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Pythagorean theorem: F = A + B . Knowing the magnitudes of A and B, we can calculate
the magnitude of F. The direction of the resultant can be obtained using trigonometry.
b. For the vector F′ = A – B we note that the subtraction can be regarded as an addition in
the following sense: F′ = A + (–B). The vector –B points due south, opposite the vector B,
so the two vectors are once again perpendicular and the magnitude of F′ again is given by
the Pythagorean theorem. The direction again can be obtained using trigonometry. Chapter 1 Problems SOLUTION a. The drawing shows the
two vectors and the resultant vector.
According to the Pythagorean theorem, we
have
F 2 = A2 + B 2 F= F
B θ 445
325
b N g+ b N g θ F′ A
2 North North A2 + B 2 F= 15 –B
A 2 (b) (a) = 551 N
Using trigonometry, we can see that the direction of the resultant is
tan θ = B
A or θ = tan −1 325
F NI=
G NJ
HK
445 36.1° north of west b. Referring to the drawing and following the same procedure as in part a, we find bg F ′2 = A2 + − B 2 tan θ = F′ = or
B
A or 445
−
b g = b N g+ b325 N g =
325
F N I = 36.1° south of west
G NJ
HK
445 A2 + − B θ = tan −1 2 2 2 551 N _____________________________________________________________________________________________ 26. REASONING The triple jump consists of a double jump in one
direction, followed by a perpendicular single jump, which we can
represent with displacement vectors J and K (see the drawing).
K
These two perpendicular vectors form a right triangle with their
resultant D = J + K, which is the displacement of the colored
D
checker. In order to find the magnitude D of the displacement, we
first need to find the magnitudes J and K of the double jump and
the single jump. As the three sides of a right triangle, J, K, and D
(the hypotenuse) are related to one another by the Pythagorean
ds
J
theorem (Equation 1.7) The double jump moves the colored
s
checker a straightline distance equal to the length of four square’s
diagonals d, and the single jump moves a length equal to two square’s diagonals.
Therefore,
J = 4d
and
K = 2d
(1) Let the length of a square’s side be s. Any two adjacent sides of a square form a right
triangle with the square’s diagonal (see the drawing). The Pythagorean theorem gives the
diagonal length d in terms of the side length s: 16 INTRODUCTION AND MATHEMATICAL CONCEPTS d = s2 + s2 = 2s2 = s 2 (2) SOLUTION First, we apply the Pythagorean theorem to the right triangle formed by the
three displacement vectors, using Equations (1) for J and K:
D= ( 4d )2 + ( 2d )2 J 2 + K2 = = 16d 2 + 4d 2 = 20d 2 = d 20 (3) Substituting Equation (2) into Equation (3) gives ( D = d 20 = s 2 ) 20 = s 40 = ( 4.0 cm ) 40 = 25 cm 27. REASONING For convenience, we can assign due east to be the positive direction and due
west to be the negative direction. Since all the vectors point along the same eastwest line,
the vectors can be added just like the usual algebraic addition of positive and negative
scalars. We will carry out the addition for all of the possible choices for the two vectors and
identify the resultants with the smallest and largest magnitudes.
SOLUTION There are six possible choices for the two vectors, leading to the following
resultant vectors: F1 + F2 = 50.0 newtons + 10.0 newtons = +60.0 newtons = 60.0 newtons, due east
F1 + F3 = 50.0 newtons − 40.0 newtons = +10.0 newtons = 10.0 newtons, due east
F1 + F4 = 50.0 newtons − 30.0 newtons = +20.0 newtons = 20.0 newtons, due east
F2 + F3 = 10.0 newtons − 40.0 newtons = −30.0 newtons = 30.0 newtons, due west
F2 + F4 = 10.0 newtons − 30.0 newtons = −20.0 newtons = 20.0 newtons, due west
F3 + F4 = −40.0 newtons − 30.0 newtons = −70.0 newtons = 70.0 newtons, due west
The resultant vector with the smallest magnitude is F1 + F3 = 10.0 newtons, due east . The resultant vector with the largest magnitude is F3 + F4 = 70.0 newtons, due west . Chapter 1 Problems 28. REASONING Both P and Q and the
vector sums K and M can be drawn with
correct magnitudes and directions by
counting grid squares. To add vectors,
place them tailtohead and draw the
resultant vector from the tail of the first
vector to the head of the last. The vector
2P is equivalent to P + P, and −Q is a
vector that has the same magnitude as Q,
except it is directed in the opposite
direction. 17 P
Q
M −Q 8.00 cm K P
The vector M runs 11 squares
P
horizontally and 3 squares vertically, and
the vector K runs 4 squares horizontally
and 9 squares vertically. These distances can be converted from grid squares to centimeters
with the grid scale: 1 square = 4.00 cm. Once the distances are calculated in centimeters, the
Pythagorean theorem (Equation 1.7) will give the magnitudes of the vectors.
SOLUTION
a. The vector M = P + Q runs 11 squares horizontally and 3 squares vertically, and these cm distances are equivalent to, respectively, 4.00 (11 squares ) = 44.0 cm and
square cm 4.00 ( 3 squares ) = 12.0 cm . Thus, the magnitude of M is...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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