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Standard XII

Physics

Center of Mass as an Average Point

Figure shows ...

Question

Figure shows a wheel barrow of mass $15 k g$ carrying a load of $30 k g f$ 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 $A B = 20 c m$ and $A C = 40 c m$ . What is the minimum effort required to keep the leg just off the ground ?

15 k g f

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25 k g f

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10 k g f

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5 k g f

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Solution

The correct option is A $15 k g f$
Load = $15 + 30 = 45 k g f$
Load arm = $20 c m$
Effort arm = $20 + 40 = 60 c m$
Effort = $20 \times 45 / 60 = 15 k g f$
An effort of $15 k g f$ equals the load. Any effort more than that will keep it off the ground.

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Figure 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. What is the minimum effort required to keep the leg just off the ground ?

The correct option is A 15kgfLoad = 15+30=45 kgfLoad arm = 20 cmEffort arm =20+40=60 cmEffort =20×45/60=15 kgfAn effort of 15 kgf equals the load. Any effort more than that will keep it off the ground.

The correct option is A $15 k g f$
Load = $15 + 30 = 45 k g f$
Load arm = $20 c m$
Effort arm = $20 + 40 = 60 c m$
Effort = $20 \times 45 / 60 = 15 k g f$
An effort of $15 k g f$ equals the load. Any effort more than that will keep it off the ground.