A uniform rope of mass m hangs freely from a ceiling. A bird of mass M climbs up the rope with an acceleration a. The force exerted by the rope on the ceiling is

Options

A. Ma + mg
B. M(a+g)+mg
C. M(a+g)
D. Dependent on the position of bird on the rope

Answer

To solve the problem, we need to analyze the forces acting on the system, which consists of a uniform rope and a bird climbing up the rope. Here's a step-by-step solution:

### Step 1: Identify the forces acting on the system
- The weight of the rope acts downward, which is given by \( mg \) (where \( m \) is the mass of the rope and \( g \) is the acceleration due to gravity).
- The weight of the bird also acts downward, which is given by \( Mg \) (where \( M \) is the mass of the bird).
- The tension \( T \) in the rope acts upward.

### Step 2: Write the equation of motion for the bird
Since the bird is climbing up with an acceleration \( a \), we can apply Newton's second law to the bird:
\[ T - Mg = Ma \]
Rearranging this gives us:
\[ T = Ma + Mg \]

### Step 3: Calculate the total force exerted by the rope on the ceiling
The force exerted by the rope on the ceiling can be calculated as the sum of the tension in the rope and the weight of the rope itself. The total force \( F \) exerted by the rope on the ceiling is given by:
\[ F = T + mg \]
Substituting the expression for \( T \) from Step 2:
\[ F = (Ma + Mg) + mg \]
This simplifies to:
\[ F = Ma + Mg + mg \]
We can factor out \( g \) from the last two terms:
\[ F = Ma + m g + M g = Ma + (M + m)g \]

### Final Expression
Thus, the force exerted by the rope on the ceiling is:
\[ F = Ma + (M + m)g \]

### Conclusion
The correct expression for the force exerted by the rope on the ceiling is \( Ma + (M + m)g \).