Problem 33 - Physics for Scientists and Engineers with Modern Physics

Question

(III) A small block of mass \(m\) rests on the rough, sloping side of a triangular block of mass \(M\) which itself rests on a horizontal frictionless table as shown in Fig. 5-41. If the coefficient of static friction is \(\mu,\) determine the minimum horizontal force \(F\) applied to \(M\) that will cause the small block \(m\) to start moving up the incline.

Step-by-Step Solution

Answer

The minimum force \(F\) required is \[F = \frac{m (g \sin \theta + \mu g \cos \theta)}{\cos \theta}\]."

Step 1: Identify the Forces

First, identify all the forces acting on the small block of mass \(m\). These include gravity \(mg\), normal force \(N\) perpendicular to the incline, and the static friction force. The static friction force will act down the incline and has a maximum value \(\mu N\) where \(\mu\) is the coefficient of static friction.

Step 2: Resolve Forces

Resolve the gravitational force \(mg\) into components parallel and perpendicular to the inclined plane. The component parallel to the incline is \(mg \sin \theta\) and the component perpendicular to the incline is \(mg \cos \theta\), where \(\theta\) is the angle of the incline.

Step 3: Establish Condition for Motion

For the small block to start moving up the incline, the force component that pushes it upward, originating from the horizontal force \(F\) applied to \(M\), must overcome the force of gravity pulling it down and the maximum static friction. This condition can be written as: \(F_\text{up incline} > mg \sin \theta + \mu mg \cos \theta\).

Step 4: Calculate the Applied Force Components

To find \(F_\text{up incline}\), consider the way horizontal force \(F\) applied to block \(M\) translates through the system. By using the geometry of the system (considering angles and applying Newton's second law), translate this applied force \(F\) to its equivalent component along the incline.

Step 5: Solve for Minimum Force F

Combine all the expressions and solve for \(F\) such that \(F > \left(mg \sin \theta + \mu mg \cos \theta\right)\). Use trigonometric identities and simplified geometry-based expressions to express \(F_\text{up incline}\) and solve for \(F\).

Step 6: Final Expression for F

After simplification, the expression for \(F\) reveals that the minimum force required is \[F = \frac{m (g \sin \theta + \mu g \cos \theta)}{\cos \theta}\].

Key Concepts

Newton's second law

Newton's second law is one of the most important principles in physics. It relates the net force acting on an object to its mass and acceleration. This law can be expressed by the equation: \[ F = ma \] where:

• \( F \) is the net force applied to the object.
• \( m \) is the mass of the object.
• \( a \) is the acceleration of the object.

In this exercise, we are interested in the force exerted on the small block of mass \( m \) sitting on an incline. The force we are solving for is the minimum horizontal force \( F \) applied to the larger block \( M \) so that the smaller block \( m \) begins to slide up the incline. The relationship given by Newton's second law helps us set up the equations needed to understand how all the forces work together. It's essential to recognize how forces along the incline will impact the small block. The net force required to cause movement is determined by balancing the gravitational force and force of static friction against the upward force applied by the block \( M \). Newton's second law ensures that we calculate the right amount of force needed to achieve motion. Newton's second law serves as the foundation for analyzing how these different force components contribute to starting the block's motion up the incline.

trigonometric functions

In physics, trigonometric functions are incredibly useful for resolving forces into their components. When dealing with objects on an incline, they help us understand how to break down forces into parts that are parallel and perpendicular to the slope. For a block on an inclined plane, when the gravitational force \( mg \) needs to be resolved, trigonometric functions are deployed as follows:

• The component of the gravitational force acting parallel to the incline is \( mg \sin \theta \).
• The component acting perpendicular is \( mg \cos \theta \).

Here, \( \theta \) represents the angle between the incline and the horizontal. By broken down into these components, it assists in the further analysis and application of Newton's laws. Utilizing trigonometric functions simplifies the complex interplay of forces on an incline. These functions allow us to address specific characteristics of each force direction. This specificity is crucial for accurately solving physics problems that involve angles and inclines. Whenever an exercise mentions an incline or an angle, you can be sure that trigonometric functions will play a role in fair understanding and clear presentation of force components.

force components

Force components are critical in breaking down complex physical interactions into more manageable parts. By doing so, one can study each component individually to understand the full picture of an object's movement. On an inclined surface, forces acting on an object can be divided into parallel and perpendicular components:

• Parallel to the incline, forces drive motion and include the component of gravitational force \( mg \sin \theta \) and any external force components involved.
• Perpendicular to the incline, forces like the normal force and gravitational component \( mg \cos \theta \) work to balance each other out.

In the exercise given, force components are used to assess when the small block starts moving up the slope. The parallel force component, which includes the force due to gravity and static friction, must be overcome by the component of the external force applied horizontally. Understanding force components is key in problems involving inclines because it simplifies complex systems into parts you can easily measure and analyze. By resolving these components properly, you can reliably predict whether an object will move—and what force is required to make it happen. This mindset is crucial for tackling many physics problems successfully.