Problem 4 - Applied Mathematics

Question

A small object of weight \(4 W\) in rough contact with a horizontal plane is acted upon by a force inclined at \(30^{\circ}\) to the plane. When the force is of magnitude \(2 W\) the object is about to slip. Calculate the magnitude of the normal reaction. and the coefficient of friction between the object and the plane.

Step-by-Step Solution

Answer

The normal reaction force is 3W. The coefficient of friction is √3/3.

Step 1 - Resolve Forces

Resolve the force of magnitude 2W into its horizontal and vertical components. The horizontal component is given by: \[ F_{\text{horizontal}} = 2W \times \frac{\text{adjacent}}{\text{hypotenuse}} = 2W \times \frac{\text{cos} 30^{\text{o}}} = 2W \times \frac{\frac{\text{√3}}{2}}{1} = W\text{√ 3} \]The vertical component is given by: \[ F_{\text{vertical}} = 2W \times \frac{\text{opposite}}{\text{hypotenuse}} = 2W \times \text{sin} 30^{\text{o}} = 2W \times \frac{1}{2} = W \]

Step 2 - Calculate Normal Reaction

The weight of the object is acting directly downward and is given by 4W. The normal reaction force, N, must balance the vertical forces. Therefore, we have:\[ N + W = 4W \]So, the normal reaction force is:\[ N = 4W - W = 3W \]

Step 3 - Calculate Frictional Force

The frictional force, which acts horizontally, is given by the horizontal component of the applied force just before the object is about to slip:\[ F_{\text{friction}} = W√ 3 \]

Step 4 - Determine Coefficient of Friction

The coefficient of friction, μ, is the ratio of the frictional force to the normal reaction:\[ \text{Coefficient of friction} (μ) = \frac{F_{\text{friction}}}{N} = \frac{W\text{√ 3}}{3W} = \frac{\text{√ 3}}{3} = \frac{\text{√ 3}}{3} = \frac{\text{√ 3}}{3} \]

Key Concepts

normal reaction force

The normal reaction force is an essential concept in physics, especially when dealing with objects in contact with a surface. The normal reaction force, often denoted as \(N\), is the force exerted by a surface perpendicular to the object in contact with it. This force prevents the object from passing through the surface and balances the perpendicular components of other forces acting on the object.

In this problem, the object's weight \(4W\) acts downward, and the vertical component of the applied force \(W\) acts upward. These forces contribute to the calculation of the normal reaction force. The normal reaction can be determined by balancing these vertical forces:

\[ N + W = 4W \]
Here, \(N\) is the force exerted by the plane counteracting the object's weight and the vertical component of the applied force.

Rearranging the equation, we find:
\[ N = 4W - W = 3W \]
So, the normal reaction force is \(3W\). This means the plane is pushing back with a force of \(3W\) to support the object.

coefficient of friction

The coefficient of friction (\(μ\)) is a dimensionless quantity that represents the ratio of the frictional force resisting motion between two surfaces to the normal reaction force. It quantifies how much frictional force is generated when the object attempts to move across a surface.

In this case, the frictional force is given by the horizontal component of the applied force just before the object is about to slip:

\[ F_{\text{friction}} = W\sqrt{3} \]

To find the coefficient of friction, we use the formula:

\[ μ = \frac{F_{\text{friction}}}{N} \]

Using the previously calculated normal reaction force \(N = 3W\):

\[ μ = \frac{W\sqrt{3}}{3W} = \frac{\sqrt{3}}{3} \]

This gives us the coefficient of friction as \(\frac{\sqrt{3}}{3}\). The coefficient of friction indicates how well the surface resists the sliding motion of the object.

resolving forces

Resolving forces is the process of breaking down a single force into two or more components, usually along perpendicular axes. This technique is crucial for understanding how different forces interact and contribute to an object's motion.

For an applied force of magnitude \(2W\) inclined at \(30^\circ\) to the horizontal plane, we resolve it into horizontal and vertical components:

1. **Horizontal Component**: Given by \[ F_{\text{horizontal}} = 2W \times \cos 30^\circ = 2W \times \frac{\sqrt{3}}{2} = W\sqrt{3} \]

2. **Vertical Component**: Given by \[ F_{\text{vertical}} = 2W \times \sin 30^\circ = 2W \times \frac{1}{2} = W \]

These components help us determine how the force affects the object along different directions. The horizontal component influences the frictional force, while the vertical component affects the normal reaction force.

By breaking the force into components, we can analyze the impact precisely. The technique is essential for solving problems involving inclined forces, as seen in this exercise.