Problem 32 - Essential Calculus Early Transcendentals

Question

Three forces act on an object. Two of the forces are at an angle of \(100^{\circ}\) to each other and have magnitudes 25 \(\mathrm{N}\) and 12 \(\mathrm{N} .\) The third is perpendicular to the plane of these two forces and has magnitude 4 \(\mathrm{N.}\) Calculate the magnitude of the force that would exactly counterbalance these three forces.

Step-by-Step Solution

Answer

The magnitude of the counterbalancing force is approximately 26.09 N.

Step 1: Resolve Forces in the Plane

First, we need to find the resultant of the two forces in the plane. These forces have magnitudes of 25 N and 12 N and act with an angle of \(100^{\circ}\) between them. Using the law of cosines, we calculate the resultant force \( R \) as follows:\[R = \sqrt{25^2 + 12^2 + 2 \times 25 \times 12 \times \cos(100^{\circ})}\]Calculate \( \cos(100^{\circ}) \), which is approximately -0.1736, and substitute it in:\[R = \sqrt{25^2 + 12^2 + 2 \times 25 \times 12 \times (-0.1736)}\]This simplifies to:\[R = \sqrt{625 + 144 - 104.16} \approx \sqrt{664.84}\]Therefore, \( R \approx 25.78 \text{ N} \).

Step 2: Consider the Perpendicular Force

The third force is perpendicular to the plane of the first two forces, with a magnitude of 4 N. This gives us a vertical component when considering the forces in three-dimensional space. We now have a system with one resultant force from the plane, \( R = 25.78 \text{ N} \), and one force perpendicular to it, \(4 \text{ N} \).

Step 3: Calculate the Counterbalancing Force

To find the magnitude of the force that will counterbalance these, we form a right triangle with the two forces, one acting in the plane (25.78 N) and the perpendicular force (4 N). The magnitude of the total force \( F \) can be found using the Pythagorean theorem:\[F = \sqrt{R^2 + 4^2} = \sqrt{25.78^2 + 4^2}\]Calculating this gives:\[F = \sqrt{664.8484 + 16} = \sqrt{680.8484}\]Thus, \( F \approx 26.09 \text{ N} \).

Key Concepts

Law of Cosines

The Law of Cosines is a crucial tool for solving problems involving triangles, especially when the triangle is not a right triangle. This law extends the Pythagorean theorem to accommodate triangles with any angle, making it particularly helpful in three-dimensional force problems.

The Law of Cosines states that for any triangle with sides of length \(a\), \(b\), and \(c\), and an angle \( \theta \) opposite side \(c\), the relationship is given by:

• \( c^2 = a^2 + b^2 - 2ab \cos(\theta) \)

This formula allows us to calculate the length of a side when we know the lengths of the other two sides and the included angle.

In our exercise, we used this formula to find the resultant force of two forces, 25 N and 12 N, acting at an angle of \(100^{\circ}\). We substituted the given values into the formula and calculated \( \cos(100^{\circ})\), which is approximately \(-0.1736\). Once you have the cosine of the angle, it becomes straightforward arithmetic to find the resultant force. This step is critical in solving many physics problems where forces are not aligned along nice, right-angle paths.

Vector Resultant

A vector resultant is the single vector that represents the sum of two or more vectors. In the context of forces, it is essential for understanding how multiple forces acting on an object can be combined into one overall effect.

When dealing with vectors, each has both a magnitude and a direction. To find the resultant of vectors, we usually use vector addition methods, such as the parallelogram law or triangle law.

In the exercise, the task was to combine two non-parallel force vectors that form a 3D system with a third perpendicular force. For the two coplanar forces of 25 N and 12 N, acting at an angle of \(100^{\circ}\), we employed the Law of Cosines to find the resultant force vector in the plane. This method simplifies the problem before considering the out-of-plane force.

By identifying this resultant first, you simplify the complexity of understanding the system, making it easier to manage the additional perpendicular force later on, in order to find the total force in three-dimensional space.

Pythagorean Theorem

The Pythagorean Theorem is a foundational principle in geometry and is particularly useful when dealing with right-angled triangles. For any right triangle, the square of the length of the hypotenuse (c) is equal to the sum of the squares of the lengths of the other two sides (a and b). Mathematically, this is expressed by the formula:

• \( c^2 = a^2 + b^2 \)

In the given problem, the Pythagorean Theorem was instrumental in determining the magnitude of the counterbalancing force. After resolving the two coplanar forces into a single resultant force using the Law of Cosines, we introduced the third force, which was perpendicular to the plane.

The system can be visualized as a right triangle, where the hypotenuse is the force needed to counterbalance. By applying the Pythagorean theorem, the hypotenuse was calculated as the combination of the resultant force in the plane (25.78 N) and the perpendicular force (4 N).

This application showcases the theorem’s simplicity and its importance in breaking down complex three-dimensional force problems into manageable right-triangle calculations. It exemplifies how foundational geometry principles can solve real-world physics problems efficiently.