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Standard VIII

Physics

Normal Force

Two persons A...

Question

Two persons A and B are carrying a plank $(M, L)$ on their heads at rest. Find the load beared by person $A$ .

\frac{m g}{2}

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\frac{3}{4} m g

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\frac{2}{3} m g

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\frac{m g}{3}

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Solution

The correct option is C $\frac{2}{3} m g$
R.E.F image
Solution For equilibrium,
$N_{1} + N_{2} = m g$
$T_{n e t} = 0$
About A,
$T_{n e t} = m g (1 / 4) - N_{2} (\frac{3 L}{4}) = 0$
$\Rightarrow N_{2} = m g / 3$
So, $N_{1} + N_{2} = m g \Rightarrow N_{1} = 2 m g / 3$
option - c is correct.

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Two persons A and B are carrying a plank (M,L) on their heads at rest. Find the load beared by person A.

The correct option is C 23mgR.E.F imageSolution For equilibrium,N1+N2=mgTnet=0 About A,Tnet=mg(1/4)−N2(3L4)=0⇒N2=mg/3So, N1+N2=mg⇒N1=2mg/3option - c is correct.

Two persons A and B are carrying a plank $(M, L)$ on their heads at rest. Find the load beared by person $A$ .

The correct option is C $\frac{2}{3} m g$
R.E.F image
Solution For equilibrium,
$N_{1} + N_{2} = m g$
$T_{n e t} = 0$
About A,
$T_{n e t} = m g (1 / 4) - N_{2} (\frac{3 L}{4}) = 0$
$\Rightarrow N_{2} = m g / 3$
So, $N_{1} + N_{2} = m g \Rightarrow N_{1} = 2 m g / 3$
option - c is correct.