Problem 5 - University Physics with Modern Physics

Question

Forces \(\vec{F_1}\) and \(\vec{F_2}\)act at a point. The magnitude of \(\vec{F_1}\) is 9.00 N, and its direction is 60.0\(^\circ\) above the \(x\)-axis in the second quadrant. The magnitude of \(\vec{F_2}\) is 6.00 N, and its direction is 53.1\(^\circ\) below the \(x\)-axis in the third quadrant. (a) What are the \(x\)- and \(y\)-components of the resultant force? (b) What is the magnitude of the resultant force?

Step-by-Step Solution

Answer

The resultant force has components \(-8.11\, N\) (x-axis) and \(3.00\, N\) (y-axis), with magnitude 8.65 N.

Step 1: Understanding the Forces

We have two forces, \(\vec{F_1}\) and \(\vec{F_2}\). The force \(\vec{F_1}\) has a magnitude of 9.00 N and is directed 60.0\(^\circ\) above the \(x\)-axis in the second quadrant. \(\vec{F_2}\) has a magnitude of 6.00 N and is directed 53.1\(^\circ\) below the \(x\)-axis in the third quadrant.

Step 2: Breaking Down \(\vec{F_1}\) into Components

To find the \(x\)- and \(y\)-components of \(\vec{F_1}\), use trigonometry:\[F_{1x} = F_1 \cdot \cos(180^\circ - 60^\circ) = 9.00 \cdot \cos(120^\circ) = 9.00 \cdot (-0.5) = -4.50 \text{ N}\]\[F_{1y} = F_1 \cdot \sin(180^\circ - 60^\circ) = 9.00 \cdot \sin(120^\circ) = 9.00 \cdot 0.866 = 7.79 \text{ N}\]

Step 3: Breaking Down \(\vec{F_2}\) into Components

For \(\vec{F_2}\), also use trigonometry:\[F_{2x} = F_2 \cdot \cos(180^\circ + 53.1^\circ) = 6.00 \cdot \cos(233.1^\circ) = 6.00 \cdot (-0.6018) = -3.61 \text{ N}\]\[F_{2y} = F_2 \cdot \sin(180^\circ + 53.1^\circ) = 6.00 \cdot \sin(233.1^\circ) = 6.00 \cdot (-0.7986) = -4.79 \text{ N}\]

Step 4: Calculating the Resultant Force Components

Add the \(x\)-components and \(y\)-components of \(\vec{F_1}\) and \(\vec{F_2}\):\[R_x = F_{1x} + F_{2x} = -4.50 + (-3.61) = -8.11 \text{ N}\]\[R_y = F_{1y} + F_{2y} = 7.79 + (-4.79) = 3.00 \text{ N}\]

Step 5: Finding the Magnitude of the Resultant Force

The magnitude of the resultant force \(R\) can be found using the Pythagorean theorem:\[R = \sqrt{R_x^2 + R_y^2} = \sqrt{(-8.11)^2 + (3.00)^2} = \sqrt{65.84 + 9.00} = \sqrt{74.84} = 8.65 \text{ N}\]

Key Concepts

Resultant Force

In physics, a resultant force is the single force that represents the vector sum of two or more individual forces acting on a point. To understand the resultant force, imagine this: if you have multiple forces pulling or pushing an object in different directions, the resultant force is what's left when you combine all those forces together.
The resultant force not only has a specific magnitude but also a unique direction. It determines the acceleration and motion of the object according to Newton's second law of motion. In the provided exercise, we were dealing with two forces, \(\vec{F_1}\) and \(\vec{F_2}\), acting at a point in different directions. By adding these forces vectorially, we found the resultant force that acts as if it were a single force exerted on the point.
To determine the magnitude and direction of the resultant force, both the individual components of the forces are considered. This involves breaking down each force into its horizontal (x-axis) and vertical (y-axis) components. Once we have these components, they are summed up separately to get the resultant components (i.e., \(R_x\) and \(R_y\)), which we then use to calculate the overall magnitude and direction of the resultant force.

Force Components

The concept of force components is crucial for solving problems involving forces acting at angles. Any force can be resolved into two perpendicular components: one along the x-axis and the other along the y-axis. These are known as the horizontal and vertical components of the force, respectively.
For instance, in the exercise, \(\vec{F_1}\) was resolved into \(F_{1x}\) and \(F_{1y}\), and \(\vec{F_2}\) was resolved into \(F_{2x}\) and \(F_{2y}\).
Here's how it works:

• The x-component of a force is found by multiplying the force's magnitude by the cosine of its angle relative to the positive x-axis.
• The y-component is similarly found by multiplying the force's magnitude by the sine of the angle.

These components are vector quantities themselves and can be positive or negative depending on the quadrant in which the force is acting. As seen in the exercise, correctly resolving these components is key to determining the overall effect resulting forces have on an object.

Trigonometric Functions

Trigonometric functions like sine and cosine play an indispensable role in physics, especially when dealing with vector quantities like forces. These functions help translate angled forces into their respective x and y components, which is essential when adding vectors.
In the exercise, each force was represented in terms of its magnitude and direction. The angles given were utilized with cosine and sine functions:

• Cosine function (cos) deals with the adjacent side (x-component) in a right triangle formed by the force and axes.
• Sine function (sin) deals with the opposite side (y-component).

This process is grounded in the fundamental principles of trigonometry, which state that for a given angle in a right triangle, the cosine of the angle is the ratio of the length of the adjacent side to the hypotenuse, while the sine is the ratio of the opposite side to the hypotenuse. Therefore, trigonometric functions bridge the angle forces make with coordinate axes, making it possible to solve for component forces operating perpendicularly on the axes.