An effort of 400 N is applied through 2 metres on a machine whose efficiency is \[80\% \] such that load is lifted by a distance \[0.4\] metre. Calculate \[MA\], \[VR\] and load lifted.
Answer
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Hint: Velocity ratio is simply the distance moved by the effort divided by the distance moved by the load. The efficiency of a machine can be given by the radio of the mechanical advantage to its velocity ratio.
Formula used: In this solution we will be using the following formulae;
\[MA = \dfrac{{{F_o}}}{{{F_i}}}\] where \[MA\] is the mechanical advantage of a machine, \[{F_o}\] is the force output (or load lifted by) of the machine and \[{F_i}\] is the force input of the machine or the effort applied to the machine.
\[VR = \dfrac{{{d_e}}}{{{d_l}}}\] where \[VR\] is the velocity ratio, \[{d_e}\] is the distance moved by the effort, and \[{d_l}\] is the distance moved by the load.
\[{E_{ff}} = \dfrac{{MA}}{{VR}}\]where\[{E_{ff}}\] is the efficiency of a machine.
Complete Step-by-Step Solution:
The velocity ratio is defined as
$\Rightarrow$ \[VR = \dfrac{{{d_e}}}{{{d_l}}}\] where \[VR\] is the velocity ratio, \[{d_e}\] is the distance moved by the effort, and \[{d_l}\] is the distance moved by the load.
According to the question, distance by effort is 2 metres and distance by load is 0.4 metres. Hence the equation becomes,
$\Rightarrow$ \[VR = \dfrac{2}{{0.4}} = 5\]
Mechanical advantage can be calculated from
\[{E_{ff}} = \dfrac{{MA}}{{VR}}\] where \[{E_{ff}}\] is the efficiency of a machine.
Hence, \[MA = {E_{ff}} \times VR\]
$\Rightarrow$ \[MA = \dfrac{{80}}{{100}} \times 5 = 4\]
Now to calculate the distance moved by load, we note that the mechanical advantage is defined as
$\Rightarrow$ \[MA = \dfrac{{{F_o}}}{{{F_i}}}\]
The load lifted is the force output. Hence, by making force output the subject of the formula, we have
$\Rightarrow$ \[{F_i} = \dfrac{{{F_o}}}{{MA}}\]
$\Rightarrow$ \[{F_o} = {F_i}MA\]
Hence, by inserting values, we get
$\Rightarrow$ \[{F_o} = 400 \times 4 = 800{\text{N}}\]
Note: For understanding, note that in a machine, though its mechanical advantage may be greater than one i.e. a force multiplier, the total energy is always conserved according to the principle of conservation of energy. However, the efficiency is less than 1 because some of the energy would be converted to heat or sound (due to friction and other resistive forces).
Formula used: In this solution we will be using the following formulae;
\[MA = \dfrac{{{F_o}}}{{{F_i}}}\] where \[MA\] is the mechanical advantage of a machine, \[{F_o}\] is the force output (or load lifted by) of the machine and \[{F_i}\] is the force input of the machine or the effort applied to the machine.
\[VR = \dfrac{{{d_e}}}{{{d_l}}}\] where \[VR\] is the velocity ratio, \[{d_e}\] is the distance moved by the effort, and \[{d_l}\] is the distance moved by the load.
\[{E_{ff}} = \dfrac{{MA}}{{VR}}\]where\[{E_{ff}}\] is the efficiency of a machine.
Complete Step-by-Step Solution:
The velocity ratio is defined as
$\Rightarrow$ \[VR = \dfrac{{{d_e}}}{{{d_l}}}\] where \[VR\] is the velocity ratio, \[{d_e}\] is the distance moved by the effort, and \[{d_l}\] is the distance moved by the load.
According to the question, distance by effort is 2 metres and distance by load is 0.4 metres. Hence the equation becomes,
$\Rightarrow$ \[VR = \dfrac{2}{{0.4}} = 5\]
Mechanical advantage can be calculated from
\[{E_{ff}} = \dfrac{{MA}}{{VR}}\] where \[{E_{ff}}\] is the efficiency of a machine.
Hence, \[MA = {E_{ff}} \times VR\]
$\Rightarrow$ \[MA = \dfrac{{80}}{{100}} \times 5 = 4\]
Now to calculate the distance moved by load, we note that the mechanical advantage is defined as
$\Rightarrow$ \[MA = \dfrac{{{F_o}}}{{{F_i}}}\]
The load lifted is the force output. Hence, by making force output the subject of the formula, we have
$\Rightarrow$ \[{F_i} = \dfrac{{{F_o}}}{{MA}}\]
$\Rightarrow$ \[{F_o} = {F_i}MA\]
Hence, by inserting values, we get
$\Rightarrow$ \[{F_o} = 400 \times 4 = 800{\text{N}}\]
Note: For understanding, note that in a machine, though its mechanical advantage may be greater than one i.e. a force multiplier, the total energy is always conserved according to the principle of conservation of energy. However, the efficiency is less than 1 because some of the energy would be converted to heat or sound (due to friction and other resistive forces).
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