A body of weight 200 N is suspended from a tree branch through a chain of mass 10 kg. The branch pulls the chain by a force equal to (if $g = 10 \, \text{m/s}^2$):
- 150 N
- 300 N
- 200 N
- 100 N
The Correct Option is B
Approach Solution - 1
The total weight acting on the chain is the sum of the weight of the body and the weight of the chain itself.
1. Weight of the body: \( 200 \, \text{N} \)
2. Weight of the chain: \( 10 \times 10 = 100 \, \text{N} \)
Thus, the total weight is:
\[\text{Total weight} = 200 + 100 = 300 \, \text{N}\]
Since the chain-block system is in equilibrium, the tension \( T \) in the chain must balance the total weight:
\[T = 300 \, \text{N}\]
Approach Solution -2
To solve this problem, we need to understand the forces involved when a body is suspended from a tree branch through a chain. The problem states that we have a body of weight 200 N suspended by a chain with a mass of 10 kg. The acceleration due to gravity, \(g\), is given as \(10 \, \text{m/s}^2\).
First, let's find the weight of the chain. The weight \((W)\) can be calculated using the formula:
\(W = m \times g\)
where \(m\) is the mass of the chain.
Substitute the given values:
\(W_{chain} = 10 \, \text{kg} \times 10 \, \text{m/s}^2 = 100 \, \text{N}\)
The total force exerted by the branch is the sum of the weight of the body and the weight of the chain:
\(F = W_{body} + W_{chain}\)
Substitute the values we know:
\(F = 200 \, \text{N} + 100 \, \text{N} = 300 \, \text{N}\)
Thus, the force with which the branch pulls the chain is \(300 \, \text{N}\).
Let's evaluate the options to confirm our solution:
- 150 N - Too low, as it doesn't account for the full weight.
- 300 N - Correct, as this is the total weight suspended.
- 200 N - This is the weight of the body alone, without the chain.
- 100 N - This is the weight of the chain alone, without the body.
The correct answer is 300 N.
Hence, the force exerted by the branch to support both the body and the chain is 300 N. This supports the correct choice in the options provided.
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