Problem 77 - Algebra and Trigonometry Real Mathematics, Real People

Question

A force of 50 pounds, exerted at an angle of \(25^{\circ}\) with the horizontal, is required to slide a desk across a floor. Determine the work done in sliding the desk 15 feet.

Step-by-Step Solution

Answer

So the work done in sliding the desk 15 feet is the multiplication of resolved horizontal force and the displacement.

Step 1: Identify the force, displacement, and angle

In the exercise, the force exerted is 50 pounds, the displacement (i.e., the distance the desk is moved) is 15 feet and the angle between the force and the horizontal direction, where displacement occurs, is \(25^{\circ}\).

Step 2: Resolve the force component parallel to displacement

The component of force in the direction of displacement can be resolved by multiplying the force by the cosine of the angle between the force and the displacement direction. Therefore, the horizontal force \( F_h \) can be calculated as \(50 \cos(25^{\circ})\).

Step 3: Calculate work done

Work done is calculated as the product of the force (in the direction of displacement) and the displacement. So, the work \( W \) done is \( F_h \times 15 \).

Key Concepts

Force and Motion

In physics, force and motion are fundamental concepts that explain how objects move. A force is any push or pull exerted on an object, and it can cause the object to start moving, stop moving, or change direction. When you slide a desk across the floor, you exert a force on it. This force impacts the desk's motion. However, not all of the force contributes effectively to the movement. Only the component of force acting in the direction of the movement does. This is why understanding how force components work is so important.

Let's break it down further:

• Motion: Motion occurs when an object changes its position over time. When you push the desk, you're applying force to change its state from stationary to moving.
• Effective Force: Not all the applied force contributes to the motion. Only the portion of the force parallel to the movement direction actually works towards moving the desk.

Understanding these concepts helps us calculate the work done on the desk as you are considering the force contributing to the motion.

Angle of Force

The angle of the force plays a crucial role in determining how effectively a force can move an object. In the original exercise, this angle is given as \(25^{\circ}\). This angle is measured between the direction of the force applied and the direction of displacement. Recognizing the impact of this angle helps determine which part of the force actually translates into motion.

Here's why it matters:

• Horizontal Component: The force's effectiveness in moving the desk depends on its horizontal component, which can be found using the cosine of the angle with the horizontal axis. This component is the one doing the actual work.
• Complementary Effect: The more the angle deviates from zero (in alignment with the displacement direction), the less efficient the force becomes in causing motion because more of it goes into other directions.

By calculating \(50 \cos(25^{\circ})\), we find the force component that truly impacts the desk's movement. This understanding allows us to accurately calculate the work done by this component.

Displacement

Displacement is a critical factor when considering work done by a force. It is the distance the object moves in a specific direction, noted as 15 feet in this problem. Displacement, unlike distance, is a vector quantity, which means it has both a magnitude and a direction — the direction in which the object is moved.

• Magnitude: In the exercise, the magnitude of displacement is 15 feet. This tells us how far the desk has been moved.
• Directional Move: Since displacement is directional, it matters which way the force is applied relative to this direction. Only force components in the direction of displacement count towards work done.

Understanding displacement helps you calculate the total work. By multiplying the effective force (derived from the horizontal component, as previously discussed) with this displacement, we find the total work exerted in moving the desk for 15 feet.