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Unformatted text preview: PROBLEM 13.30 A 10kg block is attached to spring A and connected to spring B by a cord
and pulley. The block is held in the position shown with both springs :
unstretched when the support is removed and the block is released with
no initial velocity. Knowing that the constant of each spring is 2 kN/m,
determine (a) the velocity of the block after it has moved down 50 mm,
(b) the maximum velocity achieved by the block. 1 7 l 2
Ul—Z = W024) ‘ 31940902 _ aka (x3) (Gravity) (Spring A) (Spring B) UH: (98.1 N)(0.05 m) — l(2000 N/m)(0.05 m)2 2 % (2000 N/m) (0.025 m)2 4.905 — 2.5 — 0.625 = (10) v3 i
2 l
v = 0.597 m/s , (b) Let x = Distance moved down by the 10 kg block PROBLEM 13.30 CONTINUED 0 = 98.1 — 2000 (x) — $ (2x) = 98.1 — (2000 + 500) x
x = .0392m
For x = 0.0436, U = 4.2772 — 1.9010 — 0.4752 = g (10) v2 v ax: .6198m/s m PROBLEM 13.45 A section of track for a roller coaster consists of two circular arcs AB and
CD joined by a straight portion BC. The radius of AB is 27 m and the
radius of CD is 72 m. The car and its occupants, of total mass 250 kg,
reach point A with practically no velocity and then drop freely along the
track. Determine the normal force exerted by the track on the car as the
car reaches point B. Ignore air resistance and rolling resistance. SOLUTION UA_B = W(27)(1— cos40°)
U H = (250 kg x 9.81 m/s2 )(27 m)(0.234)
UM =154951 TA + UH = TA 0 + 15495 =125v§ v2 = (15495 J)
B (125 kg) v; = 124.0 mZ/sz Newtons Law at B +/ N — Wcos40° = ; v; = 124.0 mZ/s2 (250 kg)(124.o mZ/sz)
27 In N = (250 kg x 9.81 m/s2)(cos40°) — N=1879—1148= 731N PROBLEM 13.72 A 2.5lb collar is attached to a spring and slides without friction along a
circular rod in a vertical plane. The spring has an undeformed length of
4 in. and a constant k. The collar is at rest at C and is given a slight push
to get it moving, Knowing that the maximum velocity of the collar is achieved as it passes through point A, determine (a) the spring constant k,
(b) the maximum velocity of collar. SOLUTION (a) For maximum velocity, AL = 7.61577 — 4 = 3.61577 in. = 0.30131 ﬂ a, = 's' = 0 sin9 = (3/7.61577) 3m W = 2.51b
+1 213 = 0 = 0.30131 k(3/7.61577) — 2.5 = 0
k =21.0631b/ft k =21.11b/ﬁ4
(b) Put datum at C TC = (Vcig = (Vc)e = 0, (mg = —2.5(7/12) = —1.4583 (me = %(21.063)(0.30131)2 = 0.9561 Conservation of energy: 0 = 1(3) 3‘ — 14583 + 0,956] 2 32.2 VA = 3.597 VA = 3.60 ﬁ/s4 l PROBLEM 13.132 The two blocks shown are released from rest at time t= 0. Neglecting the
masses of the pulleys and the effect of friction in the pulleys and between
the blocks and the incline, determine (a) the velocity of block A at t = 0.5 s,
(b) the tension in the cable. Constraint: v14 / =3vB ,/ . o 20
4635) A: +/ x 20(0.5) sm30 — T(0.5) = 322 VA
B: +/ x 3T(0.5) — l6(0.5) sin30° = 1—6v3 = 16 VA \ 32.2 32.2 (3)
(«0‘ 5) Substituting for T(0.5) from the equation for A into the equation for B From A:
Im ulse dia rams
p g T(0.5)=5——0.62112vA
_ 0.4969 vA
o ,4)”; 15—1.8634vA—4——3—
>30
0 2.029 = 11
31%;) VA
VA = 5.4214
R (a) VA = 5.42 ft/s 7 30° 4
\ Mo 5) T(0.5) = 5 — 0.62112(5.4214) T=3.2653 lb
T=3.27 lb 4 PROBLEM 13.170 The coefﬁcient of restitution is 0.9 between the two 60mmdiameter
billiard balls A and B. Ball A is moving in the direction shown with a
velocity of 1 m/s when it strikes ball B, which is at rest. Knowing that
after impact B is moving in the x direction, determine (a) the angle 0,
25" "'m (b) the velocity of B after impact. SOLUTION (a) Since v; is in the xdirection and (assuming no friction), the common tangent between A and B at impact must be parallel to the
yaxis Thus tan9 = 250
150 — D 6 =tan‘1i0— = 70.20°
150 — 60 (b) Conservation of momentum in x(n) direction va 0059 + m(vB)n = m(vf4)n + mv;g
(1)008 (70.20) + 0 = (v2, )” + v;
0.3387 = M)" + (v23) Relative velocities in the n direction e = 0.9 (VA 0050 — (v3)n)e = v}; —(v:1)n (0.3387 — 0)(0.9) = v}; —(VZ1)H (1) + (2)
2v; = 0.3387(1.9) v; = 0.322 m/s 4 ...
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 Winter '09
 Bhattacharya
 Statics

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