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UNITS, PHYSICAL QUANTITIES, AND VECTORS
1
VP1.7.1. IDENTIFY: 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.
SET UP: The sum of their x-components and the sum of their y-components must both be zero.
Ax + Bx + C x + Dx = 0
Ay + By + C y + Dy = 0
The magnitude of a vector is w A = Ax2 + Ay2 and the angle θ it makes with the +x-axis is
Ay
θ = arctan .
Ax
EXECUTE: We use the results of Ex. 1.7. See Fig. 1.23 in the text.
Ax = 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 + Dx = 0 → Dx = 7.99 m
Adding the y-components gives
61.40 m + (–33.68 m) + (–17.80 m) + Dy = 0 → Dy = –9.92 m
D = Dx2 + Dy2 = (7.99 m) 2 + (–9.92 m) 2 = 12.7 m
Dy
glo θ = arctan = arctan[(–9.92 m)/(7.99 m)] = –51°
Dx
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
S x = Ax − Bx + C x
S y = Ay − By + C y
The magnitude A of a vector A is A = Ax2 + Ay2 and the angle θ it makes with the +x-axis is
Ay
θ = arctan .
Ax
© 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.
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1-2 Chapter 1
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
S = S x2 + S y2 = (84.73 m)2 + (77.28 m)2 = 115 m
Sy
θ = arctan = arctan[(77.28 m)/(84.73 m)] = 42°
Sx
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 2 C are twice those of C .
EXECUTE: S x = Ax + Bx + 2C x and S y = Ay + By + 2C y
(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) T = Tx2 + Ty2 = (–7.99 m)2 + (–7.88 m) 2 = 11.2 m
Ty
θ = arctan = arctan[(–7.88 m)/(–7.99 m)] = 45°
Tx
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.
© 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.
,2
,3
UNITS, PHYSICAL QUANTITIES, AND VECTORS
1
VP1.7.1. IDENTIFY: 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.
SET UP: The sum of their x-components and the sum of their y-components must both be zero.
Ax + Bx + C x + Dx = 0
Ay + By + C y + Dy = 0
The magnitude of a vector is w A = Ax2 + Ay2 and the angle θ it makes with the +x-axis is
Ay
θ = arctan .
Ax
EXECUTE: We use the results of Ex. 1.7. See Fig. 1.23 in the text.
Ax = 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 + Dx = 0 → Dx = 7.99 m
Adding the y-components gives
61.40 m + (–33.68 m) + (–17.80 m) + Dy = 0 → Dy = –9.92 m
D = Dx2 + Dy2 = (7.99 m) 2 + (–9.92 m) 2 = 12.7 m
Dy
glo θ = arctan = arctan[(–9.92 m)/(7.99 m)] = –51°
Dx
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
S x = Ax − Bx + C x
S y = Ay − By + C y
The magnitude A of a vector A is A = Ax2 + Ay2 and the angle θ it makes with the +x-axis is
Ay
θ = arctan .
Ax
© 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.
1-1
, 4
1-2 Chapter 1
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
S = S x2 + S y2 = (84.73 m)2 + (77.28 m)2 = 115 m
Sy
θ = arctan = arctan[(77.28 m)/(84.73 m)] = 42°
Sx
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 2 C are twice those of C .
EXECUTE: S x = Ax + Bx + 2C x and S y = Ay + By + 2C y
(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) T = Tx2 + Ty2 = (–7.99 m)2 + (–7.88 m) 2 = 11.2 m
Ty
θ = arctan = arctan[(–7.88 m)/(–7.99 m)] = 45°
Tx
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.
© 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.
