Answer:
The mechanical advantage = 4/75
Explanation:
A wheel barrel is a second class as the effort is placed further from the point of fulcrum than the point of action of the load
The parameters given are;
Length of wheel barrel = 4 cm
Distance between wheel and center of load = 0.75 m
Therefore, we have the mechanical advantage, MA given as follows;
[tex]MA = \dfrac{F_{o}}{F_{i}} = \dfrac{L_{i}}{L_{o}}[/tex]
[tex]{F_{o}}[/tex] = Output force
[tex]{F_{i}}[/tex] = Input force
[tex]{L_{o}}[/tex] = Distance from output force = 0.75 m
[tex]{L_{i}}[/tex] = Distance from input force = 4 cm = 0.04 m
Therefore;
[tex]MA = \dfrac{0.04}{0.75} = \dfrac{4}{75}[/tex]
The mechanical advantage MA = 4/75 = 0.0533
Therefore, more force (75/4 × the load) will be required to lift a given load.
