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The figure below shows a wheel barrow of mass 15 kg carrying a load of 30 kgf with its centre of gravity at A. The points B and C are the centre of wheel and tip of the handle such that the horizontal distance AB = 20 cm and AC = 40 cm.

The figure below shows a wheel barrow of mass 15 kg carrying a load of 30 kgf with its centre of gravity at A. The points B and C are the centre of wheel and tip of the handle such that the horizontal distance AB = 20 cm and AC = 40 cm. Find load arm, effort arm, mechanical advantage, minimum effort required to keep the leg just off the ground. Machines, Concise Physics Class 10 Solutions.

Find:

(a) the load arm,

(b) the effort arm,

(c) the mechanical advantage, and

(d) the minimum effort required to keep the leg just off the ground.

Machines

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Answer

(a) Load arm AF = 20 cm

(b) Effort arm CF = 60 cm

(c) We know that,

Mechanical advantage M.A.=Effort armLoad arm\text{Mechanical advantage M.A.} = \dfrac{\text{Effort arm}}{\text{Load arm}} \\[0.5em]

Substituting the values in the formula we get,

M.A.=6020⇒M.A.=3M.A. = \dfrac{60}{20} \\[0.5em] \Rightarrow M.A. = 3 \\[0.5em]

(d) We know that,

Effort=LoadM.A.\text{Effort} = \dfrac{\text{Load}}{\text{M.A.}} \\[0.5em]

Given,

Load = 30 kgf + 15 kgf = 45 kgf

Substituting the values in the formula we get,

Effort=453⇒Effort=15 kgf\text{Effort} = \dfrac{45}{3} \\[0.5em] \Rightarrow \text{Effort} = 15\text{ kgf} \\[0.5em]

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