Rotational MotionHard
Question
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
Options
A.
B.
C.
D.
Solution
For the translator & rotatory equilibrium let the normal at point ′A′ & ′B′ be NA & NB
∴ NA + NB = mg (m → mass of man)
& Torque about the centre of plank - NA × 2 + NB × 2 + mg (3 - x) = 0
As x increases (NA ≡ N) decreases.
(NB - NA) × 2 = mg(3 - x)
∴ NA + NB = mg (m → mass of man)
& Torque about the centre of plank - NA × 2 + NB × 2 + mg (3 - x) = 0
As x increases (NA ≡ N) decreases.
(NB - NA) × 2 = mg(3 - x)
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