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Q5.29(c):
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 (c) the reaction of the 6th coin on the 7th coin.

Solution :

Step 1: Establishing the System Configuration

We consider a vertical stack of $10$ identical one-rupee coins, each of mass $m$, arranged such that coin 1 is at the bottom and coin 10 is at the top. The gravitational force acting on each coin is $W = mg$. We define the reaction force as the upward normal force exerted by one coin on the coin immediately above it to support the weight of the coins above that interface [Per Newton’s Third Law of Motion].

Step 2: Defining the Mechanical Equilibrium

For any coin $n$ (where $n=1$ being the bottom-most and $n=10$ being the top-most), the coin is in a state of static equilibrium [Per Newton’s First Law of Motion]. The forces acting on coin $n$ are:

  • The downward weight of coin $n$ ($mg$).
  • The downward force exerted by the coin above it ($n+1$), which is the weight of all coins from $(n+1)$ to $10$.
  • The upward reaction force $R$ exerted by the coin below it ($n-1$).

Step 3: Calculating the Force on the 7th Coin

To determine the reaction of the 6th coin on the 7th coin, we must isolate the system consisting of the 7th, 8th, 9th, and 10th coins. The total mass of this subsystem is:

$M_{total} = m_7 + m_8 + m_9 + m_{10} = 4m$

The total downward force exerted by this subsystem on the 6th coin is the weight of these 4 coins:

$F_{down} = 4mg$ [Calculated by the force of gravity equation $F = ma$, where $a=g$]

Step 4: Applying Newton’s Third Law

The 6th coin exerts an upward reaction force $R$ on the 7th coin to counteract the weight of the coins above it (7 through 10) to maintain equilibrium [Per Newton’s Third Law: For every action, there is an equal and opposite reaction].

$R = \text{Weight of coins } \{7, 8, 9, 10\}$

$R = 4mg$

Step 5: Visual Representation of Forces

Coin 7 Coin 6 R = 4mg

Step 6: Conclusion on Magnitude and Direction

The magnitude of the reaction force is the weight of all coins supported by the 6th coin. Since there are 4 coins resting on the 6th coin, the magnitude is $4mg$. The direction of the reaction force exerted by the 6th coin on the 7th coin is vertically upward.

Final Solution: The magnitude of the reaction of the 6th coin on the 7th coin is $4mg$, and the direction is vertically upward.


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