91Ó°ÊÓ

Static friction is the force that resists the initial movement of one object over another. In our problem, Block B is on top of Block A, and they are only kept from sliding over each other by this static friction. The strength of static friction is determined by the coefficient of static friction (denoted as \( \mu_s \)) and the normal force between the two surfaces. This normal force is simply the weight of Block B, which acts perpendicular to the surface between the blocks.

The maximum static friction can be calculated with the formula:

• \( f_{s, \text{max}} = \mu_s \times N \)

where \( N \) is the normal force, and \( \mu_s = 0.750 \) for our blocks. Understanding static friction is crucial because it defines the point at which Block B will start slipping off Block A if exceeded by any external force.

A pulley system is a simple mechanical device that helps lift heavy objects by redirecting forces. In our exercise, the pulley is frictionless and massless, which simplifies the calculations because it means all forces applied to it (like tension) are perfectly transferred without any loss.

The light string connected to Block A runs over this pulley and has Block C hanging from the other end. This setup allows us to infer that the tension in the string is the same on both sides of the pulley. By knowing the maximum force of static friction, we can determine how much weight can be suspended from the string without causing the blocks to move separately.

Newton's Laws of Motion are foundational principles that help us understand how objects interact with forces. In this problem, the focus is on Newton's Second Law, which states:

• \( F = m \times a \)

Where \( F \) is the total force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration. Here, we considered the system's equilibrium when released from rest, meaning the acceleration (\( a \)) might at first be zero if the blocks don't move separately.

By applying these laws, we can understand that the tension force by Block C equals the maximum static friction because the net force must be zero (no separate movement). Thanks to Newton's laws, we link the tension to the weight of Block C (\( m_C \times g \)).

Mass calculation is vital in problems like these because it's through calculations that we determine how much weight can be safely managed in the system. In our case, we solved for the maximum allowable mass of Block C using the equation derived from Newton's second law:

• \( T = m_C \times g \)
where \( T \) is the tension which equals the maximum static friction force (36.7875 N). By rearranging the equation, we solve for \( m_C \):
• \( m_C = \frac{T}{g} \)
Substituting the values gives us:
• \( m_C \approx \frac{36.7875 \, \text{N}}{9.81 \, \text{m/s}^2} \approx 3.75 \text{ kg} \)
These calculations ensure that we know the exact boundaries within which Block C's mass must fall for both blocks to stay in sync without slipping.