A man can move on a horizontal plank supporte | vedclass

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(B) Let the mass of the man be $m$ and the mass of the plank be $M$. The supports are at $x = 1 \ m$ and $x = 5 \ m$ (since $1+4=5$).Let $N_A$ be the

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(B) Let the mass of the man be $m$ and the mass of the plank be $M$. The supports are at $x = 1 \ m$ and $x = 5 \ m$ (since $1+4=5$).Let $N_A$ be the normal reaction at support $A$ and $N_B$ be the normal reaction at support $B$.Taking torque about support $B$ (at $x = 5 \ m$): $N_A 	imes 4 = mg(5 - x) + Mg(3 - 1) = mg(5 - x) + 2Mg$.Thus,$N_A = \frac{mg}{4}(5 - x) + \frac{Mg}{2}$.This is a linear equation of the form $N_A = -mx + c$,which represents a straight line with a negative slope.As the man moves from $x = 0$ to $x = 1$,$N_A$ decreases linearly.Between $x = 1$ and $x = 5$,the man is between the supports,and the equation holds.Beyond $x = 5$,the plank would tip,so the motion is restricted to the region between supports.Graph $B$ shows a linear decrease,which matches the derived relationship.

$A$ man can move on a horizontal plank supported symmetrically as shown. The variation of normal reaction on support $A$ with distance $x$ of the man from the end of the plank is best represented by:

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