Problem 44 - Thomas Calculus

Question

Force vector A kite string exerts a 12 -lb pull \((|\mathbf{F}|=12)\) on a kite and makes a \(45^{\circ}\) angle with the horizontal. Find the horizontal and vertical components of \(\mathbf{F}\) .

Step-by-Step Solution

Answer

Both components are \( 6\sqrt{2} \) lb.

Step 1: Understand the Problem

We need to find the horizontal and vertical components of the force vector \( \mathbf{F} \) exerted by the kite string, which has a magnitude of 12 lb and makes a \( 45^{\circ} \) angle with the horizontal.

Step 2: Identify Component Formulas

The horizontal component \( F_x \) of a vector \( \mathbf{F} \) can be found using \( F_x = |\mathbf{F}| \cos \theta \), and the vertical component \( F_y \) can be found using \( F_y = |\mathbf{F}| \sin \theta \). In this exercise, \( \theta = 45^{\circ} \).

Step 3: Calculate Horizontal Component

Use the formula for the horizontal component: \( F_x = 12 \cos 45^{\circ} \). Knowing \( \cos 45^{\circ} = \frac{\sqrt{2}}{2} \), calculate \( F_x = 12 \times \frac{\sqrt{2}}{2} = 6\sqrt{2} \).

Step 4: Calculate Vertical Component

Use the formula for the vertical component: \( F_y = 12 \sin 45^{\circ} \). Knowing \( \sin 45^{\circ} = \frac{\sqrt{2}}{2} \), calculate \( F_y = 12 \times \frac{\sqrt{2}}{2} = 6\sqrt{2} \).

Step 5: Conclusion

The horizontal component of \( \mathbf{F} \) is \( 6\sqrt{2} \) lb, and the vertical component is \( 6\sqrt{2} \) lb.

Key Concepts

force vectors

Force vectors are mathematical representations of forces that have both a direction and a magnitude. They are critical when dealing with problems in physics involving force, as vectors help in visualizing and calculating the influence of forces on objects. In the context of our exercise, the kite string exerts a force vector, marked as \( \mathbf{F} \), with a magnitude of 12 lbs.
Understanding a force as a vector allows us to manage complex interactions in a system. For example, force vectors help in:

• Analyzing forces acting at different angles.
• Breaking down forces into components, which simplifies the computation of their effects.
• Visualizing the direction of the force applied to a body.

In our scenario, the force vector exerted by the kite string acts at a 45-degree angle to the horizontal. By using the vector components, we can calculate how much of this force is pulling vertically versus horizontally on the kite.

trigonometry

Trigonometry is a branch of mathematics that deals with the relationships between the angles and sides of triangles. It is especially useful in calculating vector components, as it provides the formulas to find horizontal and vertical parts of a vector given an angle. In our exercise:

• We use the cosine function to find the horizontal component of the force vector.
• The sine function is used to determine the vertical component.

When dealing with angles in trigonometry:

• \( \cos \theta \) helps find the adjacent side, or in this case, the horizontal component.
• \( \sin \theta \) is used for the opposite side, corresponding to the vertical component.

Since the angle given in our problem is 45 degrees, we use the known values \( \sin 45^{\circ} = \cos 45^{\circ} = \frac{\sqrt{2}}{2} \).
These trigonometric identities simplify our calculations and make it easier to derive the force components accurately.

magnitude and direction

Magnitude and direction are pivotal in understanding vectors. The magnitude of a vector indicates its size or strength, while direction shows where the vector is pointing. In our case, the magnitude is given as 12 lbs, which tells us the strength of the pull exerted by the kite string.
The direction is often expressed in terms of angles, such as the 45-degree angle given in the exercise which refers to the orientation of the vector relative to a reference line, such as the horizontal axis. Utilizing both magnitude and direction, we calculate:

• The horizontal component, using direction (cosine) and magnitude as \( 6\sqrt{2} \) lbs.
• The vertical component, also leveraging direction (sine) and magnitude, similarly \( 6\sqrt{2} \) lbs.

The importance of magnitude and direction is that they allow us to precisely understand how a vector affects an object. In this exercise, both aspects were essential in separating the force into its directional components, aiding in comprehensive analysis and problem-solving.