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If \(\overrightarrow{\mathbf{A}}=9.0 \hat{\mathbf{i}}-8.5 \hat{\mathbf{j}}, \quad \overrightarrow{\mathbf{B}}=-8.0 \hat{\mathbf{i}}+7.1 \hat{\mathbf{j}}+4.2 \hat{\mathbf{k}},\) and \(\overrightarrow{\mathbf{C}}=6.8 \hat{\mathbf{i}}-9.2 \hat{\mathbf{j}},\) determine \((a) \overrightarrow{\mathbf{A}} \cdot(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}) ;(b)(\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}}) \cdot \overrightarrow{\mathbf{B}} ;\) \((c)(\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{A}}) \cdot \mathbf{C}\)

Short Answer

Expert verified
(a) 7.05; (b) -252.07; (c) 19.68.

Step by step solution

01

Compute \( \overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}} \)

First, we need to find the sum of \( \overrightarrow{\mathbf{B}} \) and \( \overrightarrow{\mathbf{C}} \). Add the corresponding components of each vector: \[\overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}} = (-8.0\hat{\mathbf{i}} + 7.1\hat{\mathbf{j}} + 4.2\hat{\mathbf{k}}) + (6.8\hat{\mathbf{i}} - 9.2\hat{\mathbf{j}})\]Simplify:\[(-8.0 + 6.8)\hat{\mathbf{i}} + (7.1 - 9.2)\hat{\mathbf{j}} + 4.2\hat{\mathbf{k}} = -1.2\hat{\mathbf{i}} - 2.1\hat{\mathbf{j}} + 4.2\hat{\mathbf{k}}\]
02

Calculate \( \overrightarrow{\mathbf{A}} \cdot (\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}}) \)

Perform the dot product between \( \overrightarrow{\mathbf{A}} \) and the result from Step 1, \( \overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{C}} \):\[(9.0\hat{\mathbf{i}} - 8.5\hat{\mathbf{j}}) \cdot (-1.2\hat{\mathbf{i}} - 2.1\hat{\mathbf{j}} + 4.2\hat{\mathbf{k}})\]The dot product is calculated as:\[= (9.0 \times -1.2) + (-8.5 \times -2.1) + (0 \times 4.2)\]\[= -10.8 + 17.85 + 0 = 7.05\]
03

Compute \( \overrightarrow{\mathbf{A}} + \overrightarrow{\mathbf{C}} \)

Find the sum of \( \overrightarrow{\mathbf{A}} \) and \( \overrightarrow{\mathbf{C}} \) by adding corresponding components:\[(9.0\hat{\mathbf{i}} - 8.5\hat{\mathbf{j}}) + (6.8\hat{\mathbf{i}} - 9.2\hat{\mathbf{j}})\]Simplify:\[(9.0 + 6.8)\hat{\mathbf{i}} + (-8.5 - 9.2)\hat{\mathbf{j}} = 15.8\hat{\mathbf{i}} - 17.7\hat{\mathbf{j}}\]
04

Calculate \( (\overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}}) \cdot \overrightarrow{\mathbf{B}} \)

Perform the dot product between \( \overrightarrow{\mathbf{A}}+\overrightarrow{\mathbf{C}} \) from Step 3 and \( \overrightarrow{\mathbf{B}} \):\[(15.8\hat{\mathbf{i}} - 17.7\hat{\mathbf{j}}) \cdot (-8.0\hat{\mathbf{i}} + 7.1\hat{\mathbf{j}} + 4.2\hat{\mathbf{k}})\]Calculate:\[= (15.8 \times -8.0) + (-17.7 \times 7.1) + (0 \times 4.2)\]\[= -126.4 - 125.67 + 0 = -252.07\]
05

Compute \( \overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{A}} \)

Add the vectors \( \overrightarrow{\mathbf{B}} \) and \( \overrightarrow{\mathbf{A}} \):\[(-8.0\hat{\mathbf{i}} + 7.1\hat{\mathbf{j}} + 4.2\hat{\mathbf{k}}) + (9.0\hat{\mathbf{i}} - 8.5\hat{\mathbf{j}})\]This gives:\[(1.0\hat{\mathbf{i}} - 1.4\hat{\mathbf{j}} + 4.2\hat{\mathbf{k}})\]
06

Calculate \( (\overrightarrow{\mathbf{B}}+\overrightarrow{\mathbf{A}}) \cdot \overrightarrow{\mathbf{C}} \)

Perform the dot product of the result from Step 5 with \( \overrightarrow{\mathbf{C}} \):\[(1.0\hat{\mathbf{i}} - 1.4\hat{\mathbf{j}} + 4.2\hat{\mathbf{k}}) \cdot (6.8\hat{\mathbf{i}} - 9.2\hat{\mathbf{j}})\]Calculate:\[= (1.0 \times 6.8) + (-1.4 \times -9.2) + (4.2 \times 0)\]\[= 6.8 + 12.88 + 0 = 19.68\]

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Key Concepts

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

Dot Product
The dot product is a fundamental operation in vector calculus. It provides a way to multiply two vectors, resulting in a scalar value. This calculation is immensely useful in physics and engineering as it helps determine the angle between two vectors and the concept of work done by a force.
The formula for the dot product of two vectors \( \overrightarrow{\mathbf{A}} = a_1 \hat{\mathbf{i}} + a_2 \hat{\mathbf{j}} + a_3 \hat{\mathbf{k}} \) and \( \overrightarrow{\mathbf{B}} = b_1 \hat{\mathbf{i}} + b_2 \hat{\mathbf{j}} + b_3 \hat{\mathbf{k}} \) is:
\[ \overrightarrow{\mathbf{A}} \cdot \overrightarrow{\mathbf{B}} = a_1 \times b_1 + a_2 \times b_2 + a_3 \times b_3 \]
Here's why the dot product is important:
  • It tells you how much one vector extends in the direction of another.
  • A dot product of zero signifies orthogonal vectors.
  • Helps in calculating projections of vectors, crucial in applications like force analysis.
In our problem, we calculated the dot product to find products like \( \overrightarrow{\mathbf{A}} \cdot (\overrightarrow{\mathbf{B}} + \overrightarrow{\mathbf{C}}) \), simplifying vector relationships.
Vector Addition
Vector addition is a straightforward yet vital operation in vector mathematics. It involves adding corresponding components of vectors and is foundational for problems that require combining multiple forces or movements.
To add two vectors, you sum up their respective components. For example, if \( \overrightarrow{\mathbf{D}} = d_1 \hat{\mathbf{i}} + d_2 \hat{\mathbf{j}} + d_3 \hat{\mathbf{k}} \) and \( \overrightarrow{\mathbf{E}} = e_1 \hat{\mathbf{i}} + e_2 \hat{\mathbf{j}} + e_3 \hat{\mathbf{k}} \), the addition is:
\[ \overrightarrow{\mathbf{D}} + \overrightarrow{\mathbf{E}} = (d_1 + e_1) \hat{\mathbf{i}} + (d_2 + e_2) \hat{\mathbf{j}} + (d_3 + e_3) \hat{\mathbf{k}} \]
Key points about vector addition:
  • Vectors are added component-wise.
  • The resulting vector represents the net effect of the original vectors.
  • Useful in physics for combining forces to get the resultant force.
In our task, vector addition was used to combine \( \overrightarrow{\mathbf{B}} \) and \( \overrightarrow{\mathbf{C}} \) and was pivotal to simplifying the calculations for later steps.
Physics Problem Solving
Physics often involves unraveling complex systems of forces, motion, and more, where vector calculus tools like the dot product and vector addition become invaluable.
In tackling physics problems, start by identifying all vectors involved, such as velocity, force, or displacement vectors. It's crucial to break down these vectors into their components (i, j, k).
When solving such problems, you might consider the following approach:
  • Identify vectors clearly and write them in component form.
  • Use vector operations, like addition and dot product, to combine and evaluate the vectors.
  • Interpret the results in the context of the physics problem to understand physical relationships like work or equilibrium conditions.
In the exercise we solved, each computation related back to physical concepts: finding how two vectors interact over a plane, how they combine, and the work done (or produced advantage) in their interaction. These are crucial insights for explaining many real-world scenarios in physics.

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Most popular questions from this chapter

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An \(85-g\) arrow is fired from a bow whose string exerts an average force of \(105 \mathrm{~N}\) on the arrow over a distance of \(75 \mathrm{~cm}\). What is the speed of the arrow as it leaves the bow?

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Stretchable ropes are used to safely arrest the fall of rock climbers. Suppose one end of a rope with unstretched length \(\ell\) is anchored to a cliff and a climber of mass \(m\) is attached to the other end. When the climber is a height \(\ell\) above the anchor point, he slips and falls under the influence of gravity for a distance 2\(\ell\) , after which the rope becomes taut and stretches a distance \(x\) as it stops the climber (see Fig. \(33 ) .\) Assume a stretchy rope behaves as a spring with spring constant \(k . ( a )\) Applying the work-energy principle, show that \(x = \frac { m g } { k } \left[ 1 + \sqrt { 1 + \frac { 4 k \ell } { m g } } \right]\) (b) Assuming \(m = 85 \mathrm { kg } , \quad \ell = 8.0 \mathrm { m }\) and \(k = 850 \mathrm { N } / \mathrm { m } ,\) determine \(x / \ell\) (the fractional stretch of the rope) and \(k x / m g\) (the force that the rope exerts on the climber compared to his own weight) at climber's fall has been stopped.

(II) The net force along the linear path of a particle of mass 480 g has been measured at 10.0 -cm intervals, starting at \(x = 0.0 ,\) to be \(26.0,28.5,28.8,29.6,32.8,40.1,46.6,42.2 ,\) \(48.8,52.6,55.8,60.2,60.6,58.2,53.7,50.3,45.6,45.2,43.2\) \(38.9,35.1,30.8,27.2,21.0,22.2 ,\) and \(18.6 ,\) all in newtons. Determine the total work done on the particle over this entire range.

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