To solve the problem of determining the maximum distance \( b \) between the centers \( B \) and \( C \) of the two semi-circular cylinders while ensuring equilibrium, we need to analyze the forces acting on the system and apply the principles of static equilibrium. Let's break this down step by step.
Understanding the System
We have a smooth circular cylinder (let's call it Cylinder A) resting on two semi-circular cylinders (Cylinders B and C). The weights are distributed as follows:
- Cylinder A has a weight \( Q \).
- Cylinders B and C each have a weight of \( \frac{Q}{2} \).
The coefficient of static friction between the flat faces of the semi-circular cylinders and the horizontal plane is \( 0.5 \). The friction between the cylinders themselves is neglected, which simplifies our calculations.
Forces Acting on the System
In equilibrium, the following conditions must be satisfied:
- The sum of vertical forces must equal zero.
- The sum of horizontal forces must equal zero.
- The sum of moments about any point must equal zero.
Vertical Forces
The vertical forces acting on the system include the weights of the cylinders. The total downward force due to the weights is:
Weight of Cylinder A: \( Q \)
Weight of Cylinders B and C: \( \frac{Q}{2} + \frac{Q}{2} = Q \)
Thus, the total downward force is \( Q + Q = 2Q \).
Horizontal Forces and Friction
The frictional force at the base of Cylinders B and C must be sufficient to prevent them from sliding. The maximum static friction force can be calculated using the formula:
Maximum static friction \( F_f = \mu N \)
Where \( \mu \) is the coefficient of static friction (0.5) and \( N \) is the normal force. The normal force for each semi-circular cylinder is equal to its weight:
Normal force for each semi-circular cylinder = \( \frac{Q}{2} \)
Thus, the maximum static friction for each cylinder is:
For Cylinder B: \( F_{fB} = 0.5 \times \frac{Q}{2} = \frac{Q}{4} \)
For Cylinder C: \( F_{fC} = 0.5 \times \frac{Q}{2} = \frac{Q}{4} \)
The total maximum friction force available to prevent sliding is:
Total friction force = \( F_{fB} + F_{fC} = \frac{Q}{4} + \frac{Q}{4} = \frac{Q}{2} \)
Equilibrium Condition
For the system to be in equilibrium without the middle cylinder touching the horizontal plane, the horizontal forces must balance. The weight of Cylinder A creates a moment about the line connecting the centers of Cylinders B and C. The distance \( b \) between the centers of B and C will influence this moment.
Calculating Maximum Distance \( b \)
Let’s denote the distance from the center of Cylinder A to the line connecting the centers of B and C as \( d \). The moment created by the weight of Cylinder A about this line is:
Moment = Weight \(\times\) Distance = \( Q \times d \)
For equilibrium, this moment must be balanced by the friction force acting at the base of Cylinders B and C:
Friction force \( F_f \) must equal the moment divided by the distance \( \frac{b}{2} \) (since the distance from the center of Cylinder A to either center B or C is \( \frac{b}{2} \)). Thus, we have:
F_f = \( \frac{Q \times d}{\frac{b}{2}} \)
Setting the maximum friction force equal to the moment gives:
\( \frac{Q}{2} = \frac{Q \times d}{\frac{b}{2}} \)
Solving for \( b \):
\( \frac{Q}{2} \times \frac{b}{2} = Q \times d \)
Thus, \( b = 4d \)
Final Consideration
To ensure that the middle cylinder does not touch the horizontal plane, we need to ensure that \( d \) is less than or equal to the radius \( r \) of the cylinders. Therefore, the maximum distance \( b \) can be expressed as:
Maximum \( b = 4r \)
In summary, the maximum distance \( b \) between the centers of the semi-circular cylinders while maintaining equilibrium without the middle cylinder touching the horizontal plane is \( 4r \). This ensures that the system remains stable under the given conditions.
