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 \(0.600 .\) a) How many forces are acting on the block? b) What is the normal force? c) Is this block moving? Explain.

Short answer

Answer: There are 3 forces acting on the block, the normal force is 18.45 N, and the block is not moving.

Step-by-step solution

Identify the forces acting on the block

There are three forces acting on the block: 1. Gravitational force (weight), acting vertically downward. 2. Normal force, acting perpendicular to the surface of the inclined plane. 3. Frictional force, acting parallel to the surface of the inclined plane. So, there are 3 forces acting on the block.

Resolve the gravitational force

We need to resolve the gravitational force (weight = \(mg\), where \(m\) = mass, \(g\) = gravitational acceleration) into two components: one parallel to the inclined plane and the other perpendicular. The parallel component of the gravitational force is \(mg\sin{\theta}\), where \(\theta\) = angle of inclination, and the perpendicular component of the gravitational force is \(mg\cos{\theta}\). For this problem, \(m=2.00\,\text{kg}\), \(g=9.81\,\text{m/s}^2\), and \(\theta=20.0^\circ\). Now, calculate the components of the gravitational force: Parallel component: \(F_\text{parallel} = m \cdot g \cdot \sin{\theta} = 2 \cdot 9.81 \cdot \sin{(20^\circ)} = 6.69\, \text{N}\) Perpendicular component: \(F_\text{perpendicular} = m \cdot g \cdot \cos{\theta} = 2 \cdot 9.81\cdot\cos{(20^\circ)}= 18.45\,\text{N}\)

Determine the normal force

The normal force (N) is equal in magnitude to the perpendicular component of the gravitational force, but acts in the opposite direction. Normal force is \(N=F_\text{perpendicular}=18.45\,\text{N}\).

Determine the maximum static frictional force

The maximum static frictional force (F\(_\text{max}\)) can be calculated using the formula: \(F_\text{max}=\mu_\text{s}N\), where \(\mu_\text{s}\) is the coefficient of static friction. Given, \(\mu_\text{s} = 0.600\) and \(N = 18.45\,\text{N}\), we can calculate the maximum static frictional force: \(F_\text{max} = 0.600 \times 18.45 = 11.07\,\text{N}\)

Determine if the block is moving or not

If the parallel component of the gravitational force, \(F_\text{parallel}\), is less than or equal to the maximum static frictional force, \(F_\text{max}\), then the block will not move. Otherwise, it will start moving. For this problem, \(F_\text{parallel} = 6.69\,\text{N}\), and \(F_\text{max}=11.07\,\text{N}\). Since \(F_\text{parallel} \leq F_\text{max}\), the block is not moving. To summarize: a) There are 3 forces acting on the block. b) The normal force is 18.45 N. c) The block is not moving, as the parallel component of the gravitational force is less than the maximum static friction force.

Key concepts

Static Friction

Imagine trying to push a heavy box across the floor. Initially, the box doesn't move. This resistance you feel is due to static friction, which is the force that counteracts the applied force and prevents an object from moving. In our inclined plane example, when the block rests on the slope, it's the static frictional force that holds it in place, preventing it from sliding down.

Static friction only comes into play when a force is applied to an object but hasn't overcome the force of static friction. It has a maximum value given by the equation \( F_{\text{max}} = \mu_s \times N \) where \( \mu_s \) is the coefficient of static friction and \( N \) is the normal force. This maximum static frictional force is what determines whether an object will move or continue to remain at rest under the influence of an applied force. In our scenario, as the parallel component of gravitational force (which acts as the 'pushing' force down the slope) is less than this maximum value, the block does not slide down the incline.

Normal Force

When discussing forces on an inclined plane, it's crucial to understand the concept of the normal force. This force is always perpendicular to the contact surface and is a reactionary force that surfaces exert to support the weight of an object. It is the component of the gravitational force that is 'normal', or perpendicular, to the surface of the inclined plane.

The normal force balances the perpendicular component of the gravitational force, and its magnitude can be calculated by \( N = m \cdot g \cdot \cos(\theta) \) where \( m \) represents mass, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of incline. In our block example, the normal force is what supports the block against gravity and prevents it from sinking into the plane. This force is also significantly involved in determining the maximum static friction since the two are proportional to each other.

Components of Gravitational Force

The force of gravity on an object at rest on an inclined plane can be thought of as having two distinct components. One component is parallel to the plane's surface and attempts to slide the object down the slope, while the other is perpendicular to the plane and is countered by the normal force.

The perpendicular component of the gravitational force contributes to the normal force and is given by \( F_{\text{perpendicular}} = m \cdot g \cdot \cos(\theta) \) where \( m \) is the object's mass, \( g \) is the gravitational acceleration, and \( \theta \) is the angle of the incline. The parallel component is the one that effectively tries to move the object along the surface of the incline and is calculated by \( F_{\text{parallel}} = m \cdot g \cdot \sin(\theta)\). Understanding these two components helps in analyzing the motion (or lack thereof) for an object on an inclined plane and is fundamental in solving related physics problems.