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1. The two small 0.2-kg sliders are connected by a light rigid bar and are constrained to
move without friction in the circular slot. The force P=12 N is constant in magnitude and
direction and is applied to the moving slider A. The system starts from rest in the
position shown. Determine the speed of slider A as it passes the initial position of slider
B if (a) the circular track lies in a horizontal plane and if (b) the circular track lies in a
vertical plane. The value of R is 0.8 m.
12121221 eegg VVVVTTU
smv
vm
/1.8
2
128.030sin128.030cos12
2
2
2
2.0
Nothing changes.
a) In horizontal plane
b) In vertical plane
Reference line for part (b)
2. Determine the constant force P required to cause the 0.5 kg slider to have a speed v2 = 0.8 m/s
at position 2. The slider starts from rest at position 1 and the unstretched length of the spring of
modulus k = 250 N/m is 200 mm. Neglect friction.
l1 l2
Reference line1
2
m=0.5 kg v2 = 0.8 m/s rest at position 1 and k = 250 N/m lo=200 mm
h2=0.2sin15=0.052 m
Length of cable
1
2
12121221 eegg VVVVTTU
ml 472.025.04.0 22
1
ml 32.025.02.0 22
2
Work by the cable
PPlPU 152.032.0472.021
222
2
1
2
22
2
2
2.025.02502
12.045.0250
2
1052.081.95.08.05.0
2
1152.0
2
1
2
1
2
1152.0
P
xkxkmghmvP
P=52.07 N
3. The 2 kg collar is released from rest at A and
slides down the inclined fixed rod in the vertical
plane. The coefficient of kinetic friction is 0.4.
calculate (a) the velocity v of the collar as it strikes
the spring and (b) the maximum deflection x of the
spring.
FBD of Collar
Motion
x
y
mg
N
Ff=mkN
N...NFN.N
cosmgNF
kf
y
924381930819
0600
m
(a) the velocity v of the collar as it strikes the spring
N.Ff 9243
(b) the maximum deflection x of the spring.
1
2
3Reference
12121221 eegg VVVVTTU
s/m.v
sin..v..
5562
6050819222
1509243
2
2
2
23232332 eegg VVVVTTU
0533606613800
16002
160819255622
2
19243
2
22
.x.x
xsinx..x.
x
m.x.
....x max, 09890
080
09890
8002
533680040661306613 2
21
4. The 1.2 kg slider is released from rest in position A and slides without friction
along the vertical-plane guide shown. Determine (a) the speed vB of the slider as it
passes position B and (b) the maximum deflection d of the spring.
m=1.2 kg determine vB and the maximum deflection d of the spring.
1
2
3
d
12121221 eegg VVVVTTU
a)
13131331 eegg VVVVTTU
b)
Datum
smv
mgmvVT g
/396.9
05.42
10
2
2
212
mmm
kmgmgVVV egg
2.540542.0
0240002
15.481.92.15.181.92.1
02
15.45.10
2
2
313
d
d
d
5. The light rod is pivoted at O and carries the 2- and 4-kg particles. If the rod is
released from rest at q =60o and swings in the vertical plane, calculate (a) the
velocity v of the 2 kg particle just before it hits the spring in the dashed position and
(b) the maximum compression x of the spring. Assume that x is small so that the
position of the rod when the spring is compressed is essentially horizontal.
released from rest at q =60o (a) the velocity v of the 2 kg particle just before it hits the spring in the dashed position and
(b) the maximum compression x of the spring.
12
Reference
3 (maximum compression)
22211121 0 egeg VVTVVTU
22 22
14
2
160sin45.081.9260sin3.081.94 BA vv
a) 222111 egeg VVTVVT
A
B
45.03.0 BBAA rvrv smvsrad B /16.1/58.2
2
max
350002
160sin45.081.9260sin3.081.94 xk
b) 333111 egeg VVTVVT
mmmx 07.1201207.0max
6. Two springs, each of stiffness k=1.2 kN/m, are of equal length and undeformed
when q =0. If the mechanism is released from rest in the position q =20o, determine
its angular velocity ሶ𝜃 when q =0. The mass m of each sphere is 3 kg. Treat the spheres
as particles and neglect the masses of the light rods and springs.
k=1.2 kN/m, are of equal length and undeformed when q =0. mechanism is
released from rest when q =20o, determine ሶ𝜽 when q =0. m =3 kg.
Reference line
Ref.
22211121 0 egeg VVTVVTU
1
2
l1
l2
m
ml
m
ml
056.0225.041.0
41.055sin25.02
0668.0287.0225.0
287.035sin25.02
1
2
1
1
d
d
We ignore the equal and opposite potential energy changes for masses (a) and (b).
a
b
c
22
1
2
1
2
11
056.012002
10668.01200
2
120cos25.081.93
2
1
2
120cos25.0
E
kkmgE dd
25.025.032
13
2
2
2 mgE
v
q
sradEE /22.421 q
7. The two right-angle rods with attached spheres are released from rest in the
position q = 0. If the system is observed momentarily come to rest when q = 45°,
determine the spring constant k. The spring is unstretched when q =0. Treat the
spheres as particles and neglect friction.
released from rest when q = 0. System is momentarily stationary
when q = 45°, determine the spring constant k. Spring is unstretched
when q =0.
1
2
Reference
180 mm
60 mm
a
oa 56.7160
180tan
a
22211121 0 egeg VVTVVTU
45oa
71.5645=26.56o From conservation of energy:
mx 219.006.056.26cos1897.022
2219.02
156.26sin1897.081.92218.081.922 k
Deformation of spring in Case 2:
k=155.7 N/m
8. The light quarter-circular rod is pivoted at O and carries the 3 kg particle. When the
system is released from rest at the position (1), it moves to position (2) under the action
of the constant force F=250 N applied to the cable. The spring of stiffness k=1500 N/m has
an unstretched length of 200 mm. Calculate the speed of the particle and the angular
velocity of the circular rod as the particle passes the position (2).
9. The 0.6-kg slider is released from rest at A and slides down the smooth parabolic
guide (which lies in a vertical plane) under the influence of its own weight and of the
spring of constant 120 N/m. Determine the speed of the slider as it passes point B
and the corresponding normal force exerted on it by the guide. The unstretched
length of the spring is 200 mm.
Work and Energy Principle
1
Datum2
smv
v
VVTVVT
VVVVTTU
egeg
eegg
/92.5
2.025.01202
16.0
2
12.05.025.0120
2
15.081.96.0
2
22
2
222
222111
12121221
Newton’s Second Law (Normal&Tangential Coordinates)
NN
.
......N
vmmgFNmaF
m.
dx
yd
dx
dy
dx
yd,x
dx
dyxy
k.k.kxy:guideparabolicofEquation
springnn
/
/
x
84
250
925608196020250120
2504
011
4042
25050
22
23
2
2
232
2
2
0
2
22
FBD of slider (at point B)
+t
+n
Fspring mg
N
10. Calculate the horizontal velocity v with which the 20 kg carriage must strike the spring
in order to compress it a maximum of 100 mm. The spring is known as a “hardening”
spring, since its stiffness increases with deflection as shown in the accompanying graph.
smvxxv
dxxxvVT
VVVVTTU
e
eegg
/38.203
205100010
010002010202
10
1
1.0
0
322
1
1.0
0
22
121
12121221
12
v1
