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Unformatted text preview: ° = −19.9 m R Rx = −15.0 m + 0 m − 9.83 m + 16.7 m R y = 5.47 m + 11.0 m − 6.88 m − 19.9 m = −8.1 m = −10.3 m 38 INTRODUCTION AND MATHEMATICAL CONCEPTS Note that the component Bx is zero, because B points along the y axis. Note also that the
components Cx and Cy are both negative, since C points between the −x and −y axes.
Finally, note that the component Dy is negative since D points below the +x axis.
Using equation (1), we find that
+y
Rx
2
2
2
2
+x
R = Rx + Ry = ( −8.1 m ) + ( −10.3 m ) = 13 m
θ Ry
θ = tan R
x
−1 Ry −1 −10.3 m = tan = 52° −8.1 m R Since both Rx and Ry are negative, the resultant points between the −x and −y axes. 63. SSM REASONING The performer walks out on the wire a distance d, and the vertical
distance to the net is h. Since these two distances are perpendicular, the magnitude of the
displacement is given by the Pythagorean theorem as s = d 2 + h 2 . Values for s and h are
given, so we can solve this expression for the distance d. The angle that the performer’s
displacement makes below the horizontal can be found using trigonometry.
SOLUTION
a. Using the Pythagorean theorem, we find that s = d 2 + h2 or d = s2 − h2 = 26
25
b .7 ft g− b .0 ft g =
2 2 9 .4 ft b. The angle θ that the performer’s displacement makes below the horizontal is given by
tan θ = h
d or θ = tan −1 h
25
F I = tan F .0 ft I =
GJ G ft J
HK H K
d
9.4
−1 69 ° 64. REASONING The force vector F points at an angle of θ above the +x axis. Therefore, its
x and y components are given by Fx = F cos θ and Fy = F sin θ.
SOLUTION
follows: a. The magnitude of the vector can be obtained from the y component as
Fy = F sin θ or F= Fy
sin θ = 290 N
= 370 N
sin 52 ° Chapter 1 Problems 39 b. Now that the magnitude of the vector is known, the x component of the vector can be
calculated as follows:
Fx = F cosθ = 370 N cos 52 ° = +230 N b g _____________________________________________________________________________________________ 65. SSM REASONING The force F and its two components form a right triangle. The
hypotenuse is 82.3 newtons, and the side parallel to the +x axis is Fx = 74.6 newtons.
Therefore, we can use the trigonometric cosine and sine functions to determine the angle of
F relative to the +x axis and the component Fy of F along the +y axis.
SOLUTION
a. The direction of F relative to the +x axis is specified by the angle θ as 74.6 newtons = 25.0° 82.3 newtons θ = cos −1 (1.5) b. The component of F along the +y axis is Fy = F sin 25.0° = ( 82.3 newtons ) sin 25.0° = 34.8 newtons (1.4) ______________________________________________________________________________
66. REASONING We are given that the vector sum of the three forces is zero, so F1 + F2 + F3
= 0 N. Since F1 and F2 are known, F3 can be found from the relation F3 = −(F1 + F2). We
will use the x and ycomponents of this equation to find the magnitude and direction of F3. SOLUTION The x and ycomponents of the equation F3 = −( F1 + F2) are:
xcomponent F3 x = − ( F1x + F2 x ) ycomponent F3 y = − F1 y + F2 y ( (1) ) (2) The table below gives the x and ycomponents of F1 and F2: Vector x component y component F1
F2 F1x = −(21.0 N) sin 30.0° = −10.5 N
F2x = +15.0 N F1y = −(21.0 N) cos 30.0° = +18.2 N
F2y = 0 N Substituting the values for F1x and F2x into Equation (1) gives 40 INTRODUCTION AND MATHEMATICAL CONCEPTS F3 x = − ( F1x + F2 x ) = − ( −10.5 N + 15.2 N ) = −4.5 N
Substituting F1y and F2y into Equation (2) gives ( ) F3 y = − F1 y + F2 y = − ( +18.2 N + 0 N ) = −18.2 N
The magnitude of F3 can now be obtained by employing the Pythagorean theorem: F3 = F32x + F32y = ( −4.5 N )2 + ( −18.2 N )2 = 18.7 N The angle θ that F3 makes with respect to the −x axis can be determined from the inverse
tangent function (Equation 1.6), F3 y −1 −18.2 N = tan = 76° −4.5 N F3 x ______________________________________________________________________________ θ = tan −1 67. REASONING AND SOLUTION The following figure is a scale diagram of the forces
drawn tailtohead. The scale factor is shown in the figure. The head of F3 touches the tail of
F1, because the resultant of the three forces is zero.
a. From the figure, F3 must have a θ magnitude of 78 N if the resultant
force acting on the ball is zero.
b. Measurement with a protractor
indicates that the angle θ = 34° . 60.0°
F1 = 50.0 N F3 θ 60.0°
F2 = 90.0 N Scale Factor: 20.0 N _____________________________________________________________________________________________ 68. REASONING
a. The drawing shows the person standing on the
earth and looking at the horizon. Notice the right
triangle, the sides of which are R, the radius of
the earth, and d, the distance from the person’s
eyes to the horizon. The length of the hypotenuse
is R + h, where h is the height of the person’s
eyes above the water. Since we know the lengths
of two sides of the triangle, the Pythagorean h d
R 90° R
Earth Horizon Chapter 1 Problems 41 theorem can be employed to find the length of the third side.
b. To c...
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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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