Calculating Resultant Force On A 6 Kg Body A Physics Problem

by ADMIN 61 views

Introduction

In this article, we will delve into a classic physics problem involving the calculation of the resultant force acting on a 6 kg body initially at rest. This problem involves vector addition, a fundamental concept in physics, particularly in mechanics. We are given three forces with their magnitudes and directions, and our task is to determine the net effect of these forces on the body. Understanding how to combine forces vectorially is crucial for solving a wide range of problems in physics and engineering. This article aims to provide a comprehensive explanation of the solution process, making it accessible to students and enthusiasts alike. The principles and methods discussed here have broad applications in understanding real-world phenomena where multiple forces interact. Let's break down the problem and see how we can find the resultant force.

Problem Statement

We are given a scenario where three forces act on a surge of mass 6 kg, which is initially at rest. These forces are:

  1. (CN1,035∘)\left( CN _1, 035^{\circ}\right)

  2. (8.5M,180∘)\left(8.5 M , 180^{\circ}\right)

  3. (9al,330∘)\left(9 al , 330^{\circ}\right)

Our objective is to find the resultant force acting on the body. The resultant force is the vector sum of all the individual forces acting on the body. To find it, we need to resolve each force into its horizontal (x) and vertical (y) components and then add the components separately. This method allows us to convert the vector addition problem into a series of algebraic additions, which are much easier to handle. The magnitude and direction of the resultant force can then be found from the components using trigonometric functions. The units of the forces are not explicitly provided, but it is assumed that they are in a consistent unit, such as Newtons (N). Without specific units, we will perform the calculations and express the final result in generic force units.

Resolving Forces into Components

To find the resultant force, we must first resolve each individual force into its horizontal (x) and vertical (y) components. This involves using trigonometric functions, specifically sine and cosine, to project each force vector onto the x and y axes. Let's denote the three forces as F1, F2, and F3, with magnitudes and directions as given:

  1. F1 = $, CN _1$ at 35°
  2. F2 = 8.5 M at 180°
  3. F3 = 9 al at 330°

The x-component of a force vector is given by the magnitude of the force multiplied by the cosine of the angle it makes with the x-axis. The y-component is given by the magnitude of the force multiplied by the sine of the angle. Let's calculate the components for each force:

Force 1 (F1):

  • F1x = $, CN _1 \times \cos(35^{\circ})$
  • F1y = $, CN _1 \times \sin(35^{\circ})$

Force 2 (F2):

  • F2x = $8.5 M \times \cos(180^{\circ}) = 8.5 M \times (-1) = -8.5 M$
  • F2y = $8.5 M \times \sin(180^{\circ}) = 8.5 M \times 0 = 0$

Force 3 (F3):

  • F3x = $9 al \times \cos(330^{\circ}) = 9 al \times \cos(-30^{\circ}) = 9 al \times \frac{\sqrt{3}}{2}$
  • F3y = $9 al \times \sin(330^{\circ}) = 9 al \times \sin(-30^{\circ}) = 9 al \times \left(-\frac{1}{2}\right) = -4.5 al$

By resolving each force into its x and y components, we have effectively transformed the problem into a set of simpler calculations. This approach simplifies the addition of forces, which is a vector operation, into the addition of scalar components. The next step involves summing these components to find the resultant force components.

Calculating Resultant Force Components

Now that we have resolved each force into its x and y components, the next step is to add these components together to find the resultant force components. The x-component of the resultant force (Rx) is the sum of the x-components of the individual forces, and similarly, the y-component of the resultant force (Ry) is the sum of the y-components of the individual forces. Mathematically, this can be expressed as:

  • Rx = F1x + F2x + F3x
  • Ry = F1y + F2y + F3y

Substituting the values we calculated in the previous section:

  • Rx = $, CN _1 \times \cos(35^{\circ}) + (-8.5 M) + 9 al \times \frac{\sqrt{3}}{2}$
  • Ry = $, CN _1 \times \sin(35^{\circ}) + 0 + (-4.5 al)$

Without the specific magnitudes of $, CN _1$, M, and al, we can only express the resultant force components in terms of these variables. However, let's assume some example values for demonstration purposes. Suppose $, CN _1 = 10$ units, M = 1 unit, and al = 1 unit. Then:

  • Rx = $10 \times \cos(35^{\circ}) - 8.5 + 9 \times \frac{\sqrt{3}}{2} \approx 10 \times 0.819 - 8.5 + 9 \times 0.866 \approx 8.19 - 8.5 + 7.794 \approx 7.484$
  • Ry = $10 \times \sin(35^{\circ}) - 4.5 \approx 10 \times 0.574 - 4.5 \approx 5.74 - 4.5 \approx 1.24$

Thus, the resultant force components, under these assumptions, would be approximately Rx ≈ 7.484 units and Ry ≈ 1.24 units. These components represent the net effect of the three forces in the x and y directions, respectively. The next step is to use these components to find the magnitude and direction of the resultant force.

Determining Magnitude and Direction of Resultant Force

Now that we have the x and y components of the resultant force (Rx and Ry), we can determine the magnitude (R) and direction (θ) of the resultant force. The magnitude represents the overall strength of the force, while the direction indicates the angle at which the force acts relative to the positive x-axis. The magnitude of the resultant force can be found using the Pythagorean theorem:

R=Rx2+Ry2R = \sqrt{Rx^2 + Ry^2}

And the direction (θ) can be found using the arctangent function:

θ=arctan⁡(RyRx)θ = \arctan{\left(\frac{Ry}{Rx}\right)}

Using the example values we calculated earlier (Rx ≈ 7.484 units and Ry ≈ 1.24 units), let's find the magnitude and direction:

R=(7.484)2+(1.24)2=56.01+1.5376=57.5476≈7.586R = \sqrt{(7.484)^2 + (1.24)^2} = \sqrt{56.01 + 1.5376} = \sqrt{57.5476} \approx 7.586

θ=arctan⁡(1.247.484)=arctan⁡(0.1657)≈9.41∘θ = \arctan{\left(\frac{1.24}{7.484}\right)} = \arctan{(0.1657)} \approx 9.41^{\circ}

So, the magnitude of the resultant force is approximately 7.586 units, and its direction is approximately 9.41° relative to the positive x-axis. This means that the net effect of the three forces is a force of approximately 7.586 units acting in a direction that is about 9.41° counterclockwise from the horizontal axis. It's important to note that the arctangent function gives angles in the range of -90° to +90°, so we need to consider the signs of Rx and Ry to determine the correct quadrant for the angle. In this case, both Rx and Ry are positive, so the angle is in the first quadrant, and our calculation is correct.

Conclusion

In this article, we have walked through the process of calculating the resultant force on a body acted upon by multiple forces. We began by resolving each force into its x and y components using trigonometric functions. This step is crucial because it allows us to convert vector addition into scalar addition. We then summed the x-components and y-components separately to find the components of the resultant force. Finally, we used these components to calculate the magnitude and direction of the resultant force using the Pythagorean theorem and the arctangent function. This methodology is fundamental in physics and engineering for analyzing systems where multiple forces interact.

We have demonstrated the process with a numerical example, assuming specific values for the magnitudes of the forces. However, the same approach can be applied with any set of forces, regardless of their magnitudes and directions. Understanding vector addition and force resolution is essential for solving a wide variety of problems, from simple statics problems to complex dynamics scenarios. This article has provided a step-by-step guide to help readers grasp these concepts and apply them effectively. The principles discussed here are not only applicable to textbook problems but also provide a foundation for understanding real-world phenomena where multiple forces are at play.