Problem 22 - A History of Mathematics: An Introduction

Question

Prove: The times of motion of a moveable starting from rest over equal planes unequally inclined are to each other inversely as the square root of the ratio of the heights of the. planes.

Step-by-Step Solution

Answer

Answer: The times of motion of a moveable starting from rest over equal planes that are unequally inclined are inversely proportional to the square root of the ratio of the heights of the planes, which can be represented as t1 / t2 = sqrt(h1 / h2).

Step 1: Define variables and set up the problem

Let two equal planes of length \(L\) be inclined at angles \(\alpha_1\) and \(\alpha_2\) to the horizontal. The heights of the planes are \(h_1 = L\sin\alpha_1\) and \(h_2 = L\sin\alpha_2\). The accelerations along the planes (starting from rest) are \(a_1 = g\sin\alpha_1\) and \(a_2 = g\sin\alpha_2\).

Step 2: Find the time of motion for each plane

Using \(s = \frac{1}{2}at^2\) with \(s = L\) (the length of the plane):
For plane 1: \(L = \frac{1}{2}g\sin\alpha_1 \cdot t_1^2\), so \(t_1 = \sqrt{\frac{2L}{g\sin\alpha_1}}\)
For plane 2: \(L = \frac{1}{2}g\sin\alpha_2 \cdot t_2^2\), so \(t_2 = \sqrt{\frac{2L}{g\sin\alpha_2}}\)

Step 3: Compute the ratio of times

\(\frac{t_1}{t_2} = \sqrt{\frac{\sin\alpha_2}{\sin\alpha_1}}\)
Since \(h_1 = L\sin\alpha_1\) and \(h_2 = L\sin\alpha_2\), we have \(\sin\alpha_1 = \frac{h_1}{L}\) and \(\sin\alpha_2 = \frac{h_2}{L}\).
Therefore: \(\frac{t_1}{t_2} = \sqrt{\frac{h_2/L}{h_1/L}} = \sqrt{\frac{h_2}{h_1}}\)

Step 4: Conclusion

We have \(\frac{t_1}{t_2} = \sqrt{\frac{h_2}{h_1}}\), which means \(t_1 : t_2 = \frac{1}{\sqrt{h_1}} : \frac{1}{\sqrt{h_2}}\).
This proves that the times are inversely proportional to the square root of the heights. Equivalently, \(t_1\sqrt{h_1} = t_2\sqrt{h_2}\), confirming the theorem.

Key Concepts

Equations of Motion

Understanding the equations of motion is crucial for analyzing objects in motion, particularly in physics. These equations describe how the velocity of an object changes under the influence of forces and can be used to calculate its position over time. For an object starting from rest and moving under a constant acceleration, the key equation is

\( s = \frac{1}{2}at^2 \),

where \( s \) is the distance traveled, \( a \) is the constant acceleration, and \( t \) is the time. In the context of our exercise, this equation allowed us to connect the length of the inclined plane, the gravitational acceleration component along the incline, and the time it takes for an object to move down the incline. It's a beautiful example of how certain physical concepts can be expressed concisely and used to determine other unknown variables.

Gravitational Acceleration

In physics, gravitational acceleration is the acceleration of an object due to the force of gravity. Near the surface of the Earth, this acceleration is approximately \( 9.81 \, \text{m/s}^2 \) and is symbolized by \( g \). It's interesting to note that all objects, regardless of their mass, fall at the same rate when only gravity is acting on them, neglecting air resistance.

In the exercise example, we looked at how gravity's effect changes when an object is on an inclined plane. The gravitational force exerted on the object is not directed straight down but rather along the surface of the incline. The component of gravitational acceleration that acts along the plane is what influences the object's motion, and is found using the sine function, a fundamental concept in trigonometry. Therefore, this component is calculated as
\( a = g \times \sin(\theta) \),

where \( \theta \) is the angle of inclination. Gravitational acceleration is paramount in motion physics as it determines the object's acceleration in the absence of other forces.

Trigonometry in Physics

Trigonometry is not just a branch of mathematics, but a powerful tool used extensively in physics to solve problems involving angles and dimensions, especially in cases like motion on inclined planes. The trigonometric functions—sine, cosine, and tangent—are fundamental when dealing with right triangles, which can be used to represent the geometry of an inclined plane.

In our example, we use the sine function to connect the angle of the incline with both the gravitational acceleration component along the plane and the height of the plane. The relationships
\( h = L \times \sin(\theta) \)
and
\( a = g \times \sin(\theta) \)
are derived using trigonometry, illustrating the sine function's role in dissecting the influence of gravity on a slope and how high the slope rises. Understanding how trigonometry underpins these calculations is essential for students since it provides the link between the angles of the inclined planes and their corresponding heights, aiding in solving motion problems.