/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 8 A \(2.00 \mathrm{~kg}\) object i... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(2.00 \mathrm{~kg}\) object is subjected to three forces that give it an acceleration \(\vec{a}=-\left(8.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{i}}+\left(6.00 \mathrm{~m} / \mathrm{s}^{2}\right) \hat{\mathrm{j}} .\) If two of the three forces are \(\vec{F}_{1}=(30.0 \mathrm{~N}) \hat{\mathrm{i}}+(16.0 \mathrm{~N}) \hat{\mathrm{j}}\) and \(\vec{F}_{2}=\) \(-(12.0 \mathrm{~N}) \hat{\mathrm{i}}+(8.00 \mathrm{~N}) \hat{\mathrm{j}}\), find the third force.

Short Answer

Expert verified
The third force \( \vec{F}_{3} \) is \(-34.0 \hat{\mathrm{i}} - 12.0 \hat{\mathrm{j}} \text{ N}\).

Step by step solution

01

Write the Net Force Equation

According to Newton's second law, the net force acting on an object is equal to the mass of the object multiplied by its acceleration: \( \vec{F}_{\text{net}} = m \vec{a} \). For this object, substitute the given values: \[ \vec{F}_{\text{net}} = (2.00\, \text{kg}) \times \left[-8.00\, \hat{\mathrm{i}} + 6.00\, \hat{\mathrm{j}}\right]\, \text{m/s}^{2} \] This simplifies to: \[ \vec{F}_{\text{net}} = (-16.0\, \hat{\mathrm{i}} + 12.0\, \hat{\mathrm{j}}) \text{N} \].
02

Sum the Forces in Each Direction

The net force is the vector sum of all individual forces acting on the object: \[ \vec{F}_{\text{net}} = \vec{F}_{1} + \vec{F}_{2} + \vec{F}_{3} \] Substitute the given forces: \[ (-16.0\, \hat{\mathrm{i}} + 12.0\, \hat{\mathrm{j}}) \text{N} = (30.0\, \hat{\mathrm{i}} + 16.0\, \hat{\mathrm{j}})\, \text{N} + (-12.0\, \hat{\mathrm{i}} + 8.0\, \hat{\mathrm{j}})\, \text{N} + \vec{F}_{3} \].
03

Solve for the Third Force

Combine \( \vec{F}_{1} \) and \( \vec{F}_{2} \):For the \( \hat{\mathrm{i}} \) component: \[ 30.0 - 12.0 = 18.0\, \text{N} \]For the \( \hat{\mathrm{j}} \) component: \[ 16.0 + 8.0 = 24.0\, \text{N} \] Thus, the equation becomes:\[ (-16.0\, \hat{\mathrm{i}} + 12.0\, \hat{\mathrm{j}}) \text{N} = (18.0\, \hat{\mathrm{i}} + 24.0\, \hat{\mathrm{j}}) \text{N} + \vec{F}_{3} \].For the \( \hat{\mathrm{i}} \) and \( \hat{\mathrm{j}} \) components individually: \( -16.0 = 18.0 + F_{3i} \) leading to \( F_{3i} = -34.0\, \text{N} \),\( 12.0 = 24.0 + F_{3j} \) leading to \( F_{3j} = -12.0\, \text{N} \).
04

Write the Final Answer

The third force \( \vec{F}_{3} \) is \[ \vec{F}_{3} = (-34.0\, \hat{\mathrm{i}} - 12.0\, \hat{\mathrm{j}}) \text{N} \].

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Key Concepts

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

Net Force Calculation
When dealing with forces on an object, understanding net force is crucial. Net force is the vector sum of all individual forces acting on the object, reflecting the overall impact on its motion.
According to Newton's Second Law, the formula for net force is given by:
  • \( \vec{F}_{\text{net}} = m \vec{a} \)
In this equation:
  • \( \vec{F}_{\text{net}} \) stands for the net force in newtons (N).
  • \( m \) represents the mass of the object in kilograms (kg).
  • \( \vec{a} \) is the acceleration of the object in meters per second squared (m/s\(^2\)).
In the given exercise, with a mass of 2.00 kg and an acceleration vector of \(-8.00\, \hat{\mathrm{i}} + 6.00\, \hat{\mathrm{j}}\) in m/s\(^2\), the net force becomes:\[\vec{F}_{\text{net}} = (2.00 \, \text{kg}) \times \left[-8.00\, \hat{\mathrm{i}} + 6.00\, \hat{\mathrm{j}}\right] \]Simplifies to:\[\vec{F}_{\text{net}} = (-16.0\, \hat{\mathrm{i}} + 12.0\, \hat{\mathrm{j}}) \, \text{N}\]This final result represents the net force acting on the object, summing up all contributions from the individual forces.
Vector Addition
Vector addition is a mathematical operation necessary to determine the resultant vector when combining two or more vectors. In the context of forces, it's used to find the total force acting upon an object.
This is crucial in physics as forces often have both magnitude and direction.
To add vectors, you need to add each component separately:
  • The i-components (horizontal parts of vectors) are summed.
  • The j-components (vertical parts of vectors) are summed.
In our example exercise, the forces \( \vec{F}_1 \) and \( \vec{F}_2 \) are given as:
  • \( \vec{F}_1 = (30.0\, \hat{\mathrm{i}} + 16.0\, \hat{\mathrm{j}}) \, \text{N} \)
  • \( \vec{F}_2 = (-12.0\, \hat{\mathrm{i}} + 8.0\, \hat{\mathrm{j}}) \, \text{N} \)
To find the sum of these two vectors, calculate:
  • i-component: \( 30.0 - 12.0 = 18.0 \, \text{N} \)
  • j-component: \( 16.0 + 8.0 = 24.0 \, \text{N} \)
Thus, the combined force vector becomes:\[(18.0\, \hat{\mathrm{i}} + 24.0\, \hat{\mathrm{j}}) \, \text{N}\]Using vector addition is essential in physics exercises as it simplifies finding the effects of multiple forces acting on an object.
Acceleration and Force Relationship
Newton's Second Law provides an insightful relation between acceleration and force, both of which deeply affect an object's motion. The law states that the acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass.
In mathematical terms:\[\vec{F}_{\text{net}} = m \vec{a}\]
This indicates:
  • An increase in net force, with constant mass, results in a proportional increase in acceleration.
  • A more massive object will experience less acceleration if the same net force is applied, compared to a less massive one.
In the textbook example, knowing the net force and mass allows us to calculate the acceleration vector:\[\vec{a} = \frac{\vec{F}_{\text{net}}}{m}\]For a mass of 2.00 kg with a net force of \((-16.0\, \hat{\mathrm{i}} + 12.0\, \hat{\mathrm{j}}) \, \text{N}\), the equations show an acceleration:\[\vec{a} = \frac{(-16.0\, \hat{\mathrm{i}} + 12.0\, \hat{\mathrm{j}}) \, \text{N}}{2.00 \, \text{kg}}\]This calculation reinforces the understanding of how net force and mass contribute to determining the acceleration an object experiences.

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Most popular questions from this chapter

Compute the initial upward acceleration of a rocket of mass \(1.3 \times 10^{4} \mathrm{~kg}\) if the initial upward force produced by its engine (the thrust) is \(2.6 \times 10^{5} \mathrm{~N}\). Do not neglect the gravitational force on the rocket.

A spaceship lifts off vertically from the Moon, where \(g=\) \(1.6 \mathrm{~m} / \mathrm{s}^{2}\). If the ship has an upward acceleration of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\) as it lifts off, what is the magnitude of the force exerted by the ship on its pilot, who weighs \(735 \mathrm{~N}\) on Earth?

An elevator cab that weighs \(27.8 \mathrm{kN}\) moves upward. What is the tension in the cable if the cab's speed is (a) increasing at a rate of \(1.22 \mathrm{~m} / \mathrm{s}^{2}\) and (b) decreasing at a rate of \(1.22 \mathrm{~m} / \mathrm{s}^{2} ?\)

A \(40 \mathrm{~kg}\) skier skis directly down a frictionless slope angled at \(10^{\circ}\) to the horizontal. Assume the skier moves in the negative direction of an \(x\) axis along the slope. A wind force with component \(F_{x}\) acts on the skier. What is \(F_{x}\) if the magnitude of the skier's velocity is (a) constant, (b) increasing at a rate of \(1.0 \mathrm{~m} / \mathrm{s}^{2}\), and \((\mathrm{c})\) increasing at a rate of \(2.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

The Zacchini family was renowned for their human-cannonball act in which a family member was shot from a cannon using either elastic bands or compressed air. In one version of the act, Emanuel Zacchini was shot over three Ferris wheels to land in a net at the same height as the open end of the cannon and at a range of \(69 \mathrm{~m}\). He was propelled inside the barrel for \(5.2 \mathrm{~m}\) and launched at an angle of \(53^{\circ} .\) If his mass was \(85 \mathrm{~kg}\) and he underwent constant acceleration inside the barrel, what was the magnitude of the force propelling him? (Hint: Treat the launch as though it were along a ramp at \(53^{\circ} .\) Neglect air drag.)

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