Problem 76 - Algebra and Trigonometry

Question

A man pushes a wheelbarrow up an incline of \(20^{\circ}\) with a force of 100 pounds. Express the force vector \(\mathbf{F}\) in terms of \(\mathbf{i}\) and \(\mathbf{j}\).

Step-by-Step Solution

Answer

\textbf{F} = 94 \textbf{i} + 34 \textbf{j}

Step 1 - Understand the problem

First, realize that we need to express the force vector \(\textbf{F}\) in terms of its horizontal and vertical components. Since the incline is \(20^{\text{°}}\), the force vector \(\textbf{F}\) makes an angle of \(20^{\text{°}}\) with the horizontal axis.

Step 2 - Resolve the force into components

The horizontal component \(F_x\) of the force can be found using \(F_x = F \cos \theta\), and the vertical component \(F_y\) can be found using \(F_y = F \sin \theta\), where \(F = 100\) pounds and \(\theta = 20^{\text{°}}\).

Step 3 - Calculate the horizontal component

Using \( F_x = 100 \cos(20^{\text{°}}) \), we can find the horizontal component. \(\text{Cosine of } 20^{\text{°}} \text{ is approximately } 0.94\), so \( F_x = 100 \times 0.94 = 94 \text{ pounds}\).

Step 4 - Calculate the vertical component

Using \( F_y = 100 \sin(20^{\text{°}}) \), we can find the vertical component. \(\text{Sine of } 20^{\text{°}} \text{ is approximately } 0.34\), so \( F_y = 100 \times 0.34 = 34 \text{ pounds}\).

Step 5 - Combine the components into the vector

Express the force vector \(\textbf{F}\) in terms of \(\textbf{i}\) and \(\textbf{j}\) as \(\textbf{F} = 94 \textbf{i} + 34 \textbf{j}\).

Key Concepts

force vector components

When breaking down a force acting upon an object into its individual directions, we use the concept of force vector components. This process is essential to understanding how each portion of the force affects the object's motion.

A force vector \(\textbf{F}\) can be resolved into horizontal and vertical components. In mathematical terms, this means finding how much of the force is acting in the horizontal direction \(F_x\) and how much is acting in the vertical direction \(F_y\).

The formulas for these components are tied to trigonometric functions, specifically cosine and sine. Given an angle \(\theta\) with respect to the horizontal axis:

• The horizontal component \(F_x = F \, \cos\theta\)
• The vertical component \(F_y = F \, \sin\theta\)

incline angle

Incline angles play a crucial role in solving physics problems involving forces and motion on a slope. The incline angle \(\theta\) is the angle between the inclined surface and the horizontal plane. In our example, the incline angle is \(20^{\circ}\).

To resolve forces in the context of an incline, it’s important to understand how the angle affects the force's direction. When an object moves up or down an inclined plane, the force parallel to the surface directly influences the motion.

If the incline angle increases, the horizontal component of the force becomes smaller, while the vertical component increases, maintaining the overall force applied.

The steps for resolving the force when dealing with an incline involve:

• Identifying the incline angle \(\theta\)
• Using trigonometric functions to find the force components relative to the incline

trigonometric functions

Trigonometric functions are fundamental tools in resolving vector quantities, such as forces. These functions relate the angles of a triangle to the lengths of its sides. For our purposes, the two most important functions are sine and cosine.

Here’s a quick recall of their definitions in the context of a right-angle triangle:

• Sine \(\sin\theta\) = Opposite side / Hypotenuse
• Cosine \(\cos\theta\) = Adjacent side / Hypotenuse

In our force resolution example:

• To find the horizontal component (adjacent to the angle): \(F_x = F \, \cos\theta\)
• To find the vertical component (opposite to the angle): \(F_y = F \, \sin\theta\)

Through trigonometric functions, we can systematically break down forces into their components, making complex physics problems manageable and more understandable.