[5 Mark Question Answer]

Question

(i) Define machine and write functions of a machine.
(ii) Describe mechanical advantage and its types.
(iii) (a) Derive relation between the efficiency, mechanical advantage, and the velocity ratio of a practical machine.
(b) Derive relation between ideal mechanical advantage and velocity ratio for a perfect machine

Vidyadip · Accepted Answer

(i) Machine is a device by which we can overcome large resistance or gain speed or change the direction of effort applied, by applying comparatively a small force at a convenient point, and in a desired direction.
Functions of a machine are:
(a) A machine can multiply force.
(b) A machine can increase speed.
(c) A machine can change the direction of effort applied.
(ii) Mechanical advantage means advantage given by the machine, in terms of effort applied and load lifted by it. If a machine lifts a comparatively large load, compared to effort applied, then it has a positive mechanical advantage.
Mechanical advantage can be classified under two classes
(a) Ideal mechanical advantage: Ideal mechanical advantage is the ratio between the total load (useful load + resistance due to friction and movable parts of machine) moved to the effort applied.
(b) Actual mechanical advantage : Actual mechanical advantage is the ratio between the useful load moved to the effort applied.
(iii) Consider a perfect machine (whose parts are weightless and frictionless) in which an effort ' E ' causes a displacement 'D', when total load ' $L$ ', is displaced through a distance ' $d$ '.
$\therefore$ Output of the machine $=L \times d$
Input of the machine $=E \times D$.
By the principle of a perfect machine,
Output of the machine $=$ Input of the machine
$\begin{aligned} & L \times d=E \times D \\ \Rightarrow & \frac{L}{E}=\frac{D}{d} \\ \therefore & I M A = V R \end{aligned}$
However, if $L=l+x$, where $T$ is the useful load and $x$ is the resistance due to the movable parts of the machine and friction, etc., then from the above expression :
$\frac{L}{E}=\frac{D}{d} \Rightarrow \frac{l+x}{E}=\frac{D}{d}$
$\Rightarrow \frac{l}{E}+\frac{x}{E}=\frac{D}{d}$
$AMA +\frac{x}{E}= VR$
$\therefore \frac{x}{E}= VR - AMA$
$x= E ( VR - AMA )$
But $ V R=I M A$
$\therefore x=E($ IMA - AMA $)$

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