A body of weight w1 is suspended from t...

A body of weight `w_1` is suspended from the ceiling of a room through as chain of weight `w_2`. The ceiling pulls the chain by a force

`w_1`

`w_2`

`w_1+w_2`

`(w_1+w_2)/2`

Text Solution

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The correct Answer is:

To solve the problem, we need to determine the force that the ceiling exerts on the chain when a body of weight \( w_1 \) is suspended from it through a chain of weight \( w_2 \). ### Step-by-Step Solution: 1. **Identify the System**: We have a body of weight \( w_1 \) suspended from a chain that has a weight \( w_2 \). The entire system can be treated as a single unit where both the body and the chain contribute to the total weight that the ceiling must support. **Hint**: Consider the entire system as a single entity to simplify calculations. 2. **Calculate the Total Weight**: The total weight that the ceiling has to support is the sum of the weight of the body and the weight of the chain. Therefore, the total weight \( W \) is given by: \[ W = w_1 + w_2 \] **Hint**: Remember that the ceiling must support both the body and the chain together. 3. **Force Exerted by the Ceiling**: The force that the ceiling exerts on the chain must equal the total weight of the system (the body plus the chain). According to Newton's second law, the upward force exerted by the ceiling (let's denote it as \( F \)) must balance the downward gravitational force (the total weight). Thus, we have: \[ F = w_1 + w_2 \] **Hint**: The force exerted by the ceiling is equal to the total weight acting downwards. 4. **Conclusion**: Therefore, the force that the ceiling pulls the chain with is \( w_1 + w_2 \). **Final Answer**: The ceiling pulls the chain with a force of \( w_1 + w_2 \).

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A heavy rope is suspended from the ceiling of a room. If phi is the density of the rope, L be its original length and Y be its. Young's modulus, then increase DeltaL in the length of the rope due to its own weight is

`DeltaL=(phi gL^(2))/(Y)`

`DeltaL=(2phi g L^(2))/(Y)`

`DeltaL=(phi g L^(2))/(2Y)`

`DeltaL=(phigL)/(Y)`

One end of a wire of length L and weight w is attached rigidly to a point in roof and a weight w_(1) is suspended from its lower end. If A is the area of cross-section of the wire then the stress in the wire at a height (3L)/(4) from its lower end is

`(w_(1))/(A)`

`(w_(1)+(w)/(4))/(A)`

`(w_(1)+(3w)/(4))/(A)`

`(w_(1)+w_(2))/(A)`

One end of uniform wire of length L and of weight W is attached rigidly to a point in the roof and a weight W_(1) is suspended from its lower end. If s is the area of cross section of the wire, the stress in the wire at a height ( 3L//4 ) from its lower end is

`(W_(1))/s`

`[W_(1)+W/4]s`

`[W_(1)+(3W)/4]//s`

`(W_(1)+W)/s`