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Chapter 5: Problem 72

Three forces act on a particle that moves with unchanging velocity \(\vec{v}=(2 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}-(7 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) Two of the forces are \(\vec{F}_{1}=(2 \mathrm{~N}) \hat{\mathrm{i}}+\) \((3 N) \hat{j}+(-2 N) \hat{k}\) and \(\vec{F}_{2}=(-5 N) \hat{i}+(8 N) \hat{j}+(-2 N) \hat{k} .\) What is the third force?

Short Answer

Expert verified
The third force is \((3 \text{ N}) \hat{i} - (11 \text{ N}) \hat{j} + (4 \text{ N}) \hat{k}\).

Step by step solution

01

Understanding the Problem

We need to find the third force acting on a particle with a constant velocity. Since the velocity is unchanging, the net force acting on the particle must be zero, according to Newton's First Law. Therefore, the sum of the three forces must equal zero.
02

Set Up the Equation for Net Force

Given that the sum of the forces must be zero, we can write: \( \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} = \vec{0} \). We need to solve for \( \vec{F}_{3} \). This means \( \vec{F}_{3} = -(\vec{F}_{1} + \vec{F}_{2}) \).
03

Calculate \(\vec{F}_{1} + \vec{F}_{2}\)

We add the given forces component-wise. For the \(\hat{i}\) component: \(2 + (-5) = -3\). For the \(\hat{j}\) component: \(3 + 8 = 11\). For the \(\hat{k}\) component: \(-2 + (-2) = -4\). Thus, \(\vec{F}_{1} + \vec{F}_{2} = (-3 \text{ N}) \hat{i} + (11 \text{ N}) \hat{j} + (-4 \text{ N}) \hat{k}\).
04

Find \(\vec{F}_{3}\)

Since \(\vec{F}_{3} = - (\vec{F}_{1} + \vec{F}_{2})\), we take the negative of each component: \(\vec{F}_{3} = (3 \text{ N}) \hat{i} + (-11 \text{ N}) \hat{j} + (4 \text{ N}) \hat{k}\).
05

Verify the Solution

To ensure correctness, add \(\vec{F}_{1}\), \(\vec{F}_{2}\), and the calculated \(\vec{F}_{3}\) to confirm they sum to zero: Each component should be resolved as zero. For \(\hat{i}\), \(2 - 5 + 3 = 0\); for \(\hat{j}\), \(3 + 8 - 11 = 0\); and for \(\hat{k}\), \(-2 - 2 + 4 = 0\). Thus, the solution is consistent.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Constant Velocity
When an object is moving with a constant velocity, it means that its speed and direction remain unchanged. This idea is tied to the concept of uniform motion. According to Newton's First Law, if an object is moving with a constant velocity, the net external force acting on it must be zero.
In other words, there is no unbalanced force that could change its state of motion. Constant velocity signifies equilibrium among forces.
  • The speed is consistent over time.
  • The direction doesn't change.
  • No acceleration occurs.
Properties of constant velocity are crucial when solving physics problems because they point out that forces are balanced. In the given exercise, the constant velocity implies that the sum of all forces acting on the particle is zero. This is a great clue to use when determining unknown forces.
Net Force
Net force is a concept that describes the total force acting on an object when all individual forces are combined. If the net force is zero, the object remains at rest or continues to move at a constant velocity. This is due to Newton's First Law, also known as the Law of Inertia, which states that an object will remain in its state of motion unless acted upon by a net external force.
  • Net force is the vector sum of all forces acting on an object.
  • When the net force is zero, the status of motion remains unchanged.
  • A non-zero net force results in acceleration.
In the exercise, to find the third force, we ensure that the vector sum of all forces equals zero to satisfy the condition of constant velocity. This tells us how the forces balance out to maintain equilibrium.
Vector Addition
Vector addition is the process of combining several vectors into a single resultant vector. In physics problems, vectors like forces are added component-wise. That means carefully adding corresponding components in the same direction, often represented by unit vectors i, j, and k.
Understanding vector addition is crucial for determining net forces or resultant forces in multiple dimensions.
  • Vectors have both magnitude and direction.
  • Each vector can be divided into components using i, j, and k notation.
  • Vectors are added by summing each component separately.
In the solution provided, vector components are added to find the resultant of two given forces. Once determined, we used the negative of this vector to deduce the third force needed to ensure the net force is zero. The accuracy of vector addition is vital in preserving equilibrium in the system.

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Most popular questions from this chapter

A "sun yacht" is a spacecraft with a large sail that is pushed by sunlight. Although such a push is tiny in everyday circumstances, it can be large enough to send the spacecraft outward from the Sun on a cost-free but slow trip. Suppose that the spacecraft has a mass of \(900 \mathrm{~kg}\) and receives a push of \(20 \mathrm{~N}\). (a) What is the magnitude of the resulting acceleration? If the craft starts from rest, (b) how far will it travel in 1 day and (c) how fast will it then be moving?

A \(1400 \mathrm{~kg}\) jet engine is fastened to the fuselage of a passenger jet by just three bolts (this is the usual practice). Assume that each bolt supports one-third of the load. (a) Calculate the force on each bolt as the plane waits in line for clearance to take off. (b) During flight, the plane encounters turbulence, which suddenly imparts an upward vertical acceleration of \(2.6 \mathrm{~m} / \mathrm{s}^{2}\) to the plane. Calculate the force on each bolt now.

A block with a weight of \(3.0 \mathrm{~N}\) is at rest on a horizontal surface. A \(1.0 \mathrm{~N}\) upward force is applied to the block by means of an attached vertical string. What are the (a) magnitude and (b) direction of the force of the block on the horizontal surface?

An elevator cab and its load have a combined mass of 1600 \(\mathrm{kg}\). Find the tension in the supporting cable when the cab, originally moving downward at \(12 \mathrm{~m} / \mathrm{s}\), is brought to rest with constant acceleration in a distance of \(42 \mathrm{~m}\).

A nucleus that captures a stray neutron must bring the neutron to a stop within the diameter of the nucleus by means of the strong force. That force, which "glues" the nucleus together, is approximately zero outside the nucleus. Suppose that a stray neutron with an initial speed of \(1.4 \times 10^{7} \mathrm{~m} / \mathrm{s}\) is just barely captured by a nucleus with diameter \(d=1.0 \times 10^{-14} \mathrm{~m}\). Assuming the strong force on the neutron is constant, find the magnitude of that force. The neutron's mass is \(1.67 \times 10^{-27} \mathrm{~kg}\).
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