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Unformatted text preview: PROBLEM 16.1 ' Two identical 0.4kg slender rods AB and BC are welded together to form
an Lshaped assembly. The assembly is guided by two small wheels that
roll freely in inclined parallel lets cut in a vertical plate. Knowing that
0 = 30°, determine (a) the acceleration of the assembly, (b) the reactions atA and C. SOLUTION 217x = (0.8 kg)(9.81m/s2)(cos60°) = (0.8 kg)a a=£
2 a = 4.91m/s2 ‘8: 30° 4
+) EMC = (—0.8 kg)(g)(45 mm) + RA (0.866)(180 mm) — RA(0.5)(180 mm)
= (—0.8 kg)(a)(—0.866)(135 mm) — (0.8 kg)(a)(0.5)(45 m)
R A (65.88 mm) = —(39.528 kgmm)a —(39.528 kgmm)[—9;28—l m/sz) R =————~——————~_ 2.943k m/ 2
A (65.88 m) g S or RA = 2,94N 7 60° 4
EFy = RA + RC  (0.8g)cos30° = 0
RA + RC = 6.7964 RC = 6.7965 + 2.943 = 9.7395 N
or RC = 9.74N 4 60° 4 PROBLEM 16.5 Knowing that the coefﬁcient of static friction between the tires and the
road is 0.80 for the automobile shown, determine the maximum possible ‘
acceleration on a level road, assuming (a) fourwheel drive, (b) rear
wheel drive, (c)‘frontwheel drive, SOLUTION (a) F ourwheel drive: a = 0.80g = 0.80(9.81m/s2) = 7.848 m/s2 (b) Rearwheel drive:
. [I : W3
L M
qwﬁﬂe ”A +‘) 2MB = 2(MB)eﬁ: (1 m)W — (1.5 m)NA = —(0.5 m)ma—
NA = 0.4W + 0.2mZz' FA = ,ukNB = 0.80(0.4W + 0.2ma) = 0.32mg + 0.16mi? .‘t. ZFx = 2(Fx) FA = m eff I
0.32mg + 0.16m? = m5
0.32g = 0.845 = ﬂ(9.81m/sz) = 3.7371 m/s,2 0.84
or a = 3.74 m/s2 4 PROBLEM 16.5 CONTINUED (c) F rantwheel drive: . (2.5 m)NB — (1.5 m)W = —(0.5 m)ma NB = 0.6W — 0.2m5 _ > FE. = ,ukNB = 0.80(0.6W — 0.2ma) = 0.48mg  0.16mi: 1:. Xi; = 2(Fx)eﬁ: FB = ma 0.48mg  0.16m?! = ma—
0.48g = 1.165 a = 9'ﬁ(9.81m/52) = 4.0593 m/s2  1.16
or E = 4.06 m/s2 —> 4 ‘,/ PROBLEM 16.14 Bars AB and BE, each of mass 4 kg, are welded together and are pin
connected to two links AC and BD. Knowing that the assembly is
released from rest in the position shown and neglecting the masses of the links, determine (a) the acceleration of the assembly, (b) the forces in the
links. SOLUTION \+ zFiBG = (78.48 N)(0.5) (8 kg)(a) a = 4.905 m/s2 a = g +/ ZFHBG = FB — (78.48 N)(0.866) = 0
FB = 67.966N a = 4.91m/sz‘q 30°< FA = 0, FB = 68.0 N compression 4 Ar .1— or “1'" ”r" .4." .1 w.’ a.” ‘2‘;— PROBLEM 1&39: «A cylinder of radius r and mass m rests onktwo small casters A and B as
showmlnitially, the~cylinder is atvrest and is set in motion by rotating
caster B. clockwise athighspeed so that slipping occurs between the
cylinder and casterB. Denoting by, yk the coefﬁcient of kinetic friction
and neglecting the moment of inertia of the free caster A, derive an expression for the angular acceleration of the cylinder.  4:) 2M6 = ,ukNBr = igmrzay ___ 2/1st
mr a \+ EFX = —NBsin2¢2 + ,ukNBCOSZq) +.mgsin¢ = 0 mg sin (27
NB = —.—————
sm 2¢ —— ,uk cos 2¢ ora= 2xtkgsin¢ .
r(sin2¢) — pk cosZ¢) 3‘ PROBLEM 16.52 A S—m beam of mass 200' kg is lowered by means of two cables
unwinding from overhead cranes. As the beam approaches the ground, the crane operators apply brakes to slow the unwinding motion. Knowing
that the deceleration of cable A 'is 5 111/52 and the deceleration of cable B
is 0.5 m/sz, determine the tension in each cable. SOLUTION Kinematics: Kinetics: a3 = aA + 5a f\
10.5111/52 =15 m/s2 + (5 m)a :> a = 0.9 rad/s2)
a = aA 1‘ +2.50: 1 = 5 m/s2 — (2.5 m)(0.9 rad/s2) = 2.75 m/s2 ‘) 1 2 1 2 2
I:— =— k = 1. k '
12 L 12(200 g)(5m) 4 667 gm 12MB = 2(MB)eﬂ:
TA (5 m) — (200 kg)(g;)(2.5 m) = (200 kg)(a)(2.5 m) + (416.67 kgm2)a
Substituting g = 9.81 m/sz, a = 2.75 m/sz, a = 9.9 rad/s2
TA(5 m) = (4905 +1375 + 375)kgm2/52, TA = 1331 N or TA =1331N 4 : TA + TB — (20.0 kg)(g) =«(200 kg)c_1 sz = 2(Fy) eff 1331 N + TB  (200 kg)(9.81 111/52) = (200 kg)(2.75 111/52) TB =1181N
or TB =1181N 4 ”\ PROBLEM 16.67 Solve Prob. 16.66 assuming that the sphere is replaced by a uniform thin
hoop of radius r and mass m. SOLUTION Use equations derived for P16.66 Hoop (a) Eq. (3): (b) And (c) same as above: ...
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 Summer '06
 

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