Vector Sum Calculation Of Four Displacements In Physics

In physics, understanding vector addition is crucial for analyzing motion and forces. This article delves into the process of finding the vector sum of multiple displacements, providing a step-by-step guide to solving such problems. We will specifically address the scenario of finding the resultant displacement from four given displacements: 60 mm north, 30 mm west, 40 mm at 60° north of east, and 50 mm at 30° west of south. This comprehensive guide aims to provide a clear and concise understanding of vector addition, equipping you with the skills to tackle similar problems with confidence. We will break down the problem into manageable steps, starting with the resolution of vectors into their components, followed by the addition of components, and finally, the determination of the magnitude and direction of the resultant vector. By the end of this guide, you will have a solid grasp of the principles of vector addition and its practical applications in physics. Understanding vector addition is not just about solving textbook problems; it's about developing a fundamental understanding of how physical quantities combine and interact. This knowledge is essential for understanding more advanced topics in physics, such as mechanics, electromagnetism, and even quantum mechanics. So, let's embark on this journey of unraveling the intricacies of vector addition and discover how it helps us make sense of the world around us.

Understanding Vectors and Displacements

Before diving into the calculations, let's establish a solid understanding of vectors and displacements. A vector is a physical quantity that has both magnitude and direction. This is in contrast to scalar quantities, which have only magnitude (e.g., temperature, mass). Displacement, a vector quantity, refers to the change in position of an object. It's not just about how far an object has moved (distance), but also in what direction. For example, a displacement of 60 mm north indicates a movement of 60 millimeters in the northward direction. Vectors are often represented graphically as arrows, where the length of the arrow represents the magnitude and the arrowhead indicates the direction. Understanding the graphical representation of vectors is crucial for visualizing and manipulating them. When dealing with multiple displacements, we need to find the resultant displacement, which is the single displacement that has the same effect as all the individual displacements combined. This is where the concept of vector addition comes into play. Vector addition is not the same as scalar addition because we need to consider both the magnitudes and directions of the vectors. We can't simply add the magnitudes together; we need to use techniques that account for the directional nature of vectors. This article will guide you through one such technique: the component method. The component method involves breaking down each vector into its horizontal and vertical components, adding the corresponding components together, and then constructing the resultant vector from the summed components. This method provides a systematic and accurate way to find the vector sum of any number of displacements. In the following sections, we will apply this method to the specific problem at hand, but the underlying principles can be applied to a wide range of vector addition problems.

Resolving Vectors into Components

The first step in finding the vector sum is to resolve each displacement vector into its horizontal (x) and vertical (y) components. This process involves using trigonometric functions (sine and cosine) to determine the projections of the vector onto the x and y axes. Let's consider the four displacements given in the problem:

1. 60 mm north: This displacement is purely vertical, so its x-component is 0 mm and its y-component is 60 mm.
2. 30 mm west: This displacement is purely horizontal, but in the negative x-direction, so its x-component is -30 mm and its y-component is 0 mm.
3. 40 mm at 60° north of east: This displacement has both horizontal and vertical components. To find them, we use the following trigonometric relationships:
   • x-component = magnitude * cos(angle) = 40 mm * cos(60°) = 40 mm * 0.5 = 20 mm
   • y-component = magnitude * sin(angle) = 40 mm * sin(60°) = 40 mm * 0.866 ≈ 34.64 mm
4. 50 mm at 30° west of south: This displacement also has both horizontal and vertical components. Note that since the angle is measured from the south, the vertical component will be negative, and since it is west of south, the horizontal component will also be negative.
   • x-component = - magnitude * sin(angle) = -50 mm * sin(30°) = -50 mm * 0.5 = -25 mm
   • y-component = - magnitude * cos(angle) = -50 mm * cos(30°) = -50 mm * 0.866 ≈ -43.30 mm

By resolving each vector into its components, we have transformed the problem from one of adding vectors in different directions to one of adding scalar components along the x and y axes. This simplifies the addition process significantly. It's important to pay close attention to the signs of the components, as they indicate the direction of the component along the respective axis. A positive x-component means the component points to the right, while a negative x-component means it points to the left. Similarly, a positive y-component means the component points upwards, while a negative y-component means it points downwards. In the next step, we will add these components together to find the x and y components of the resultant displacement.

Adding the Components

Now that we have resolved each displacement vector into its x and y components, the next step is to add the corresponding components together. This involves summing all the x-components to find the x-component of the resultant displacement and summing all the y-components to find the y-component of the resultant displacement. Let's tabulate the components for clarity:

To find the x-component of the resultant displacement (Rx), we sum the x-components of the individual displacements:

Rx = 0 + (-30) + 20 + (-25) = -35 mm

Similarly, to find the y-component of the resultant displacement (Ry), we sum the y-components of the individual displacements:

Ry = 60 + 0 + 34.64 + (-43.30) = 51.34 mm

Therefore, the resultant displacement has an x-component of -35 mm and a y-component of 51.34 mm. These components represent the horizontal and vertical contributions to the overall displacement. The negative x-component indicates a displacement to the west, while the positive y-component indicates a displacement to the north. These two components now define the resultant displacement vector. To fully describe the resultant displacement, we need to find its magnitude and direction. The magnitude tells us the overall distance of the displacement, while the direction tells us the angle of the displacement relative to a reference direction (usually the positive x-axis or the north direction). In the next section, we will use the Pythagorean theorem and trigonometric functions to determine the magnitude and direction of the resultant displacement from its components.

Finding the Magnitude and Direction of the Resultant Vector

With the x and y components of the resultant displacement calculated (Rx = -35 mm, Ry = 51.34 mm), we can now determine the magnitude and direction of the resultant vector. The magnitude (R) represents the overall length of the displacement, and the direction (θ) represents the angle the displacement makes with a reference axis (in this case, we'll use the positive x-axis). To find the magnitude, we use the Pythagorean theorem:

R = √(Rx² + Ry²) = √((-35 mm)² + (51.34 mm)²) = √(1225 + 2635.7956) mm = √3860.7956 mm ≈ 62.14 mm

So, the magnitude of the resultant displacement is approximately 62.14 mm. This tells us the total distance the object has been displaced as a result of the four individual displacements. To find the direction, we use the arctangent function (tan⁻¹), which relates the angle to the ratio of the opposite side (Ry) to the adjacent side (Rx) in a right triangle:

θ' = tan⁻¹(|Ry / Rx|) = tan⁻¹(|51.34 mm / -35 mm|) = tan⁻¹(1.4669) ≈ 55.71°

Note that we used the absolute values of Rx and Ry because the arctangent function only gives angles in the range -90° to +90°. The angle θ' we calculated is the reference angle, which is the angle the resultant vector makes with the x-axis in the quadrant where it lies. Since Rx is negative and Ry is positive, the resultant vector lies in the second quadrant. Therefore, to find the actual angle θ with respect to the positive x-axis, we need to add 180° to the reference angle:

θ = 180° - θ' = 180° - 55.71° ≈ 124.29°

Alternatively, we can express the direction as an angle with respect to the north direction. Since 124.29° is in the second quadrant, it is equivalent to 124.29° - 90° = 34.29° west of north. Therefore, the resultant displacement is approximately 62.14 mm at 34.29° west of north. This fully describes the resultant displacement, giving both its magnitude and direction. By following these steps, we have successfully found the vector sum of the four given displacements. This process can be applied to any number of displacements in any direction.

Conclusion

In conclusion, we have successfully determined the vector sum of four displacements: 60 mm north, 30 mm west, 40 mm at 60° north of east, and 50 mm at 30° west of south. The resultant displacement was found to be approximately 62.14 mm at 34.29° west of north. This was achieved by systematically resolving each displacement into its x and y components, adding the components, and then calculating the magnitude and direction of the resultant vector. This problem highlights the importance of understanding vector addition in physics. Unlike scalar quantities, which can be added directly, vectors require consideration of both magnitude and direction. The component method provides a reliable and efficient way to add vectors, especially when dealing with multiple displacements or forces acting at different angles. Mastering the technique of resolving vectors into components and then recombining them is a fundamental skill for any physics student. It allows us to break down complex problems into simpler parts, making them easier to analyze and solve. The concepts and techniques discussed in this article are not limited to displacement vectors; they can be applied to any type of vector, including velocity, acceleration, force, and momentum. Understanding vector addition is crucial for understanding a wide range of physical phenomena. By practicing these techniques and applying them to different scenarios, you can develop a deeper understanding of the physical world and enhance your problem-solving skills. The ability to accurately add vectors is not just a theoretical exercise; it has practical applications in fields such as navigation, engineering, and computer graphics. Whether you are calculating the trajectory of a projectile, designing a bridge, or creating a video game, a solid understanding of vector addition is essential. We hope this comprehensive guide has provided you with a clear and concise understanding of vector addition and its applications. Remember to practice these techniques regularly to solidify your understanding and build confidence in your problem-solving abilities.