Two cylinders A and B are at rest in a box as shown in the figure. Cylinder A has a diameter of 3d and mass 120 kg. Cylinder B has a diameter of 2d and mass 36 kg . The width of the box is 4.5d. If all the contact surfaces are assumed to be smooth, then find the normal force exerted by the left wall on the cylinder B.

Options

A. 1200N
B. 1600N
C. 2000 N
D. nothing can be said

Answer

To solve the problem of finding the normal force exerted by the left wall on cylinder B, we can follow these steps:

### Step 1: Understand the Setup
We have two cylinders, A and B, resting in a box. Cylinder A has a diameter of 3d and a mass of 120 kg, while cylinder B has a diameter of 2d and a mass of 36 kg. The width of the box is 4.5d.

### Step 2: Identify the Forces Acting on Cylinder B
Since the surfaces are smooth, the only forces acting on cylinder B are:
- The normal force exerted by the left wall on cylinder B (let's denote it as N1).
- The normal force exerted by cylinder A on cylinder B (let's denote it as N3).
- The weight of cylinder B acting downwards (W = m_B * g).

### Step 3: Determine the Angles
To find the relationship between the forces, we need to find the angle θ formed by the line connecting the centers of the two cylinders and the vertical.
- The horizontal distance from the left wall to the center of cylinder B is d (half of its diameter).
- The vertical distance from the top of cylinder B to the center of cylinder A is 1.5d (half of cylinder A's diameter minus half of cylinder B's diameter).

Using the Pythagorean theorem:
- The hypotenuse (h) = √(d² + (1.5d)²) = √(d² + 2.25d²) = √(3.25d²) = √(13/4)d = (√13/2)d.

### Step 4: Calculate Cosine and Sine of θ
Using the triangle formed:
- cos(θ) = adjacent/hypotenuse = d / ((√13/2)d) = 2/√13.
- sin(θ) = opposite/hypotenuse = 1.5d / ((√13/2)d) = 3/√13.

### Step 5: Apply Equilibrium Conditions
Since the system is at rest, we can apply the equilibrium conditions:
1. In the vertical direction:
\[ N3 \cdot \sin(θ) = W_B \]
where \( W_B = m_B \cdot g = 36 \cdot 10 = 360 \, \text{N} \).

Substituting the values:
\[ N3 \cdot \frac{3}{\sqrt{13}} = 360 \implies N3 = \frac{360 \cdot \sqrt{13}}{3} = 120\sqrt{13} \, \text{N}. \]

2. In the horizontal direction:
\[ N1 = N3 \cdot \cos(θ). \]
Substituting the values:
\[ N1 = 120\sqrt{13} \cdot \frac{2}{\sqrt{13}} = 240 \, \text{N}. \]

### Step 6: Conclusion
The normal force exerted by the left wall on cylinder B is \( N1 = 240 \, \text{N} \).