Solutions Manual for

Solutions Manual for University Physics with Modern Physics, 15e Edition By Hugh Young, Roger Freedman (All Chapters Download link at the end of this file) 1 / 4

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VP1.7.1. I

DENTIFY: We know that the sum of three known vectors and a fourth unknown vector is zero. We want to find the magnitude and direction of the unknown vector.S ET UP: The sum of their x-components and the sum of their y-components must both be zero.

xxx x

ABCD+++ =

yyy y

ABCD+++ =

The magnitude of a vector is w 22 x y AAA=+ and the angle θ it makes with the +x-axis is arctan y x A A θ= .

EXECUTE: We use the results of Ex. 1.7. See Fig. 1.23 in the text.

A x = 38.37 m, Bx = –46.36 m, Cx = 0.00 m, Ay = 61.40 m, By = –33.68 m, Cy = –17.80 m Adding the x-components gives 38.37 m + (–46.36 m) + 0.00 m + D x = 0 → Dx = 7.99 m Adding the y-components gives 61.40 m + (–33.68 m) + (–17.80 m) + D y = 0 → Dy = –9.92 m

22 x y

DDD=+ =

22 (7.99 m) + (–9.92 m) = 12.7 m glo arctan y x D D θ= = arctan[(–9.92 m)/(7.99 m)] = –51° Since D  has a positive x-component and a negative y-component, it points into the fourth quadrant making an angle of 51° below the + x-axis and an angle of 360° – 51° = 309° counterclockwise with the + x-axis.

EVALUATE: The vector D

 has the same magnitude as the resultant in Ex. 1.7 but points in the opposite direction. This is reasonable because D  must be opposite to the resultant of the three vectors in Ex. 1.7 to make the resultant of all four vectors equal to zero.

VP1.7.2. IDENTIFY: We know three vectors A

 , B  , and C  and we want to find the sum S  where S  = A

 –

B 

• C

. The components of – B  are the negatives of the components of B  .

SET UP: The components of S

 are

x xxx

S ABC=−+

yyyy

SABC=−+

The magnitude A of a vector A  is 22 x y AAA=+ and the angle θ it makes with the +x-axis is arctan y x A A θ= .

UNITS, PHYSICAL QUANTITIES, AND VECTORS

1

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EXECUTE: Using the components from Ex. 1.7 we have

Sx = 38.37 m – (–46.36 m) + 0.00 m = 84.73 m

Sy = 61.40 m – (–33.68 m) + (–17.80 m) = 77.28 m

22 x y

SSS=+ =

22 (84.73 m) + (77.28 m) = 115 m arctan y x S S θ= = arctan[(77.28 m)/(84.73 m)] = 42° Since both components of S  are positive, S  points into the first quadrant. Therefore it makes an angle of 42° with the + x-axis.

EVALUATE:

Figure VP1.7.2 The graphical solution shown in Fig. VP1.7.2 shows that our results are reasonable.

VP1.7.3. IDENTIFY: We know three vectors A

 , B  , and C  and we want to find the sum T  where

T

= A

 + B



• 2C

.

SET UP: Find the components of vectors A

 , B  , and C  and use them to find the magnitude and direction of T . The components of 2C  are twice those of C  .

EXECUTE: 2

x xx x SAB C=++ and 2 yyy y

SAB C=++

(a) Using the components from Ex. 1.7 gives

Tx = 38.37 m + (–46.36 m) + 2(0.00 m) = –7.99 m

Ty = 61.40 m + (–33.68 m) + 2(–17.80 m) = –7.88 m (b) 22 x y

TTT=+ =

22 (–7.99 m) + (–7.88 m) = 11.2 m arctan y x T T θ= = arctan[(–7.88 m)/(–7.99 m)] = 45° Both components of T  are negative, so it points into the third quadrant, making an angle of 45° below the – x-axis or 45° + 180° = 225° counterclockwise with the +x-axis, in the third quadrant. 3 / 4

Units, Physical Quantities, and Vectors 1-3 © Copyright 2020 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist.No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.

E

VALUATE:

Figure VP1.7.3 The graphical solution shown in Fig. VP1.7.3 shows that this result is reasonable.VP1.7.4. IDENTIFY: The hiker makes two displacements. We know the first one and their resultant, and we want to find the second displacement.

SET UP: Calling A

 the known displacement, R  the known resultant, and D  the unknown vector, we know that A

 + D

 = R  . We also know that R = 38.0 m and R  makes an angle R

θ= 37.0° + 90° =

127° with the + x-axis. Fig. VP1.7.4 shows a sketch of these vectors.Figure VP1.7.4 EXECUTE: From Ex. 1.7 we have Ax = 38.37 m and Ay = 61.40 m. The components of R  are

Rx = R cos 127.0° = (38.0 m) cos 127.0° = –22.87 m

Ry = R sin 38.0° = (38.0 m) sin 127.0° = 30.35 m

x xx RAD=+ and yyy RAD=+ Using these components, w e find the components of D  .

38.37 m + Dx = –22.87 m → Dx = –22.87 m 61.40 m + Dy = –31.05 m → Dy = –31.05 m

22 x y

DDD=+ =

22 (–61.24 m) + (–31.05 m) = 68.7 m arctan y D D θ= = arctan[(–31.05 m)/(–61.24 m)] = 27°

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