Laws of Motion

Question : 29

Total: 40

Ten one-rupee coins are put on top of each other on a table. Each coin has a mass m. Give the magnitude and direction of
(a) the force on the 7‌th coin (counted from the bottom) due to all the coins on its top,
(b) the force on the 7‌th coin by the eighth coin,
(c) the reaction of the 6‌th coin on the 7th coin.

Solution:

Mass of each coin =m
(a) If F7 be the force on 7‌th coin (counted from the bottom) experienced due to all coins above it, then F7 = weight of three coins above it = 3 mg N (downwards)
(b)F87 = force on 7th coin by 8‌th coin, then the 8‌th coin has to support the weight of the two coins above it. So, the 8th coin shall exert the force F87 such that F87 = weight of 8th coin + weight of two coins above the 8‌th coin =mg+2mg=3mg(N) and it acts downwards.
(c) The sixth coin experiences force equal to weight of the four coins above it. Hence reaction due to 6‌th coin on 7‌th coin = 4mg N and it acts vertically upwards.

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