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College Physics (Urone)

Found in: Page 121

Book edition 1st Edition

Author(s) Paul Peter Urone

Pages 1272 pages

ISBN 9781938168000

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Short Answer

Show that the sum of the vectors discussed in Example 3.2 gives the result shown in Figure 3.24.

The location of the dock is \(52.94\;{\rm{m}}\), \({89.9^ \circ }\) north of west.

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Vectors

Vectors are physical quantities that have magnitude and direction.

The two vectors cannot be added by using simple algebraic rules. They can be added by using the triangle law of vector addition.

Step 2: Representation of vectors

The vectors \({\rm{A}}\) and \({\rm{B}}\) are represented as,

Representation of vectors \({\rm{A}}\) and \({\rm{B}}\)

Step 3: Given data:

The magnitude of the vector \({\rm{A}}\)is, \(A = 27.5\;{\rm{m}}\).
The magnitude of the vector \({\rm{B}}\)is, \(B = 30\;{\rm{m}}\).

Step 4: Horizontal component of vectors

The horizontal component of the vector \({\rm{A}}\) is,

\({A_x} = {\rm{A}}\cos \left( {{{66}^ \circ }} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{A_x} &= \left( {27.5\;{\rm{m}}} \right) \times \cos \left( {{{66}^ \circ }} \right)\\ &= 11.185\;{\rm{m}}\end{aligned}\)

The horizontal component of the vector \({\rm{B}}\) is,

\({B_x} = {\rm{B}}\cos \left( {{{112}^ \circ }} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{B_x} &= \left( {30\;{\rm{m}}} \right) \times \cos \left( {{{112}^ \circ }} \right)\\ &= - 11.238\;{\rm{m}}\end{aligned}\)

The resultant of horizontal components vectors \({\rm{A}}\) and \({\rm{B}}\) is,

\({R_x} = {A_x} + {B_x}\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{R_x} &= \left( {11.185\;{\rm{m}}} \right) + \left( { - 11.238\;{\rm{m}}} \right)\\ &= - 0.053\;{\rm{m}}\end{aligned}\)

Step 5: Vertical component of vectors

The vertical component of the vector \({\rm{A}}\) is,

\({A_y} = A\sin \left( {{{66}^ \circ }} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{A_y} &= \left( {27.5\;{\rm{m}}} \right) \times \sin \left( {{{66}^ \circ }} \right)\\ &= 25.123\;{\rm{m}}\end{aligned}\)

The vertical component of the vector \({\rm{B}}\) is,

\({B_y} = B\sin \left( {{{112}^ \circ }} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{B_y} &= \left( {30\;{\rm{m}}} \right) \times \sin \left( {{{112}^ \circ }} \right)\\ &= 27.816\;{\rm{m}}\end{aligned}\)

The resultant of vertical components vectors \({\rm{A}}\) and \({\rm{B}}\) is,

\({R_y} = {A_y} + {B_y}\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}{R_y} &= \left( {25.123\;{\rm{m}}} \right) + \left( {27.816\;{\rm{m}}} \right)\\ &= 52.939\;{\rm{m}}\end{aligned}\)

Step 6: Resultant vector

The magnitude of the resultant vector is,

\(R = \sqrt {R_x^2 + R_y^2} \)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}R &= \sqrt {{{\left( { - 0.053\;{\rm{m}}} \right)}^2} + {{\left( {52.939\;{\rm{m}}} \right)}^2}} \\ \approx 52.94\;{\rm{m}}\end{aligned}\)

The direction of the resultant vector is,

\(\theta = {\tan ^{ - 1}}\left( {\frac{{{R_y}}}{{{R_x}}}} \right)\)

Substitute the values in the above expression, and we get,

\(\begin{aligned}{}\theta &= {\tan ^{ - 1}}\left( {\frac{{52.939\;{\rm{m}}}}{{ - 0.053\;{\rm{m}}}}} \right)\\ &= - {89.9^ \circ }\end{aligned}\)

Hence, the location of the dock is \(52.94\;{\rm{m}}\), \({89.9^ \circ }\) north of west.

Most popular questions for Physics Textbooks

Suppose you take two steps A and B (that is, two nonzero displacements). Under what circumstances can you end up at your starting point? More generally, under what circumstances can two nonzero vectors add to give zero? Is the maximum distance you can end up from the starting point A+B the sum of the lengths of the two steps?

A basketball player is running at $5.00$ m/s directly toward the basket when he jumps into the air to dunk the ball. He maintains his horizontal velocity. (a) What vertical velocity does he need to rise $0.750$ m above the floor? (b) How far from the basket (measured in the horizontal direction) must he start his jump to reach his maximum height at the same time as he reaches the basket?

A cloud of dirt falls from the bed of a moving truck. It strikes the ground directly below the end of the truck. What is the direction of its velocity relative to the truck just before it hits? Is this the same as the direction of its velocity relative to ground just before it hits? Explain your answers.

Construct Your Own Problem Consider an airplane headed for a runway in a cross wind. Construct a problem in which you calculate the angle the airplane must fly relative to the air mass in order to have a velocity parallel to the runway. Among the things to consider are the direction of the runway, the wind speed and direction (its velocity) and the speed of the plane relative to the air mass. Also calculate the speed of the airplane relative to the ground. Discuss any last minute maneuvers the pilot might have to perform in order for the plane to land with its wheels pointing straight down the runway.

For a fixed initial speed, the range of a projectile is determined by the angle at which it is fired. For all but the maximum, there are two angles that give the same range. Considering factors that might affect the ability of an archer to hit a target, such as wind, explain why the smaller angle (closer to the horizontal) is preferable. When would it be necessary for the archer to use the larger angle? Why does the punter in a football game use the higher trajectory?

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