Problem 81 - Calculus Early Transcendentals: Pearson New International Edition

Question

A weight of 30 pounds is suspended by three wires with resulting tensions \(3 \mathbf{i}+4 \mathbf{j}+15 \mathbf{k},-8 \mathbf{i}-2 \mathbf{j}+10 \mathbf{k}\), and \(a \mathbf{i}+b \mathbf{j}+c \mathbf{k}\). Determine \(a, b\), and \(c\) so that the net force is straight up.

Step-by-Step Solution

Answer

The values are \(a = 5\), \(b = -2\), \(c = 5\).

Step 1: Identify Forces

The problem gives us the tensions in the three wires: \(\mathbf{T}_1 = 3 \mathbf{i} + 4 \mathbf{j} + 15 \mathbf{k}\), \(\mathbf{T}_2 = -8 \mathbf{i} - 2 \mathbf{j} + 10 \mathbf{k}\), and the unknown tension \(\mathbf{T}_3 = a \mathbf{i} + b \mathbf{j} + c \mathbf{k}\). We are to find \(a\), \(b\), and \(c\) such that the net force is only in the \(\mathbf{k}\) direction.

Step 2: Write the Equation for Net Force

Since the weight is 30 pounds, the net force acting on the system that sums the tensions must support this weight straight up. Thus, the sum of the forces \(\mathbf{T}_1 + \mathbf{T}_2 + \mathbf{T}_3\) should have components only in the \(\mathbf{k}\) direction, with \(\mathbf{i}\) and \(\mathbf{j}\) components equal to zero.

Step 3: Calculate the Sum of Forces

Sum the given forces: \[\mathbf{T}_{total} = (3 - 8 + a) \mathbf{i} + (4 - 2 + b) \mathbf{j} + (15 + 10 + c) \mathbf{k}.\] This simplifies to \[\mathbf{T}_{total} = (a - 5) \mathbf{i} + (b + 2) \mathbf{j} + (25 + c) \mathbf{k}.\]

Step 4: Set Conditions for Forces

For the net force to be straight up, set \(a - 5 = 0\) and \(b + 2 = 0\). This eliminates any \(\mathbf{i}\) and \(\mathbf{j}\) components.

Step 5: Solve for Unknowns

Solving the system of equations:- From \(a - 5 = 0\), we get \(a = 5\).- From \(b + 2 = 0\), we get \(b = -2\).

Step 6: Determine the Value of c

Since the force is upwards, the net force in the \(\mathbf{k}\) direction should total 30 pounds upwards. Therefore, we have: \[25 + c = 30.\]Solving this gives: \[c = 5.\]

Step 7: Verify the Result

Verify the results: \(a = 5\), \(b = -2\), and \(c = 5\) satisfy the conditions for balancing the system with a net force only in the \(\mathbf{k}\) direction with magnitude 30.

Key Concepts

Tension in Cables

Tension in cables is a fundamental concept in physics, often associated with objects being suspended by ropes or wires. In this scenario, an object weighing 30 pounds is hung by three cables. Each cable exerts a force, known as tension, which helps keep the object in place. Imagine these forces like invisible arms pulling in different directions to hold the weight steady.

Each tension can be represented as a vector in space. A vector has both magnitude (strength) and direction. The problem provides two of these tensions as vectors, while the third is unknown. Our task is to identify the missing components of the third vector that ensure the forces balance out properly.

• The first tension vector is given as \(3\mathbf{i} + 4\mathbf{j} + 15\mathbf{k}\).
• The second tension vector is \(-8\mathbf{i} - 2\mathbf{j} + 10\mathbf{k}\).
• The third vector \(a \mathbf{i} + b \mathbf{j} + c \mathbf{k}\) needs components \(a\), \(b\), and \(c\) yet to be found.

In physics problems, understanding these vectors helps in solving for tensions when objects are in equilibrium.

Net Force Calculation

Net force calculation is a critical step in analyzing systems where multiple forces are acting simultaneously. For the weight to move upwards only, the horizontal forces (represented by \(\mathbf{i}\) and \(\mathbf{j}\) components) must balance out to zero. In this exercise, we calculate the net forces by summing up all the tensions.

The net force formula is derived from vector addition:

• \((3 - 8 + a) \mathbf{i} + (4 - 2 + b) \mathbf{j} + (15 + 10 + c) \mathbf{k}\)

This simplifies to:

• \((a - 5) \mathbf{i} + (b + 2) \mathbf{j} + (25 + c) \mathbf{k}\)

The conditions for the weight to rise straight up require that the horizontal components be zero:

• \(a - 5 = 0\) which gives \(a = 5\)
• \(b + 2 = 0\) which gives \(b = -2\)

Ensuring the net force is in the vertical direction allows the determination of \(c\), ensuring the object's stability.

Equilibrium of Forces

Achieving equilibrium of forces in a physical system means that all acting forces balance out, and the object remains at rest or moves with a constant velocity. For this system, the aim is for the total resultant force to act only in the upward direction, a concept crucial in many engineering applications.

When the weight is supported by cables, equilibrium implies that:

• The sum of the horizontal forces: \(a - 5\) and \(b + 2\) must be zero.
• The vertical sum \(25 + c\) must equal the weight's force upwards, which is 30 pounds.

To maintain balance, the forces in the \(\mathbf{i}\) and \(\mathbf{j}\) directions cancel out, leading us to set \(a = 5\) and \(b = -2\). The remaining task is to solve for \(c\). Since the total upward force must equal the object's weight, we have:

• \(25 + c = 30\)

Solving gives \(c = 5\), ensuring the equilibrium is satisfied with purely vertical motion.