is smaller than the ideal mechanical advantage.


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