Beer and Johnston chap 15 slide

1. PROBLEM 15.71
A 5-m beam AE is being lowered by means of two overhead cranes. At
E the instant shown it is known that the velocity of point D is 1 m/s
downward and the velocity of point E is 1.5 m/s downward. Determine
(a) the instantaneous center of rotation of the beam, (b) the velocity of
point A.
SOLUTION
~""" D 1.5 WI
E:
Q"D II WI5
/
-
11lE'1 1.5 /s
Jk YCJ)
OJ = VE - VD = 1.5 - 1.0 1
lED 1.5 = 3"rad/s)
lCE = vD = !.:Q -
OJ 1- -
3
m
3
(a) lAC = 1.5 + 2 - 3 = 0.5m C lies 0.500 m to the right of A ....
(b) VA = 0.1667m/s t....
VA= lACOJ (0.5)(*) = 0.1667 m/s
=
-

2. PROBLEM 15.110
The motion of the 3-in.-radius cylinder is controlled by the cord shown.
Knowing that end E of the cord has a velocity of 12 in./s and an
E
acceleration of 19.2 in./s2, both directed upward, determine the
acceleration (a) of point A, (b) ofpointB.
SOLUTION
Velocity .analysis. Point A is the instantaneous center of rotation of the cylinder. Vc = vE = 12 in./s.
Vc= 2rco co= -Vc =- 12 = 2 rad/s "'- )
2r (2)(3) )
Acceleration analysis. rol = (3)(2)2 = 12 in./s2
aA = [a -
A ] aC = [( ac)( t ] + [(ac t -- ] = t ]
[aE + [( ac t -- ]
ac = aA + [(aCIA)( t ] +[(aCIA)n
-- ] t
~
[aEtJ+[(ac)n- ] = [a -
A ]+[2ra tJ+[2rco2-- ]
[19.2
tJ+[(ac)n
- ] = [aA - ]+[6a t ]+[24 --] (1)
From (1), Components + t: 19.2 = 6a, a = 3.2 rad/s2 )
aG = aA + (aGIA)(+ (aGIAt
[aGtJ= [aA - ] + [ra 1] = [rco2-- ]
[aGt] = [aA - ]+ [9.6 in./s2 t ] + [12 in./s2 -- ]
From which aG = 9.6 in./s2 and aA = 12 in./s2 aG = 9.6 in./s2 t
(a) aA = 12.00 in./s2 - ....
(b) as = aG + (aSIG)( + (aSIG t L
= [9.6 t ] + [ra -- ] + [rco2! ] = [9.6t ] + [9.6- ] + [12! ]
= [9.6in./s2 ] + [2.4
-- in./s2 !] as = 9.90 in./s2 '7 14.0° ....