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Chapter 6: Problem 27

Three forces act on an object, considered to be a particle, which moves with constant velocity \(v=(3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}) \mathrm{m} / \mathrm{s} . \quad\) Two of the forces are \(\overrightarrow{\mathbf{F}}_{1}=(3 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) \mathrm{N} \quad\) and \(\overrightarrow{\mathbf{F}}_{2}=(4 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mathrm{N} .\) Find the third force.

Short Answer

Expert verified
The third force acting on the object is \(\overrightarrow{\mathbf{F}}_{3} = (-7 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 5 \hat{\mathbf{k}}) \mathrm{N}\).

Step by step solution

01

Calculate the sum of the known forces

Add the two forces \(\overrightarrow{\mathbf{F}}_{1}\) and \(\overrightarrow{\mathbf{F}}_{2}\) that are acting on the object: \(\overrightarrow{\mathbf{F}}_{1} + \overrightarrow{\mathbf{F}}_{2} = (3\hat{\mathbf{i}} + 5\hat{\mathbf{j}} - 7\hat{\mathbf{k}}) + (4\hat{\mathbf{i}} - 7\hat{\mathbf{j}} + 2\hat{\mathbf{k}})\) Combine the vector components: \((3\hat{\mathbf{i}} + 4\hat{\mathbf{i}}) + (5\hat{\mathbf{j}} - 7\hat{\mathbf{j}}) + (-7\hat{\mathbf{k}} + 2\hat{\mathbf{k}})= 7\hat{\mathbf{i}} -2\hat{\mathbf{j}} -5\hat{\mathbf{k}}\)
02

Calculate the third force

Since the sum of the forces must be zero, we need to calculate the third force as the negative of the sum of the forces in step 1: \(\overrightarrow{\mathbf{F}}_{3} = - (7\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 5\hat{\mathbf{k}})\) Apply the negative sign to each component of the force: \(\overrightarrow{\mathbf{F}}_{3} = -7\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 5\hat{\mathbf{k}}\) The third force acting on the object is: \(\overrightarrow{\mathbf{F}}_{3} = (-7 \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + 5 \hat{\mathbf{k}}) \mathrm{N}\)

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Key Concepts

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

Vector Addition
Understanding vector addition is essential when dealing with forces, as forces are vector quantities, which means they have both magnitude and direction. When multiple forces are acting on an object, the total effect is the vector sum of these forces. To perform vector addition, one must combine the like components of each vector.

Vectors are often represented in Cartesian coordinates with the unit vectors \(\hat{\mathbf{i}}\), \(\hat{\mathbf{j}}\), and \(\hat{\mathbf{k}}\) indicating direction along the x, y, and z axes, respectively. When adding vectors like \(\overrightarrow{\mathbf{F}}_{1}\) and \(\overrightarrow{\mathbf{F}}_{2}\), the corresponding components in each direction are added to form a resultant vector.
Constant Velocity
The concept of constant velocity implies that an object is moving at a steady speed in a straight line, with no acceleration. According to Newton's first law of motion, also known as the law of inertia, an object in motion will stay in motion with constant velocity unless acted upon by an external force. This ties into the central exercise, where an object maintains constant velocity \(\mathrm{m} / \mathrm{s}\), indicating that the forces acting upon it are in equilibrium鈥攎eaning the net force is zero. This construct serves both as a foundation for understanding motion and as a tool for solving problems concerning the forces acting on an object.
Equilibrium of Forces
Equilibrium of forces occurs when all the forces acting upon an object are balanced, resulting in no net force and, consequently, no change in the state of motion of that object. Following Newton's laws, we know an object at rest or moving with constant velocity will remain so if the sum of the forces acting on it is zero. In our exercise, this equilibrium is demonstrated by the necessity for a third force \(\overrightarrow{\mathbf{F}}_{3}\) that exactly opposes the vector sum of \(\overrightarrow{\mathbf{F}}_{1}\) and \(\overrightarrow{\mathbf{F}}_{2}\). This balancing force ensures the object continues with a constant velocity, which is a state of dynamic equilibrium.

To maintain equilibrium, \(\overrightarrow{\mathbf{F}}_{3}\) is calculated as the negative of the sum of the other two forces. This equates to taking the sum and changing the sign of each component, thus negating the vector, to achieve a net force of zero.

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Most popular questions from this chapter

Modern roller coasters have vertical loops like the one shown here. The radius of curvature is smaller at the top than on the sides so that the downward centripetal acceleration at the top will be greater than the acceleration due to gravity, keeping the passengers pressed firmly into their seats. (a) What is the speed of the roller coaster at the top of the loop if the radius of curvature there is \(15.0 \mathrm{m}\) and the downward acceleration of the car is \(1.50 \mathrm{g}\) ? (b) How high above the top of the loop must the roller coaster start from rest, assuming negligible friction? (c) If it actually starts \(5.00 \mathrm{m}\) higher than your answer to (b), how much energy did it lose to friction? Its mass is \(1.50 \times 10^{3} \mathrm{kg}\)

A particle of mass \(m\) is located at the origin. It is at rest and in equilibrium. A time-dependent force of \(\overrightarrow{\mathbf{F}}(t)\) is applied at time \(t=0,\) and its components are \(F_{x}(t)=p t\) and \(F_{y}(t)=n+q t\) where \(p, q,\) and \(n\) are constants. Find the position \(\overrightarrow{\mathbf{r}}(t)\) and velocity \(\overrightarrow{\mathbf{v}}(t)\) as functions of time \(t\)

A piece of bacon starts to slide down the pan when one side of a pan is raised up \(5.0 \mathrm{cm} .\) If the length of the pan from pivot to the raising point is \(23.5 \mathrm{cm},\) what is the coefficient of static friction between the pan and the bacon?

To simulate the apparent weightlessness of space orbit, astronauts are trained in the hold of a cargo aircraft that is accelerating downward at \(g\). Why do they appear to be weightless, as measured by standing on a bathroom scale, in this accelerated frame of reference? Is there any difference between their apparent weightlessness in orbit and in the aircraft?

Two friends are having a conversation. Anna says a satellite in orbit is in free fall because the satellite keeps falling toward Earth. Tom says a satellite in orbit is not in free fall because the acceleration due to gravity is not \(9.80 \mathrm{m} / \mathrm{s}^{2} .\) Who do you agree with and why?
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