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:
\[T = 300 \, \text{N}\]
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:
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.
Two blocks of masses \( m \) and \( M \), \( (M > m) \), are placed on a frictionless table as shown in figure. A massless spring with spring constant \( k \) is attached with the lower block. If the system is slightly displaced and released then \( \mu = \) coefficient of friction between the two blocks.
(A) The time period of small oscillation of the two blocks is \( T = 2\pi \sqrt{\dfrac{(m + M)}{k}} \)
(B) The acceleration of the blocks is \( a = \dfrac{kx}{M + m} \)
(\( x = \) displacement of the blocks from the mean position)
(C) The magnitude of the frictional force on the upper block is \( \dfrac{m\mu |x|}{M + m} \)
(D) The maximum amplitude of the upper block, if it does not slip, is \( \dfrac{\mu (M + m) g}{k} \)
(E) Maximum frictional force can be \( \mu (M + m) g \)
Choose the correct answer from the options given below:
A wooden block of mass M lies on a rough floor. Another wooden block of the same mass is hanging from the point O through strings as shown in the figure. To achieve equilibrium, the coefficient of static friction between the block on the floor and the floor itself is 
Let one focus of the hyperbola \( H : \dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 \) be at \( (\sqrt{10}, 0) \) and the corresponding directrix be \( x = \dfrac{9}{\sqrt{10}} \). If \( e \) and \( l \) respectively are the eccentricity and the length of the latus rectum of \( H \), then \( 9 \left(e^2 + l \right) \) is equal to:
