PROBLEM 11.98 A helicopter is flying with a constant horizontal velocity of 180 km/h and is directly above Point A when a loose part begins to fall. The part lands 6.5 s later at Point B on an inclined surface. Determine (a) the distance d between Points A and B, (b) the initial height h.
SOLUTION Place origin of coordinates at Point A. Horizontal motion:
(vx )0 = 180 km/h = 50 m/s x = x0 + (vx )0 t = 0 + 50t m
At Point B where t B = 6.5 s, (a)
Distance AB. 325 cos10°
From geometry
d=
Vertical motion:
y = y0 + ( v y ) 0 t −
At Point B (b)
xB = (50)(6.5) = 325 m
d = 330 m 1 2 gt 2
1 − xB tan10° = h + 0 − (9.81)(6.5) 2 2 h = 149.9 m
Initial height.
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PROBLEM 11.106 At halftime of a football game souvenir balls are thrown to the spectators with a velocity v0. Determine the range of values of v0 if the balls are to land between Points B and C.
SOLUTION The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below Point A. The coordinates of Point A are x0 = 0, y0 = 2m. The components of initial velocity are (vx )0 = v0 cos 40° m/s and (v y )0 = v0 sin 40°. Horizontal motion:
x = x0 + (vx )0 t = (v0 cos 40°)t
Vertical motion:
y = y0 + ( v y ) 0 t =
From (1), Then
1 2 gt 2 1 = 2 + (v0 sin 40°) = − (9.81)t 2 2
v0 t =
(2)
x cos 40°
(3)
y = 2 + x tan 40° − 4.905t 2 t2 =
Point B:
(1)
2 + x tan 40° − y 4.905
(4)
x = 8 + 10 cos 35° = 16.1915 m y = 1.5 + 10sin 35° = 7.2358 m 16.1915 v0 t = = 21.1365 m cos 40° 2 + 16.1915 tan 40° − 7.2358 t2 = 4.905 21.1365 v0 = 1.3048
t = 1.3048 s v0 = 16.199 m/s
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PROBLEM 11.106 (Continued)
Point C:
x = 8 + (10 + 7) cos 35° = 21.9256 m y = 1.5 + (10 + 7)sin 35° = 11.2508 m v0 t =
21.9256 = 28.622 m cos 40°
t2 =
2 + 21.9256 tan 40° − 11.2508 4.905
v0 =
28.622 1.3656
t = 1.3656 s v0 = 20.96 m/s 16.20 m/s < v0 < 21.0 m/s
Range of values of v0.
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PROBLEM 11.118 The three blocks shown move with constant velocities. Find the velocity of each block, knowing that the relative velocity of A with respect to C is 300 mm/s upward and that the relative velocity of B with respect to A is 200 mm/s downward.
SOLUTION From the diagram Cable 1:
y A + yD = constant
Then
v A + vD = 0
Cable 2:
(1)
( yB − yD ) + ( yC − yD ) = constant vB + vC − 2vD = 0
Then
(2)
Combining Eqs. (1) and (2) to eliminate vD , 2v A + vB + vC = 0
(3)
Now
v A/C = v A − vC = −300 mm/s
(4)
and
vB/A = vB − v A = 200 mm/s
(5)
(3) + (4) − (5)
Then
(2v A + vB + vC ) + (v A − vC ) − (vB − v A ) = (−300) − (200) v A = 125 mm/s
or vB − (−125) = 200
and using Eq. (5)
v B = 75 mm/s
or −125 − vC = −300
Eq. (4)
vC = 175 mm/s
or
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PROBLEM 11.137 An outdoor track is 420 ft in diameter. A runner increases her speed at a constant rate from 14 to 24 ft/s over a distance of 95 ft. Determine the magnitude of the total acceleration of the runner 2 s after she begins to increase her speed.
SOLUTION We have uniformly accelerated motion v22 = v12 + 2at Δs12
Substituting or
(24 ft/s)2 = (14 ft/s)2 + 2at (95 ft) at = 2 ft/s 2
Also
v = v1 + at t
At t = 2 s:
v = 14 ft/s + (2 ft/s 2 )(2 s) = 18 ft/s v2
Now
an =
At t = 2 s:
an =
Finally
a 2 = at2 + an2
At t = 2 s:
a 2 = 22 + 1.542862
ρ (18 ft/s) 2 = 1.54286 ft/s 2 210 ft
a = 2.53 ft/s 2
or
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PROBLEM 11.146 Three children are throwing snowballs at each other. Child A throws a snowball with a horizontal velocity v0. If the snowball just passes over the head of child B and hits child C, determine the radius of curvature of the trajectory described by the snowball (a) at Point B, (b) at Point C.
SOLUTION The motion is projectile motion. Place the origin at Point A. Horizontal motion:
v x = v0
x = v0 t
Vertical motion:
y0 = 0,
(v y ) = 0
v y = − gt
1 y = − gt 2 2
2h , g
t=
where h is the vertical distance fallen.
| v y| = 2 gh
Speed:
v 2 = vx2 + v 2y = v02 + 2 gh
Direction of velocity. cos θ =
v0 v
Direction of normal acceleration. an = g cos θ =
Radius of curvature: At Point B,
ρ=
gv0 v 2 = v ρ
v3 gv0
hB = 1 m; xB = 7 m tB =
(2)(1 m) = 0.45152 s 9.81 m/s 2
xB = v0t B
v0 =
xB 7m = = 15.504 m/s t B 0.45152 s
vB2 = (15.504) 2 + (2)(9.81)(1) = 259.97 m 2 /s 2 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 215
PROBLEM 11.146 (Continued)
(a)
Radius of curvature at Point B.
ρB = At Point C
(259.97 m 2 /s 2 )3/ 2 (9.81 m/s 2 )(15.504 m/s)
ρ B = 27.6 m
hC = 1 m + 2 m = 3 m
vC2 = (15.504) 2 + (2)(9.81)(3) = 299.23 m 2 /s 2
(b)
Radius of curvature at Point C.
ρC =
(299.23 m 2 /s 2 )3/2 (9.81 m/s 2 )(15.504 m/s)
ρC = 34.0 m
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PROBLEM 11.155 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Venus: g = 29.20 ft/s 2 , R = 3761 mi.
SOLUTION From Problems 11.153 and 11.154,
an =
gR 2 r2
For a circular orbit,
an =
v2 r
v= R
Eliminating an and solving for v, For Venus,
g r
g = 29.20 ft/s 2 R = 3761 mi = 19.858 × 106 ft. r = 3761 + 100 = 3861 mi = 20.386 × 106 ft
Then,
v = 19.858 × 106
29.20 = 23.766 × 103 ft/s 20.386 × 106 v = 16200 mi/h
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