Problem 43 - College Physics

Question

A block of mass \(25.0 \mathrm{~kg}\) slides down a frictionless surface inclined at \(30^{\circ} .\) To ensure that the block does not accelerate, what is the smallest force that you must exert on it and what is its direction?

Step-by-Step Solution

Answer

Apply a force of 122.5 N up the incline.

Step 1: Identify the Forces

To solve this problem, first identify all the forces acting on the block. These include the force of gravity (weight), the normal force, and the force that we are trying to find, which will keep the block from accelerating.

Step 2: Calculate the Gravitational Force

The gravitational force acting on the block is the weight of the block, given by the formula \( F_g = m imes g \), where \( m = 25.0 \text{ kg} \) and \( g = 9.8 \, \text{m/s}^2 \). Calculate the weight: \( F_g = 25.0 \, \text{kg} imes 9.8 \, \text{m/s}^2 = 245 \, \text{N} \).

Step 3: Resolve the Gravitational Force

The weight of the block acts vertically downward, but we need to resolve it into components parallel and perpendicular to the plane. The parallel component is given by \( F_{g, \parallel} = F_g \times \sin(30^") \), and the perpendicular component is \( F_{g, \perp} = F_g \times \cos(30^") \).

Step 4: Calculate the Parallel Component

Calculate the parallel component of the gravitational force using \( F_{g, \parallel} = 245 \, \text{N} \times \sin(30^") = 122.5 \, \text{N} \). This force tends to pull the block down the plane.

Step 5: Determine the Required Force

To keep the block from sliding down, you must apply an equal and opposite force to this parallel component. Thus, the required force \( F \) must be \( 122.5 \, \text{N} \) and directed up the plane.

Step 6: State the Direction of the Force

The applied force of magnitude \( 122.5 \, \text{N} \) should be directed up along the incline to cancel out the parallel component of the gravitational force.

Key Concepts

Force Calculation

In physics, calculating forces is essential to understanding how objects interact with each other. When dealing with forces, we use Newton's second law, which states that force is the product of mass and acceleration, expressed as \( F = ma \). In the context of our problem, we first calculate the gravitational force, also known as weight, using the formula \( F_g = mg \), where \( m \) is the mass of the object and \( g \) is the acceleration due to gravity. For the block, with a mass of 25 kg and a gravitational pull of 9.8 m/s², the gravitational force is \( F_g = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 245 \, \text{N} \). This is the starting point for finding any other forces acting on the block on an inclined plane. Understanding force calculation helps us determine the magnitude and direction of additional forces required to affect the motion or maintain the static position of an object.

Inclined Plane

An inclined plane, often referred to as a ramp, is a flat supporting surface tilted at an angle. It is a fundamental concept in physics as it reduces the amount of force needed to lift an object. When an object is placed on an inclined plane, gravity acts on it, and this force can be resolved into two components:

• Parallel to the surface of the incline
• Perpendicular to the surface of the incline

In problems like this, the inclined plane plays a crucial role in illustrating how forces need to be handled differently than on a horizontal surface. The angle of the incline affects the magnitude of these components. The steeper the incline, the greater the force component that acts parallel to the surface, which acts to pull the object down the plane. This requires us to exert a force in the opposite direction to keep the object from sliding down.

Gravitational Force

Gravitational force is the force with which the Earth attracts an object towards itself. It acts downwards, towards the center of the Earth. In many physics problems, it is crucial to resolve this force into components to better understand and manage it in relation to other forces. For inclined planes, resolving gravitational force involves breaking it down into:

• The component perpendicular to the incline: \( F_{g, \perp} = F_g \times \cos(\theta) \)
• The component parallel to the incline: \( F_{g, \parallel} = F_g \times \sin(\theta) \)

In our exercise, with the incline set at \( 30^\circ \), the parallel component \( F_{g, \parallel} \) is what causes the block to slide down the plane at 122.5 N. Understanding gravitational force this way helps us apply the correct opposing force to keep the block stationary.

Newton's Laws

Newton's laws of motion are the foundation for understanding classical mechanics. In our example, Newton's first law, also known as the law of inertia, states that an object at rest stays at rest unless acted upon by a net external force. Newton's second law is key in our scenario on the inclined plane, as it states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration \( F = ma \). For the block on the incline, we determined that the net external force (the applied force) must equal the gravitational force component pulling the block down, but in the opposite direction, to maintain equilibrium. This is because we want the block not to accelerate, adhering to the first law.Newton's laws not only help us calculate the forces needed but also predict the behavior of objects under the influence of these forces, ensuring their applicability in various physical situations.