Problem 45 - Algebra and Trigonometry

Question

Work The force \(\mathbf{F}=4 \mathbf{i}-7 \mathbf{j}\) moves an object 4 \(\mathrm{ft}\) along the \(x\) -axis in the positive direction. Find the work done if the unit of force is the pound.

Step-by-Step Solution

Answer

The work done is 16 foot-pounds.

Step 1: Understand the Concept of Work

Work is done when a force causes an object to move in the direction of the force. The formula for work in physics is given by \( W = \mathbf{F} \cdot \mathbf{d} \), where \( \mathbf{F} \) is the force vector and \( \mathbf{d} \) is the displacement vector.

Step 2: Determine the Displacement Vector

The problem states the object moves 4 feet along the \( x \)-axis in the positive direction. Therefore, the displacement vector \( \mathbf{d} \) is \( 4 \mathbf{i} \).

Step 3: Calculate the Dot Product

The work done \( W \) is the dot product of the force and displacement vectors, \( \mathbf{F} \cdot \mathbf{d} \). Here, \( \mathbf{F} = 4 \mathbf{i} - 7 \mathbf{j} \) and \( \mathbf{d} = 4 \mathbf{i} \). Thus, the dot product is computed as \( (4 \mathbf{i} - 7 \mathbf{j}) \cdot 4 \mathbf{i} = 16 \).

Step 4: Evaluate the Work Done

Since the dot product is 16 and the unit of force is the pound, the work done is 16 foot-pounds. Note that the \( \mathbf{j} \) component does not contribute to the work since it is perpendicular to the direction of the displacement.

Key Concepts

Force Vector

In physics, a force vector represents the magnitude and direction of a force applied to an object. A vector is essentially an arrow that indicates the direction in which the object is pushed or pulled. The length of the vector represents the force's magnitude, usually measured in units like pounds or newtons.

In the exercise, the force vector is given as \(\mathbf{F} = 4 \mathbf{i} - 7 \mathbf{j}\). This indicates that the force has a component of 4 units in the positive x-direction and -7 units in the y-direction. Here, \(\mathbf{i}\) and \(\mathbf{j}\) are unit vectors along the x-axis and y-axis respectively. These components help to determine how the force acts in a two-dimensional plane.

• The positive \(\mathbf{i}\) component means the force pulls rightward on the x-axis.
• The negative \(\mathbf{j}\) component means the force pulls downward on the y-axis.

Understanding the components of a force vector is crucial in calculating work, as only the component parallel to the displacement direction contributes to the work done.

Displacement Vector

The displacement vector describes the object's change in position. It is a vector that points from the initial position to the final position and is especially useful in determining how far something has moved and in which direction. The magnitude of this vector is the straight-line distance between the two positions.

In our specific problem, the displacement vector is \(\mathbf{d} = 4 \mathbf{i}\). This means the object moves 4 units along the x-axis in a positive direction, with no movement in the y-direction. This vector is crucial for computing the work done since work depends on the distance and direction of displacement with respect to the applied force.

• The displacement vector points entirely along the x-axis.
• Displacement of 4 feet implies movement to the right on this axis.

When evaluating work, the alignment between the force vector and displacement vector directly influences the calculation. Only the vector component of displacement parallel to the force vector affects the work done.

Dot Product

The dot product is a mathematical operation used to determine the magnitude of work done in physics by combining two vectors: the force vector and the displacement vector. This product measures how much of the force is applied in the direction of movement. The dot product is calculated as \(\mathbf{A} \cdot \mathbf{B} = A_xB_x + A_yB_y + A_zB_z\) for vectors \(\mathbf{A}\) and \(\mathbf{B}\).

In our example, the vectors are \(\mathbf{F} = 4 \mathbf{i} - 7 \mathbf{j}\) and \(\mathbf{d} = 4 \mathbf{i}\). Thus, the calculation simplifies to:

• \(4 \mathbf{i} \cdot 4 \mathbf{i} = 16\) (as the y-components do not contribute).
• \( -7 \mathbf{j} \cdot 0 \mathbf{j} = 0\) (because there is no displacement in the y-direction).

This gives a total of 16, indicating that the work done by the force is 16 foot-pounds. The perpendicular components don't contribute to the work since the displacement is entirely along the x-axis.