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Unformatted text preview: Problem 14.9 Aerodynamic drag exerts a force on the
skier in Problem 14.8 of magnitude 0.6v2, where v is the
magnitude of his velocity. Determine the skier’s velocity
when he has gone 20 in down the slope, and compare
your answer with the answer to part (b) of Problem 14.8. Strategy: After using Newton’s second law to deter
mine the skier’s acceleration in the direction parallel to
the slope, use the chain rule to express it as the product
of the velocity and the derivative of the velocity with
respect to position. Solution: m = 80 kg g = 9.31 m/s ZFX : —f— D+mgsin25° =ma 2F}, = N —mgcos25° = 0 D = 0.6112 f = MN = 0.08 N .‘.f = 0.08 mg cos 25° Substituting for D and F in the ﬁrst equation, we get ma = ~0.08 mg cos 25° + mg sin 25° — 0.61)2 Rewriting, we have a:———~——v.._ dt _ dx dt _ dx
Also a = Cg sin 25° — 0.08 gcos 25°) — v2
m Let b2 = (g sin 25° — 0‘08 gcos25°) (‘2 = 0.6/m Calculating, b = 1.85, c = 0.0866 v = 13.4 m/s 1
20¢2 z —5 In —40€2 = ln 0 Solving, [Gr—vi Problem 14.16 A 2—kg cart containing 8 kg of water y r
is initially stationary (Fig. a). The center of mass of the
“object” consisting of the cart and water is at x = 0. The
cart is subjected to the timedependent force shown in
Fig. b, where F0 = 5 N and to = 2 5. Assume that no
water spills out of the cart and that the horizontal forces
exerted on the wheels by the ﬂoor are negligible. (a) Do you know the acceleration of the cart during
the period 0 < t < to? (b) Do you know the acceleration of the center of mass
of the “object” consisting of the cart and water
during the period 0 < t < to? (c) What is the x—coordinate of the center of mass of
the “object” when t > 2to? (b) Solution: (a) No, the internal dynamics make it impossible to determine the
acceleration of just the cart.
(b) Yes, the entire system (cart + water) obeys Newton‘s 2“d Law. . _ _ 5N _ 2
ZF.(5N)—(10kg)a => 11— 10kg—0.5n1/s (c) The center of mass moves as a “super particle”. F0r0<t<t0 5N
N:10k =—=o.5m/2
5 ( g)a=>a 10kg s v = (0.5 m/s2)t, s = (0.25 m/Sz)t2
Att=tU:2s, v=l.0m/s, s=1.0m
For to < t < 2m, — 5 N = (10 kg)a, a = —0.5 m/52, v = —(0.5 m/s2)(t — to) + 1.0 m/s 3 = —(0.25 m/s2)(t — to)2 + (1.0 m/s)(t — to) + 1.0 m Problem 14.22 Consider the 11,000—kg airplane shown
in Problem 14.2]. At the instant shown, the airplane’s
velocity is v = 300i (m/s). The rate of change of the
magnitude of the velocity is dv/dt = 5 m/sz. The radius
of curvature of the airplane’s path is 4500 m, and the
y axis points toward the concave side of the path.
The thrust is T = 120,000 N. Determine the lift L and
drag D. Solution:
ax = 5 m/s2
m = [1000 kg
g = 9.81 m/s2
V = 300 m/s
T = 120000 N ‘ i
p = 4500 m  2sz TcoslS°~D—mgsin15°=max
ZFy: L+Tsinls°—mgcos15°:mu_V “y = Vz/P Solving, D = 33.0 kN, L := 293 kN Path Horizontal Problem 14.36 The 100lb crate is initially stationary.
The coefﬁcients of friction between the crate and the
inclined surface are ,ux = 0.2 and ,uk = 0.16. Determine how far the crate moves from its initial position in 2 s
if the horizontal force F = 90 1b. Solution: Denote w : IOO lb, g = 32.17 ft/sz, F = 901b, and
0 = 30°. Choose a coordinate system with the positive x axis parallel
to the inclined surface. (See free body diagram next page.) The normal
force exerted by the surface on the box is N = F sin0 + Wcos0 =
131.6 lb. The sum of the nonfriction forces acting to move the box
is Fe 2 Fcost9  Wsin0 = 27.9 lb. Slip only occurs if chl a lNusl
which 27.9 > 26.32 (lb), s0 slip occurs. The direction of slip is determined from the sign of the sum of
the non friction forces: Fe > 0, implies that the box slips up the
surface, and F6 < 0 implies that the box slips down the surface
(if the condition for slip is met). Since Fc > 0 the box slips up
the surface. After the box slips, the sum of the forces on the box
parallel to the surface is 2 FX : Fc — sgn(Fc)p.kN, where sgn(Fc) = F W
176'. From Newton‘s second law, 2 Fx = (~) (1, from which a =
c 3 %(Fc —sgn(Fc);tkN) : 2.125 ft/sz. The velocity is v(t) =at ft/s, since 11(0) 2 0. The displacement is s = gtz ft, since 3(0) = 0. The position after 2 s is s (2) = 4.43 ft up the inclined surface. Problem 1456* In Problem 1455, how long does the
collar take to go from A to B if the coefﬁcient of kinetic
friction between the collar and the bar is M = 0.2? Strategy: This problem is almost the same as prob lem 14.55. The major difference is that now we must
calculate the magnitude of the. normal force, N, and then
must add a term uk N to the forces tangent to the bar (in
the direction from B toward A ——opposing the motion).
This will give us a new acceleration, which will result
in a longer time for the collar to go from A to B. Solution: We use the unit vector e” from Problem 14.55. The
free body diagram for the collar is shown at the right. There are four
forces acting on the collar. These are the applied force F, the weight
force W = —mgj = —58.86 j (N), the force N which acts normal to
the smooth bar, and the friction force I“ = ~MNeA3. The normal
force must be equal and opposite to the components of the forces F
and W which are perpendicular (not parallel) to AB. The friction force
is parallel to AB. The magnitude of IF + W is calculate by adding
these two known forces and then ﬁnding the magnitude of the sum.
The result is that IF + W = 57.66 N. From Problem 14.55, we know
that the component of IF + WI tangent to therbar is FAB = 20.76 N.
Hence, knowing the total force and its component tangent to the bar, we
can ﬁnd the magnitude of its component normal to the bar. Thus, the
magnitude of the component of IF + W normal to the bar is 53.79 N.
This is also the magnitude of the normal force N. The equation of
motion for the collar is ZFWHM = F + W + N — uklNleAB = ma.
In the direction tangent to the bar, the equation is (F + W) ~eA3 —
[LkINI 2 mm. The force tangent to the bar is FAB = (F + W) ~eAB — [LklNl
= 10.00 N. The acceleration of the 6kg collar caused by this force
is a, = 1.667 m/sz. We now only need to know how long it takes the
collar to move a distance of 0.3 m, starting from rest, with this aCCel
eration. The kinematic equations are u, 2 cm, and s, = a,r2/2. We set
s, = 0.3 m and solve for the time. The time required is t I: 0.600 s Problem 14.76 In Problem 14.75, determine the mag—
nitude of the velocity of the mass and the angle 9 if the
tension in the string is 50 lb. Solution: EFT : Tcose —mg :0 (T2 — In2g2)L
Tm Solving we ﬁnd 6 = cos‘1 , v 2 Using the problem numbers we have 1 slug 32.2 ms2
_ ~I ______ = o
0 — cos ( 50 lb ) 49.9 _ [(50 lb)2 _ (l slug 32.2 ft/S2)2]4 ft _
v _ W _10.8 ft/s Problem 14.93 The combined mass of the motorcy
cle and rider is 160 kg. The motorcycle starts from rest
at t = 0 and moves along a circular track with 400m
radius. The tangential component of acceleration as a
function of time is a, = 2 + 0.2t m/sz. The coefﬁcient of
static friction between the tires and the track is ,us 2 0.8.
How long after it starts does the motorcycle reach the
limit of adhesion, which means its tires are on the verge
of slipping? How fast is the motorcycle moving when
that occurs? Strategy: Draw a freebody diagram showing the tan—
gential and normal components of force acting on the
motorcycle. ” Solution:
m = 160 kg
R = 400 m Along track motion:
a, 2 2 + 0.21 m/s2
v, = V =2t+0.1t2 m/s 3 =t2 +0.1:3/3 m Forces at impending slip F + f1 = ukN at impending slip F+f=‘/F2+f2 since f_LF Force eqns.
ZFl:F=ma, R=400m
ZFnzfzmvz/R m=l60 kg
ZFZ:N—mg=0 g=9.81m/32
“5:08
F2+f2=usN
a, =2+0.2t
v=2t+0.lt2 Six eqns, six unknowns (F, f, v, (1,, N, t) Solving, we have t=14.4s F=781N v = 49.6 m/s f = 983 N
N = 1570 N a, 2 4.88 m/s2 (at t = 14.4 S) [as]:=0.25 = [(4 cos 60(20) + 2(—80sin 0)(t)(20)(t)],=0_25 = 5214 ft/sz.
rad From Newton‘s second law for transverse component F9 —
far],=o.25 = [— 80 sin 0 — (1600 cos 0)t2 mg cos 0 =
( as, from vnhich F9 = —0.2025 lb
(4 cos 6)(20t)2]:=o,25 = —209 ft/sz. Problem 14.108* In Problem 14.107, suppose that the _ \
unstretched length of the spring is r0. Determine the ‘
smallest value of the spring constant k for which the pin will remain on the surface of the cam. Solution: The spring force holding the pin on the surface of the
cam is Fr = k(r — r0) = k(m(2 —— cos6) — m) = kro(l — cos 6). This
force acts toward the origin, which by deﬁnition is the same direction
as the radial acceleration, from which Newton‘s second law for the \
pin is 2 F = ker — cos 6) = —ma,.. From the solution to Problem
l4.l07, kr0(l  cos 6) = —2mrwﬁ(cos0 — 1). Reduce and solve: k =
meﬁ. Since cos() 5 l, ker — cos 6) 3 0, and mewgmosd — 1) 5
0. If k > 2mg, Deﬁne Fcq = ker  cos a) + 2mrw3(cos6 — 1). If
ch > 0 the spring force dominates over the range of 0, so that the
pin remains on the cam surface. If k < meS, ch < 0 and the radial
acceleration dominates over the range of 6, so that the pin will leave the cam surface at some value of (9. Thus is the minimum value of the spring constant required to keep the pin in contact with
the cam surface. ...
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This note was uploaded on 03/20/2008 for the course EM 311 M taught by Professor N/a during the Spring '08 term at University of Texas.
 Spring '08
 N/A
 Dynamics

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