Problem 72 - University Physics with Modern Physics

Question

Block \(A\) in Fig. P5.72 weighs 60.0 \(\mathrm{N} .\) The coefficient of static friction between the block and the surface on which it rests is \(0.25 .\) The weight \(w\) is 12.0 \(\mathrm{N}\) and the system is in equilibrium. (a) Find the friction force exerted on block \(A .\) (b) Find the maximum weight \(w\) for which the system will remain in equilibrium.

Step-by-Step Solution

Answer

(a) The friction force is 12.0 N. (b) The maximum weight is 15.0 N.

Step 1: Understanding the Problem

We have a system with block \(A\) weighing 60.0 N and a weight \(w\) of 12.0 N. The coefficient of static friction between block \(A\) and the surface is 0.25. We need to find the friction force exerted on block \(A\) when the system is in equilibrium and determine the maximum weight \(w\) that keeps the system in equilibrium.

Step 2: Identify the Forces

For block \(A\), the forces acting along the horizontal are the tension from the cord (equal in magnitude to the weight \(w = 12.0\, \mathrm{N}\)) and the friction force. The forces acting vertically are the weight of block \(A\) and the normal force. We use the following equations to set up the force equilibrium: \( T = w\) and \( f = T\).

Step 3: Calculate the Frictional Force

Since the system is in equilibrium, the friction force \( f \) must balance the tension from the block which is equal to the weight \(w = 12.0 \mathrm{N}\). Therefore, the friction force is \(12.0\, \mathrm{N}\).

Step 4: Calculate the Maximum Static Friction

The maximum static friction \( f_{\text{max}} \) can be calculated using the equation \( f_{\text{max}} = \mu_s \cdot N \), where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. For block \( A \), \( N = 60.0 \mathrm{N} \). So, \( f_{\text{max}} = 0.25 \times 60.0 = 15.0 \mathrm{N} \).

Step 5: Determine the Maximum Weight for Equilibrium

The maximum weight \( w \) for maintaining equilibrium is equal to the maximum static friction force because it provides the maximum tension that can be countered by friction. Hence, the maximum weight \( w \) is \( 15.0 \mathrm{N} \).

Key Concepts

Equilibrium

Equilibrium occurs when all the forces acting on a system are balanced, leading to a state where there is no net force or motion. In physics, equilibrium situations often involve balancing various types of forces like gravitational, normal, tension, and friction forces.
In the context of the given problem, the system involving block \(A\) and the weight \(w\) achieves equilibrium when the force pulling block \(A\) due to \(w\) exactly balances the static friction force acting against it. This means that the net force in both horizontal and vertical directions is zero.
The horizontal force equilibrium ensures that the tension in the connecting cord equals the static frictional force:

• The tension \(T\) induced by the weight \(w = 12.0 \, \mathrm{N}\) needs to be countered by friction to prevent movement.
• In equilibrium, the friction force \(f\) matches the pulling force: \(f = T\).

When designing systems or solving problems, confirming that these conditions hold helps guarantee that the object or system is in equilibrium, ensuring stability without acceleration.

Normal Force

The normal force is a contact force exerted by a surface perpendicular to the object resting on it. It opposes the object's weight due to gravity and plays a crucial role in friction-related problems.
For block \(A\), resting on a flat surface, the normal force \(N\) is equivalent to the weight of block \(A\), since there are no vertical movements or additional forces acting upwards or downwards. Thus, for block \(A\):

• Normal force \(N\) has a magnitude of \(60.0 \, \mathrm{N}\), equal to the weight of the block.

The normal force is vital because it directly influences the maximum static friction possible. Static friction is calculated using the coefficient of static friction \(\mu_s\) and the normal force:
\[ f_{\text{max}} = \mu_s \cdot N \]
Without the normal force, determining the maximum resistance to motion by static friction would be impossible. Thus, it's important to recognize and quantify the normal force correctly to solve equilibrium or friction problems effectively.

Friction Force

Friction is the resistive force that opposes the motion of an object. In static scenarios like this, friction prevents motion until a force threshold is breached. The static friction force you'll encounter depends on two factors:

• The coefficient of static friction \(\mu_s\).
• The normal force \(N\) exerted by the surface.

In the exercise with block \(A\), the static friction keeps it in place despite the lateral pulling force from weight \(w\). Here's how it works:

• The static friction force (\(f\)) exerted by the surface counteracts the tension in the connection caused by weight \(w = 12.0 \, \mathrm{N}\).
• At equilibrium, the static friction force equals the horizontal tension, making \(f = 12.0 \, \mathrm{N}\).
• To find out how much weight can be added while maintaining this equilibrium, calculate the maximum static friction: \(f_{\text{max}} = 0.25 \times 60.0 = 15.0 \, \mathrm{N}\).

The calculated \(f_{\text{max}}\) determines the highest value of \(w\) that maintains equilibrium, ensuring friction counteracts the motion effectively. Understanding these principles helps manage friction-related tasks and systems efficiently, ensuring stable and predictable outcomes.