Problem 61 - Contemporary Precalculus

Question

Deal with an object on an inclined plane. The situation is similar to that in Figure \(9-20\) of Example \(12,\) where \(\|\overline{T P}\|\) is the component of the weight of the object parallel to the plane and \(\|\overline{T Q}\|\) is the component of the weight perpendicular to the plane. An object weighing 50 pounds lies on an inclined plane that makes a \(40^{\circ}\) angle with the horizontal. Find the components of the weight parallel and perpendicular to the plane.

Step-by-Step Solution

Answer

Answer: The components of the 50-pound weight on an inclined plane at an angle of 40° are approximately 32.14 pounds for the parallel component and 38.27 pounds for the perpendicular component.

Step 1: Draw and label the diagram

Draw a right triangle, with the hypotenuse representing the weight (W) of the object, one leg parallel to the inclined plane (T_P), and the other leg perpendicular to the inclined plane (T_Q). Label the angle between the hypotenuse and T_P as 40°.

Step 2: Apply trigonometric ratios to find the components

Since we have a right triangle, we can use the sine and cosine functions to find the components of the weight. - The sine of the angle will give us the ratio of the weight component parallel to the inclined plane (T_P) to the total weight (W). - The cosine of the angle will give us the ratio of the weight component perpendicular to the inclined plane (T_Q) to the total weight (W).

Step 3: Calculate the parallel component (T_P)

Using the sine function, we can find the parallel component of the weight: $$T_P = W \cdot \sin{(\theta)}$$ Where W is the total weight (50 pounds) and θ is the angle of inclination (40°). $$T_P = 50 \cdot \sin{(40°)}$$ $$T_P \approx 32.14 \,\text{pounds}$$ The parallel component of the weight is approximately 32.14 pounds.

Step 4: Calculate the perpendicular component (T_Q)

Using the cosine function, we can find the perpendicular component of the weight: $$T_Q = W \cdot \cos{(\theta)}$$ $$T_Q = 50 \cdot \cos{(40°)}$$ $$T_Q \approx 38.27 \,\text{pounds}$$ The perpendicular component of the weight is approximately 38.27 pounds.

Step 5: State the final answer

The components of the 50-pound weight on an inclined plane at an angle of 40° are approximately: - Parallel component (T_P): 32.14 pounds - Perpendicular component (T_Q): 38.27 pounds

Key Concepts

Trigonometric Functions

In physics, especially when dealing with inclined planes, trigonometric functions are key tools that help us solve problems related to forces. These functions allow us to break down forces into components which makes it easier to analyze the effects on an object.

There are three primary trigonometric functions often used: sine, cosine, and tangent; for inclined plane problems, sine and cosine are most relevant. They relate the angles and sides of a right triangle formed by the object, the inclined surface, and the components of the force acting on it.

• **Sine Function**: This function is used to calculate the component of a force that runs parallel to the inclined plane. It is defined as the ratio of the length of the side opposite the angle to the hypotenuse. In the context of weight, it calculates the parallel force: \[ \sin{(\theta)} = \frac{\text{opposite}}{\text{hypotenuse}} \]
• **Cosine Function**: This function helps in finding the component of the weight that is perpendicular to the inclined surface. It is the ratio of the adjacent side length to the hypotenuse: \[ \cos{(\theta)} = \frac{\text{adjacent}}{\text{hypotenuse}} \]

Thus, by applying these trigonometric functions, we break down the weight vector into its components parallel and perpendicular to the incline, aiding in the comprehensive analysis of physical dynamics.

Weight Components

When an object rests on an inclined plane, the gravitational force acting on it can be divided into two components. These components are vital in understanding how the object will behave on the slope.

• **Parallel Component**: Denoted as \(T_P\), this part of the weight pulls the object down the slope. Its magnitude is calculated by multiplying the total weight by the sine of the incline angle:
  \[T_P = W \cdot \sin{(\theta)}\]
  This component is responsible for the tendency of the object to slide down.
  
• **Perpendicular Component**: Represented as \(T_Q\), this force acts at a right angle to the plane. It is found using the cosine function: \[T_Q = W \cdot \cos{(\theta)}\]
  This component counteracts any motion normal to the plane and is crucial in calculating frictional forces.

Understanding these components is essential, as they determine the stability of objects on slopes and affect factors such as friction, which influences whether an object will remain stationary or begin to slide.

Right Triangle

The right triangle is an essential geometric shape in solving inclined plane problems in physics. It helps us understand the relationship between weight, its components, and the angle of the incline.

A right triangle is defined by having one of its angles exactly equal to 90 degrees. In the context of an inclined plane:

• **Hypotenuse**: This is the force of gravity acting on the object, often depicted as the weight. It forms the longest side of the triangle.
  
• **Base and Height**: These represent the component forces: one runs parallel to the plane; the other, perpendicular. Using trigonometry, they align with the sine and cosine of the angle of the incline, allowing us to calculate the components accurately.

The angle of inclination directly helps calculate these components, as the rest of the triangle's geometry depends on it. It serves as a practical guide for breaking down force vectors and thereby simplifies the process of determining the effects of forces on the inclined surface.