To solve this problem, we need to analyze the forces acting on the pendulum and the block as the trolley accelerates. The key here is to understand how the acceleration of the trolley affects the pendulum's position and the frictional forces at play. Let's break it down step by step.
Understanding the Forces Involved
When the trolley accelerates to the right, the pendulum will experience a force due to this acceleration. The pendulum will not hang straight down; instead, it will incline at an angle due to the combination of gravitational force and the inertial force caused by the trolley's acceleration.
Identifying the Forces
For the pendulum, we have two main forces acting on it:
- Gravitational Force (Weight): This acts vertically downward and is equal to \( mg \), where \( m \) is the mass of the pendulum bob and \( g \) is the acceleration due to gravity.
- Inertial Force: As the trolley accelerates, the pendulum experiences an inertial force to the left, which can be expressed as \( ma \), where \( a \) is the acceleration of the trolley.
Setting Up the Equations
Let’s denote the angle of inclination of the pendulum from the vertical as \( \theta \). The forces can be resolved into components along the direction of the pendulum string and perpendicular to it. The tension in the string will balance these forces.
The components of the forces can be expressed as follows:
- The component of the gravitational force acting along the string: \( mg \cos(\theta) \)
- The component of the gravitational force acting perpendicular to the string: \( mg \sin(\theta) \)
- The inertial force acting horizontally: \( ma \)
Applying Newton's Second Law
For the pendulum to be in equilibrium (not falling), the net force in the horizontal direction must equal the net force in the vertical direction. Thus, we can set up the following relationship:
In the horizontal direction:
\( ma = mg \sin(\theta) \)
In the vertical direction:
\( T = mg \cos(\theta) \)
Relating the Forces
From the first equation, we can express the acceleration \( a \) in terms of \( g \) and \( \theta \):
\( a = g \tan(\theta) \)
Next, we need to consider the frictional force that prevents block B from falling. The maximum static frictional force can be calculated using the coefficient of friction \( \mu \):
\( f_{\text{max}} = \mu N \)
Where \( N \) is the normal force acting on block B, which is equal to \( mg \cos(\theta) \). Thus:
\( f_{\text{max}} = \mu mg \cos(\theta) \)
Equating Forces for Equilibrium
For block B to remain in contact without falling, the inertial force must not exceed the maximum static frictional force:
\( ma \leq \mu mg \cos(\theta) \)
Substituting \( a = g \tan(\theta) \) into this inequality gives:
\( mg \tan(\theta) \leq \mu mg \cos(\theta) \)
We can cancel \( mg \) from both sides (assuming \( m \neq 0 \)):
\( \tan(\theta) \leq \mu \cos(\theta) \)
Finding the Angle of Inclination
Now, substituting \( \mu = 0.5 \):
\( \tan(\theta) \leq 0.5 \cos(\theta) \)
Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we can rewrite the inequality:
\( \sin(\theta) \leq 0.5 \)
This implies that:
\( \theta \leq \arcsin(0.5) \)
Calculating this gives:
\( \theta \leq 30^\circ \)
Conclusion
Therefore, the maximum angle of inclination of the pendulum to the vertical, while ensuring that block B does not fall, is \( 30^\circ \). This analysis shows how the forces interact in a dynamic system and how friction plays a crucial role in maintaining equilibrium.
