Question

A \(2.00-\mathrm{kg}\) block is on a plane inclined at \(20.0^{\circ}\) with respect to the horizontal. The coefficient of static friction between the block and the plane is \(0.60 .\) a) How many forces are acting on the block? b) What is the normal force? c) Is this block moving? Explain.

Short answer

There are 3 forces acting on the block: gravitational force, normal force, and frictional force. b) What is the normal force? The normal force is 18.51 N. c) Is this block moving? Explain. The block is not moving because the maximum static friction force (11.11 N) is greater than the parallel force component of gravity acting on the block (6.68 N).

Step-by-step solution

Identify forces acting on the block

To determine the number of forces acting on the block, we must first recognize the types of forces involved. There are three forces acting on the block: gravitational force (weight), the normal force (perpendicular to the inclined plane), and frictional force (opposing any motion). a) So, there are 3 forces acting on the block.

Resolve gravitational force into components

To analyze the normal force and frictional force, we need to resolve the gravitational force of the block into components parallel and perpendicular to the inclined plane: 1. Parallel component: \(F_{g \parallel} = mg \sin{(\theta)}\) 2. Perpendicular component: \(F_{g \perp} = mg \cos{(\theta)}\) Here, \(m = 2.00\,\text{kg}\), \(g = 9.81\,\text{m/s}^2\), and \(\theta = 20.0^{\circ}\). First, let's find \(F_{g\parallel}\) and \(F_{g\perp}\): 1. \(F_{g \parallel} = (2.00\,\text{kg})(9.81\,\text{m/s}^2)\sin{(20.0^{\circ})} = 6.68\,\text{N}\) 2. \(F_{g \perp} = (2.00\,\text{kg})(9.81\,\text{m/s}^2)\cos{(20.0^{\circ})} = 18.51\,\text{N}\)

Calculate the normal force

The normal force (\(F_N\)) acts perpendicular to the inclined plane and has the same magnitude as \(F_{g\perp}\), but in the opposite direction. Thus, b) The normal force is \(F_N = 18.51\,\text{N}\).

Determine the maximum static friction force

The maximum static friction force (\(F_{s\,max}\)) can be calculated using: \(F_{s\,max} = \mu_{s}F_N\) where, \(\mu_{s}=0.60\) (coefficient of static friction) and \(F_N = 18.51\,\text{N}\) (normal force). Now, let's calculate \(F_{s\,max}\): \(F_{s\,max} = (0.60)(18.51\,\text{N}) = 11.11\,\text{N}\)

Determine if the block is moving

To determine if the block will move, compare \(F_{s\,max}\) and \(F_{g\parallel}\). If the maximum static friction force (\(F_{s\,max}\)) is greater than or equal to the parallel gravitational force component (\(F_{g\parallel}\)), the block remains stationary. If \(F_{s\,max} < F_{g\parallel}\), the block will slide down the incline. We have calculated \(F_{g\parallel} = 6.68\,\text{N}\) and \(F_{s\,max} = 11.11\,\text{N}\). Here, \(F_{s\,max} > F_{g\parallel}\). c) So, the block is not moving, as the maximum static friction force is greater than the parallel force component of gravity acting on the block.

Key concepts

Static Friction

Static friction plays a crucial role in determining whether an object on an inclined plane will start moving or remain at rest. When a block is placed on a slope, static friction opposes the tendency of the block to slide down. This frictional force is essential in balancing forces when the object is stationary.
An important aspect of static friction is that it has a maximum value, beyond which the object will start to move. This maximum value is determined by the formula:

• \( F_{s\,max} = \mu_{s}F_N \)

Where \( \mu_{s} \) is the coefficient of static friction and \( F_N \) is the normal force. In this exercise, \( \mu_{s} \) is given as 0.60. By knowing the maximum static friction force, \( F_{s\,max} \), we can assess whether the gravitational force parallel to the inclined plane, \( F_{g\parallel} \), is sufficient to overcome it. If \( F_{g\parallel} \) is greater than \( F_{s\,max} \), the block will move. In our case, \( F_{s\,max} = 11.11\,\text{N} \), which is greater than \( F_{g\parallel} = 6.68\,\text{N} \), indicating that the block remains stationary.

Normal Force

The normal force is another important aspect of physics when analyzing objects on inclined planes. It is the force perpendicular to the contact surface, counteracting the component of the gravitational force that presses the block into the surface. Normally, it can be seen as the response of a surface to the object's weight acting on it.
Using the formula:

• \( F_{g\perp} = mg \cos{(\theta)} \)

and understanding that the normal force is equal in magnitude to \( F_{g\perp} \) but acts in the opposite direction, we can compute it. In the problem, we calculated \( F_{g\perp} \) to be \( 18.51\,\text{N} \). Therefore, the normal force \( F_N \), which supports the block against the surface, is equally \( 18.51\,\text{N} \). This force ensures the block does not penetrate the plane and influences the static friction force.

Gravitational Force Components

Gravitational force acting on an object can be broken down into two components when placed on an inclined surface: parallel and perpendicular to the plane. The full weight of the object is divided based on the incline’s angle, affecting both the normal and frictional forces.
For the parallel component, the force is directed along the incline, prompting potential movement. It can be calculated using the formula:

• \( F_{g\parallel} = mg \sin{(\theta)} \)

For the perpendicular component, which contributes to the normal force, we use:

• \( F_{g\perp} = mg \cos{(\theta)} \)

In our scenario, the block has a mass \( m \) of \( 2.00\,\text{kg} \), gravity \( g \) is \( 9.81\,\text{m/s}^2 \), and the incline angle \( \theta \) is \( 20.0^{\circ} \). Thus, we found \( F_{g\parallel} \) to be \( 6.68\,\text{N} \) and \( F_{g\perp} \) as \( 18.51\,\text{N} \). Understanding these components helps in analyzing whether forces are balanced or if motion will occur.