__Question 5.29__:*m*. Give the magnitude and direction of

(a) the force on the 7

^{th}coin (counted from the bottom) due to all the coins on its top,

(b) the force on the 7

^{th}coin by the eighth coin,

(c) the reaction of the 6th coin on the 7

^{th}coin.

__Solution__:**(a)**Force on the seventh coin is exerted by the weight of the three coins on its top.

*m*g

Weight of three coins = 3

*m*g

Hence, the force exerted on the 7

^{th}coin by the three coins on its top is 3

*m*g. This force acts vertically downward.

**(b)**Force on the seventh coin by the eighth coin is because of the weight of the eighth coin and the other two coins (ninth and tenth) on its top.

Weight of the eighth coin =

*m*g

Weight of the ninth coin =

*m*g

Weight of the tenth coin =

*m*g

Total weight of these three coins = 3

*m*g

Hence, the force exerted on the 7

^{th}coin by the eighth coin is 3

*m*g. This force acts vertically downward.

**(c)**The 6

^{th}coin experiences a downward force because of the weight of the four coins (7

^{th}, 8

^{th}, 9

^{th}, and 10

^{th}) on its top.

Therefore, the total downward force experienced by the 6

^{th}coin is 4

*m*g.

As per Newton’s third law of motion, the 6

^{th}coin will produce an equal reaction force on the 7

^{th}coin, but in the opposite direction. Hence, the reaction force of the 6

^{th}coin on the 7

^{th}coin is of magnitude 4

*m*g. This force acts in the upward direction.

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