MECHANICAL ENERGY
Machines

A machine is any device or tool which allows a force or effort applied at one point to overcome a resisting force (or load) at another point.

In physics, a machine is a device that uses energy to perform a specific task. Machines typically consist of components such as levers, gears, and pulleys, which work together to transform and transmit energy. The efficiency of a machine is often measured by the ratio of useful work output to the energy input.

Terms used in machines
  1. Load: This is the resisting force that must be overcome by the machine.

  2. Effort: This is the force applied by a person or mechanism to a machine to overcome a load.

  3. Mechanical Advantage: The mechanical advantage or force ratio of a machine is the ability of a machine to overcome a large load through a small effort. It is a measure of amplification of force of a machine. $$ M.A = \frac{Load(L)}{effort(E)} $$ $$ M.A = \frac{\text{output force}}{\text{input force}} $$ Note: Mechanical advantage is affected by friction which reduces the efficiency of the machine

  4. Velocity Ratio: This can be defined as the ratio of distance moved by effort and load. $$ V.R = \frac{\text{distance moved by effort}}{\text{distance moved by load}} $$ $$ V.R = \frac{e}{l} $$ Note: Velocity ratio is independent of friction but dependent on the shape / geometry of the machine.

  5. Workdone by machine: This is the workdone by the machine to move a load(L) a distance (l). $$ \text{workdone by machine} = L × l $$
  6. Workdone on the machine: This is the workdone on the machine or the workdone by a person using the machine to overcome a load. It is the product of effort (E) and distance covered (e). $$ \text{workdone on machine} = E × e $$
  7. Efficiency: The efficiency of a machine is a measure of how well it converts input energy into useful output energy. It is the ratio of the useful work output to the total input energy and is mathematically expressed as: \[ \eta = \frac{\text{Useful Work Output}}{\text{Total Input Energy}} \times 100\% \] $$ \eta = E_f = \frac{W_o}{W_i} × 100\% $$ where; $$ W_o = \text{workdone on load} $$ $$ W_i = \text{workdone by effort} $$ $$ \eta = \frac{L × l}{E× e} $$ $$ \eta = \frac{L}{E} × \frac{l}{e} $$ But, $$ M.A = \frac{L}{E} $$ $$ \frac{1}{V.R} = \frac{l}{e} $$ $$\therefore E_f = \frac{MA}{VR} × 100\% $$
Ideal machine

An ideal or perfect machine has 100% efficiency. Here, Work input is equal to work output and no there is no loss in energy. For an ideal or frictionless machine, $$ W_i = W_o $$ $$ M.A = V.R $$ However, practical machines have an efficiency less than 100% due to workdone by the machine to overcome friction in the moving parts of the machine. Hence for a practical machine, $$ W_o = W_i - W_{friction} $$ where,
\( W_{friction} \) = workdone against friction

Calculations

Example 1: A machine has a velocity of 6 and an efficiency of 75%. Calculate the effort needed to raise a load of 90N. (NECO)

Solution

$$ E_f = \frac{M.A}{V.R} × 100\% $$ $$ V.R = 6 $$ $$ E_f = 75% $$ $$ L = 90N $$ $$ \frac{75}{100} = \frac{M.A}{6} $$ $$ M.A = \frac{6 × 75}{100} $$ $$ M.A = 4.5 $$ $$ \text{But} $$ $$ M.A = \frac{L}{E} $$ $$ E = \frac{L}{M.A} $$ $$ E = \frac{90}{4.5} $$ $$ Effort = 20N $$

Example 2: The efficiency of a machine is 80%. Calculate the workdone by a person using the machine to raise a load of 300kg through a height of 4m. (g = 10m/s²) (WAEC)

Solution

The efficiency of a machine is given by: $$ Eff= \frac{\text{work output}}{\text{work input}} × 100 $$ $$ Eff = 80% $$ $$ m = 300kg $$ $$ h = 4m $$ $$ \text{work input} = \text{workdone by person} $$ $$ {W_i} = ? $$ $$\text{workdone by machine} = \text{work output}$$ $$ {W_o} = mgh $$ $$ {W_o} = 300 × 10 × 4 = 12000J $$ $$ Eff = \frac{W_o}{W_i} × 100 $$ $$ 80 = \frac{12000}{W_i} × 100 $$ $$ {W_i} = \frac{12000 × 100} {80} $$ $$\text{Workdone by person} = 15000J $$

Example 3: A simple machine with an efficiency of 75% lifts a load of 5000N when a force of 500N is applied to it. Calculate the velocity ratio of the machine. (WAEC)

Solution

$$ E_f = 75\% $$ $$ L = 5000N $$ $$ E = 500N $$ $$ V.R = ? $$ $$ M.A = \frac{L}{E} $$ $$ M.A = \frac{5000}{500} $$ $$ M.A = 10 $$ $$ E_f = \frac{M.A}{V.R} × 100\% $$ $$ \frac{75}{100} = \frac{10}{V.R} $$ $$ V.R = \frac{1000}{75} $$ $$ V.R = 13.33 $$

Example 4: A load of mass 120kg is raised vertically upwards through a height of 2m in 30s by a machine whose efficiency is 100%. Calculate the power generated by the machine. [g = 10m/s²] (WAEC)

Solution

$$ M = 120kg $$ $$ h = 2m $$ $$ t = 30s $$ $$ E_f = \frac{P_o}{P_i} × 100% $$

Since the machine is 100%, $$\therefore P_o = P_i $$ $$ P_o = \frac{workdone}{time} $$ $$ P_o = \frac{mgh}{t} $$ $$ P_o = \frac{120 × 10 × 2}{30} $$ $$ \text{power output} = 80w $$

Simple Machines Calculator





Formulas:
  1. For 1: \( M.A = \frac{L}{E}\)

  2. For 2: \( V.R = \frac{M.A × 100}{Eff}\)

  3. For 3: \( Eff = \frac{M.A × 100}{V R} \)

  4. For 4: \( E = \frac{L × 100}{V R × Eff} \)

  5. For 5: \( {W_i}= \frac{mgh × 100}{Eff} \)
This calculator assumes g = 10m/s². Click a formula to use the calculator. Input the values for the parameters to solve using the calculator. Ensure to use standard units when inputting.

Summary