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Unformatted text preview: ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. sturges 17—17* A 3lb collar slides on a frictionless vertical guide as shown.
A constant horizontal force of 12 lb is applied to the end of the light
inextensible cord that is attached to the
collar. If the collar is released from
rest when d = 32 in, determine the work
done on the collar
By the gravitational force as the
collar rises 12 in.
By the tension in the cord as the
collar rises 12 in.
Solution
a. weight force W is constant
and acts vertically downward. As the
3—lb collar rises 12 in. = 1 ft, the
work done by the weight force is
U = Fd = —3(1) W
= 3 lb'ft b. Initially, the length of cord between the collar
and the pulley is L = / 322 + 242 = 40 in. 1
After the collar has risen 12 in., the length of cord
between the collar and the pulley is
L2 = V 202 + 242 = 31.2410 in.
Therefore, the 12—lb force pulls the cord 8.759 in. to the right and does work viz 12(8.759) 105.108 lb'in. = 8.76 lbft 849 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 1751 The two blocks shown weigh NA = 60 lb and W3 = 40 lb. If the blocks are released from rest from the position shown, determine the velocity of block B after it has moved 10 ft. Solution The length of the cord is constant, = + =
L 5A ZSB constant Differentiating the length with respect to time gives = 2
VA VB
where VA = 5A and VB = —53 (VA, 5A, and 53 are all positive downward while VB is positive upward). The tension is an internal force — its work
will cancel out when the work—energy equations
for the two blocks are added together. When
block B rises 10 ft, block A will fall a distance of 20 ft, and their weights will do work on the system U = Fd = 60(20) — 40(10) = 800 lb°ft Therefore, the workenergy equation I} + U = T gives f
1 60 2 1 4o 2
+ + = —————— —————
(0 0) 80° 2 [32.2 ](2VB) + 2 [32.2 ]VB
VB = 13.56 ft/s T . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. Ans.
VA = 27.1 ft/s ~L 889 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 17—52* The pair of blocks shown are connected by a light inextensible bar. Both the horizontal 3kg
and vertical guide slots are frictionless. If the blocks are released from rest from the position shown, determine the velocity of the 3kg block when a. It is at the same level as the 2—kg block.
b. It is 150 mm below the 2kg block. Solution
Since the length of the bar is constant, differentiating
2 2
x + y = 2x; + 2y} VB yVA/x where VA s y (T) and VB = x (é). Neither NA, Nb, nor the weight of
the 2kg block do work. Only the
work done by the weight of the 3kg
block is needed. When A moves down a distance Ay,
U = 3(9.81) Ay J Therefore, the work—energy equation T. + U = T gives 1 f
0 + 29.43 Ay [(1/2)(3)v: + (1/2)(2)v:] When y = 0, x = VA = 2.17 m/s ¢ When y = 150 mm, x = 212.4 mm, Ay = 0.39 m, VB 0 + 29.43(o.39) = 1.5V: + (0.7062vA)2 VA = 2.40 m/s l ENGINEERING MECHANICS — Dynamics W.F. Riley & L.D. Sturges 1755* The two blocks A and B shown weigh NA = 30 lb and NE 20 lb, respectively. The kinetic coefficient of friction between block A and the
inclined surface is pk = 0.15 and the horizontal surface supporting block B is smooth. when the blocks are in the position shown, block B is moving to the right with a velocity of 5 ft/s. Determine the distance traveled by block 3 before it comes to rest. Solution
The length of the cord is constant, L = 25A + 53 = constant Differentiating the length with respect
to time gives = . +
0 25A 53 B = zvh where VA = 5A (VA is positive to the right and EA is positive to the left) and VB = (both are positive to the right). 3 Neither block has any motion normal to
the surface. Newtons second law gives +5 2F = man NA  30(4/5) = o n
= 4 b
NA 2 l §k= 0.15N = 3.6 lb
The tension is an internal force  its work will cancel out when the
workenergy equations for the two blocks are added together. When block
block B slides to the right a distance b, block A slides up the inclined
surface b/2; the work done by the constant weights and friction force is U = FU = —3.6(b/2) — [30(3/5)](b/2) = —lO.80b lb'ft Therefore, the work—energy equation Ti + U = Tf 1 30 2 1 20 2
2 [ 32.2 + 2 [ 32.2 ](5)  10.80b _ 0 gives b = 0.988 ft ................... . . . . ............ Ans. 893 ENGINEERING MECHANICS  Dynamics W.F. Riley & L.D. Sturges 17—74* A 0.5—kg mass slides on a frictionless
rod in a vertical plane as shown. The undeformed
length of the spring is £0 = 250 mm, the spring
modulus is 5 = 600 N/m, and the distance d = 800 mm. If the slider is released from rest when b = 300 mm, determine the speed of the slider at positions A and B. Solution The normal force N does no work; the
spring and weight forces have potentials.
The zero of gravitational potential energy
will be set at the position where b = 0.
In the initial position, b = 0.3 m, the
length of the spring is C = 0.5 m, the
stretch of the spring is 5 = C — £0 = 0.53 0.25 m, and the initial potential energy
is
V = V + V = 0 5(9 Bl)(O 3) + —l— 600 O 25 2  17 2785 J
igs.z()(.)—.
WhenthesliderisatA,b=0m,£=0.4m,5:0.15m, the final potential energy is 2
vs = o + —:2L(600)(0.15) = 6.75 J and the work—energy equation Ti + vi + U = Tf + V? gives
1 o + 17.2735 + o = —2(0.5)v: + 6.75 V When the slider is at B, b the final potential energy is + v5 O.5(9.81)(O.4) + %(600)(0.15)2 = 8.712 J and the work—energy equation 13 + V: + U = Tf + V? gives 0 + 17.2785 + O = %—(O.5)v: + 8.712 VB = 5.85 m/s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ... Ans. 918 ...
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This homework help was uploaded on 03/25/2008 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.
 Spring '08
 McVay

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