This preview shows pages 1–7. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 15—4* A lSkg block of ice slides on a horizontal surface for 20 m
before it stops. If the initial speed of the block was 15 m/s, determine a. The force of friction
between the block of
ice and the surface. b. The kinetic coefficient of
friction Mk between the block of ice and the surface. Solution
The freebody diagram includes the weight of the block W = 15(9.81) = 147.15 N, the normal force N and friction force F exerted on the block by the surface. The equations of motion are
+4 SE; = max: F = 15a
+T 2F =ma: N—147.15=O
Y Y
‘ where ay = 0 since there is no motion in the vertical direction.
§ Therefore
N = 147.15 N
15a = F Rewriting the xcomponent of acceleration using the chainrule of differentiation
dv dv dx dv
a=—=————=V
dt dx dt dx and integrating 15V dv F dx 2
15V /2 Fx + C = 1687.5 — Fx where the constant of integration has been chosen so that v = 15 m/s when x = 0. Then, if v = 0 when x = 20 m,
F = 84.4 N ............................................. Ans. and the kinetic coefficient of friction is u F/N = 0.573 ............... . . . . . . . . ................. Ans. k ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1525* Two blocks A and B are connected by a flexible cable as shown.
The kinetic coefficient of friction
between block A and the inclined
surface is Mk = 0.15 and the system is
released from rest. If the weight of
body A is 50 lb and block B strikes
the horizontal surface 3 s after being
released, determine a. The acceleration of body 8. b. The weight of body B. c. The tension in the cable while
the blocks are in motion. Solution Since the two blocks move in separate
directions, separate freebody diagrams
must be drawn for each block. Using
coordinates along and normal to the
surface for block A, the equations of motion are 50 2 = : — — ‘ o = —
+& F mAa T BA 50 Sin 40 32.2 aA (a)
0
+ E = :  = 0
5 FAY mAaAy NA 50 cos 40 (b)
+T Z = :  . =
Fay mBaBy T 32 2mB mBaB rn (c) where the y—component of acceleration of
block A is zero since it does not move in
the direction normal to the surface, only
the y—component of motion gives useful information for block B, and a = —a we: "13% A B since both aA and a3 are positive upward. Therefore N = 38.3022 lb = constant
F = 0.15NA = 5.7453 lb = constant (Problem 15—25 continues ...) velocity and VB YB so that VB
block B hits 0
H and subtracting Eq. a from Eq. ENGINEERING MECHANICS  Dynamics 5.7453 + 32.1394  32.2m c gives = mBa — 1.5528(—aB) B 37.8847  32.2m B — constant + . 5 8
m3 1 5 2 position gives a t + C
B l 2
+
aBt /2 Integrating the (constant) acceleration to get the = a t ft/s B C2 = O and yB = 10 the floor (yB aB(3)2/2 + 10 2
—2.2222 ft/s 1.3789 slug 32.2mB 41.3 lb 44.4 2
+
aBt /2 10 ft where the constants of integration have been chosen ft when t = 0. Then, if
= 0) when t = 3 s,
2
= 2.22 ft/s ¢ . . . . . . . . . . . . . . . . . . .. Ans
lb . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans W.F. Riley & L.D. Sturges (Problem 1525 — cont.) ‘ ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges
15—34 A 55kg skydiver jumps from
a stationary balloon at a height of
3000 m. The drag force exerted on
her body by the air when she is in
a spreadeagle position can be
expressed as FD = 0.180v2 where F
is in Newtons and V is in meters
per second. Determine her terminal
velocity and the time required to
reach 95 percent of her terminal velocity. Solution The freebody diagram of the skydiver's
body includes her weight W and the air drag Fb. Only the yhcomponent of the equations of motion gives useful information 2
+T 2F = ma : 0.180V — 55(9.8l) = 55
y y where y, v, and a are all positive upward.
Rewriting the acceleration 2
305.556a = V  2997.5 and integrating to get the velocity gives I 205.556 dV I dt V  2997.5 1
5.5810 tanh (v/54.75) = t + c = t where the constant of integration is zero since the skydiver starts from
rest. Rewriting gives v = 54.75 tanh (t/5.5810) m/s
and as t 6 m V‘9 = 54.8 m/s ................................... Ans. Vterm
The skydiver will reach 95 percent of the terminal velocity when v = 0.95(54.75) = 54.75 tanh (t/5.5810) t = 10.22 s ........................................ Ans. ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 15—54* A 100g apple falls from a tree branch 3 m above the ground.
As it falls, a strong cross wind exerts a constant horizontal force
of 0.05 N on the apple. Determine the horizontal distance traveled by the apple as it falls to the ground. Solution
The freebody diagram of the apple
includes its weight W and the wind force. The equations of motion are +9 2F ma : 0.05
x x +T 2F ma : —0.1(9.81)
y y Therefore, the x— and y components of acceleration are 2 2
a = 0.5 m/s a = 9.81 m/s X Y
and they are both constant. Integrating the (constant) accelerations to get the velocity and position of the apple gives
V 0.5t m/s v —9.81t m/s
X Y
2 2
x 0.25t m y 3  4.905t m where the constants of integration have been chosen so that 3 3 m when t = 0. The apple will hit the ground when
2
y 3 — 4.905t = 0
t 0.78206 s
at which time its xposition is x = 0.1529 m ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1557 A circular disk rotates in a horizontal
plane. A.3—1b block rests on the disk 8 in. from
the axis of rotation. The static coefficient of
friction between the block and the disk is 0.50.
If the disk starts from rest with a constant
angular acceleration of 0.5 rad/sz, determine the length of time required for the block to begin to slip. Solution The freebody diagram of the block
includes its weight an a normal force t
N exerted on the block by the surface,
and a friction force exerted on the
block by the surface. The friction
force is represented in terms of its
n and tcomponents. The equations of
motion are 3
«+2 = : =—
Fn man Fn 32.2 m
3 2
+ EFt  mat. Ft  32.2 rw
+T 2F = ma : N — 3 = 0
z z
where the z—component of acceleration
is zero since the block has no
vertical motion,
2
a = 0.5 rad/s = constant
w = 0.5t rad/s
Therefore
N = 3 lb , .
(3,J¢ mew)
F; = 0.03106 lb
2
Ft = 0.01553t lb
2 2 2 2 2 2
F = (0.03106) + (0.01553t ) S (0.50N) = (1.5 lb) and the block will begin to slip when t = 9.83 s ........................................ Ans. ENGINEERING MECHANICS  Dynamics
1569 A small (W = 2 lb) chunk of ice
is initially at rest at the very top of
the hemispherical observatory. If a gust of wind gives the ice an initial speed of v = l ft/s, determine the 0
angle 9 where the ice will lose contact
with the roof. Assume that friction between the ice and the roof of the observatory is negligible. Solution The free—body diagram of the ice
includes its weight W and the normal
force N exerted on it by the cylindrical
roof. Using components along and normal to the surface, the equations of motion are
+2 ZFt = mat: mg sin 9 = my
+Y 2F = ma : mg cos 9  N =
n n Rewriting the tangential component 2
mv /R W.F. Riley & L.D. of acceleration and integrating to get the velocity of the ice as a function of 9 I vR d9 = I gR sin 6 d9 I V dv 2
V /2 —gR cos 9 + C = 0.5 + gR(l — cos 6) where the constant of integration has been chosen so that 0 V = 1 ft/s when 6 = 0 . Then,
N = mg cos 6 — m[ :0 + 29(1  cos 6] and the ice will leave the surface when A7: 0 6 = 48.100 the normal force is Sturges ...
View Full
Document
 Spring '09
 Dr.

Click to edit the document details