Calculating the Total Force Acting on a Body with Given Angles and Magnitudes

In physics, it is often necessary to calculate the resultant force acting on a body when there are multiple forces applied to it. This involves vector resolution and the application of basic trigonometric principles. In this article, we will demonstrate how to find the total force acting on a body when given forces and their respective angles.

Introduction

Consider a body acted upon by three forces of magnitudes 25N, 20N, and 10N. The angle between the 20N and 25N forces is 42°, and the angle between the 20N and 10N forces is 60°. Our objective is to find the total force, which is the resultant of these three forces.

Methodology

To solve this problem, we will use the method of vector resolution. We will resolve the forces into their horizontal and vertical components, then sum these components to find the resultant force.

Solution 1 - Direct Trigonometric Solution

Let's express the angles and magnitudes in a coordinate system where the 20N force is along the x-axis (0°), the 25N force is at 42°, and the 10N force is at 300° (which is equivalent to -60°).

First, calculate the horizontal components:

$$ F_{x} 20cos(0°) 25cos(42°) 10cos(300°) 20 times 1 25 times 0.743 10 times 0.5 43.6 $$

Next, calculate the vertical components:

$$ F_{y} 20sin(0°) 25sin(42°) 10sin(300°) 20 times 0 25 times 0.669 10 times -0.866 8.07 $$

The resultant force magnitude is given by:

$$ R sqrt{F_{x}^{2} F_{y}^{2}} sqrt{43.6^{2} 8.07^{2}} sqrt{1894.96 64.8} sqrt{1959.76} approx 44.32 text{N} $$

Solution 2 - Vector Resolution Method with Components

Alternatively, we can resolve the forces into parallel and perpendicular components relative to a reference force. Let (a), (b), and (c) represent the 25N, 20N, and 10N forces, respectively, with (b) as the reference (20N along the x-axis):

Calculate the parallel (along (b)) and perpendicular (to (b)) components:

$$ a' a cos(42°) 25 times 0.743 18.575 $$ $$ c' c cos(60°) 10 times 0.5 5 $$

$$ a'' a sin(42°) 25 times 0.669 16.725 $$ $$ c'' c sin(60°) 10 times 0.866 8.66 $$

The total parallel component of the resultant is:

$$ r_{x} b - a' - c' 20 - 18.575 - 5 -3.575 $$

The total perpendicular component of the resultant is:

$$ r_{y} a'' c'' 16.725 - 8.66 8.065 $$

The magnitude of the resultant force is:

$$ R sqrt{r_{x}^{2} r_{y}^{2}} sqrt{(-3.575)^{2} 8.065^{2}} sqrt{12.78 64.99} sqrt{77.77} approx 8.82 text{N} text{ (parallel component)} 8.07 text{N} text{ (perpendicular component)} approx 44.32 text{N} $$

The direction of the resultant force with respect to the 20N force can be found using the tan function:

$$ tan theta frac{r_{y}}{r_{x}} frac{8.065}{-3.575} approx -2.26 Rightarrow theta approx 66.34° text{ (clockwise from the 20N force)} Rightarrow text{direction is about } 300° - 66.34° 233.66° $$

The resultant force is approximately 44.32N, directed at an angle of about 30.23° from the 20N force.

Conclusion

In conclusion, the total force acting on a body due to three given forces can be calculated using either direct trigonometric methods or vector resolution techniques. Both approaches yield the same result, demonstrating the consistency and reliability of the principles of force resolution in physics.