Question

A boy lifts a load of 40 kgf through a vertical height of 2m in 5s by using a single fixed pulley when he applies an effort of 48 kgf. Calculate: (i) the mechanical advantage, and (ii) the efficiency of the pulley. Why is the efficiency of the pulley not 100%? (iii) The energy gained by the load in 5s, and (iv) the power developed by the boy in raising the load.

Answer 1

The pulley experiences some friction losses when it rotates due to its bearings. Since it is a single fixed pulley, the pulley stays stationary and only the load and the boy moves, both by an equal amount.

Formula used:
-Mechanical Advantage ${\text{(M}}{\text{.A}}{\text{.) = }}\dfrac{{Load}}{{Effort}}$
-Efficiency $= \dfrac{{{\text{Mechanical advantage}}}}{{{\text{Velocity}}\,{\text{ratio}}}}$
-Energy gain $= F \times d$, where $F$ is the force and $d$ is the displacement of the object.
-Power developed $= \dfrac{{{\text{Energy gain}}}}{t}$, where $t$ is the time taken by the object to cover distance $d$ .

Complete step by step solution:
We’ve been given:
Load is $(L)\, = \,40\,kgf$
Vertical displacement is $d\, = \,2\,m$
Time taken is $t\, = \,5s$
Effort is $E\, = \,48\,kgf$
i) Mechanical advantage is defined as the ratio of load to effort. So
Mechanical Advantage is given by
$\Rightarrow {\text{(M}}{\text{.A}}{\text{.) = }}\dfrac{L}{E}$
$\Rightarrow \dfrac{{40}}{{48}}$
$\Rightarrow 0.833$
ii) Before calculating the efficiency, we need to find the velocity ratio (V.R.) of load to effort. Since it is a single pulley system, the load and the effort move with similar velocity. So velocity ratio (V.R.) = 1.The efficiency can then be calculated as
$\Rightarrow {\text{Efficiency}}\, = \dfrac{{{\text{M}}{\text{.A}}{\text{.}}}}{{{\text{V}}{\text{.R}}{\text{.}}}}$
$\Rightarrow \dfrac{{0.833}}{1}$
$\Rightarrow 0.833\,\,{\text{(or}}\,{\text{83}}{\text{.3\% )}}$
The efficiency of the pulley is not ${\text{100 }}\%$ because part of the energy applied is wasted in overcoming friction forces in the pulley system. As a result, only $83.3{\text{ }}\%$ of the total energy is used in lifting the load while the rest $16.7{\text{ }}\%$ is used in overcoming the friction in the pulley.
iii) The energy gained by the load in 5s is equal to Load multiplied by Displacement of load in 5s.
Hence, energy gained by the load in 5s is
$\Rightarrow {\text{Energy gained}} = 40\,kgf \times 2m$
$\Rightarrow 80\,kgf \times m$
iv) The power developed by the boy is given by $\dfrac{{{\text{Effort done by the boy x Displacement of effort}}}}{{{\text{Time}}}}$
Since the effort is displaced by the same amount as the load, displacement of effort is 2m. So,
$\Rightarrow {\text{Power }} = \,\dfrac{{48 \times 2}}{5}$
$\Rightarrow 19.2\,kgf \times m{s^{ - 1}}$.

Note:
Alternatively, the power developed by the boy can also be calculated in the following manner:
Since energy gained by load ${\text{ = 0}}{\text{.833 }} \times {\text{energy spent by boy}}$ ,
$\Rightarrow {\text{Energy spent by boy = }}\dfrac{{{\text{Energy gained by load}}}}{{0.833}}$.
And as ${\text{Power = }}\dfrac{{{\text{Energy}}}}{{{\text{Time}}}}$, the power developed by the boy will be
$\Rightarrow {\text{Power}} = \dfrac{{{\text{Energy spent by boy}}}}{{{\text{time}}}}$
$\Rightarrow \dfrac{{{\text{Energy}}\,{\text{gained}}\,{\text{by}}\,{\text{load}}}}{{{\text{0}}{\text{.833}} \times {\text{time}}}}$
$\Rightarrow \dfrac{{80}}{{0.833 \times 5}}$
$\Rightarrow 19.2\,kgf \times m{s^{ - 1}}$.