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

/1.82128.030sin128.030cos12

2

22

2.0

Nothingchanges.

a)Inhorizontalplane

b)Inhorizontalplane

Referencelineforpart(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

Referenceline1

2

m=0.5kgv2 =0.8m/srestatposition1 andk=250N/mlo=200mm

h2=0.2sin15=0.052m

Lengthofcable

1

2

12121221 eegg VVVVTTU

ml 472.025.04.0 221

ml 32.025.02.0 222

Workbythecable

PPlPU 152.032.0472.021

222

21

222

22

2.025.0250212.045.0250

21052.081.95.08.05.0

21152.0

21

21

21152.0

P

xkxkmghmvP

P=52.07N

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.

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 of the spring.

m=1.2 kg determine vB and the maximum deflection of the spring.

1

2

3x

12121221 eegg VVVVTTU

a)

13131331 eegg VVVVTTU b)

Datum

smv

mgmvVT g

/396.9

05.4210

2

2212

mmm

xkmgmgVVV egg

2.540542.0

024000215.481.92.15.181.92.1

0215.45.10

2

2313

5. 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

/38.203205100010

01000201020210

1

1.0

0

3221

1.0

0

22121

12121221

12v1

6. The light rod is pivoted at O and carries the 2‐ and 4‐kg particles. If the rod is

released from rest at =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 =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.

1 2

Reference

3 (maximum compression)

22211121 0 egeg VVTVVTU

22 2214

2160sin45.081.9260sin3.081.94 BA vv

a) 222111 egeg VVTVVT

A

B

45.03.0 BBAA rvrv smvsrad B /16.1/58.2

2max

350002160sin45.081.9260sin3.081.94 xk

b) 333111 egeg VVTVVT

mmmx 07.1201207.0max

7. Two springs, each of stiffness k=1.2 kN/m, are of equal length and undeformed

when =0. If the mechanism is released from rest in the position =20o, determine

its angular velocity when =0. The massm 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 =0. mechanism is

released from rest when =20o, determine when =0.m =3 kg.

Referenceline

Ref.

22211121 0 egeg VVTVVTU

1

2

l1

l2

m

mlm

ml

056.0225.041.0

41.055sin25.020668.0287.0225.0

287.035sin25.02

1

2

1

1

Weignoretheequalandoppositepotentialenergychangesformasses(a)and(b).

a

b

c

221

21

211

056.01200210668.01200

2120cos25.081.93

21

2120cos25.0

E

kkmgE

25.025.03213

2

22 mgE

v

sradEE /22.421

8. The two right‐angle rods with attached spheres are released from rest in the

position = 0. If the system is observed momentarily come to rest when = 45°,

determine the spring constant k. The spring is unstretched when =0. Treat the

spheres as particles and neglect friction.

released from rest when = 0. System is momentarily stationary

when = 45°, determine the spring constant k. Spring is unstretched

when =0.

1

2

Reference

180mm

60mm

oa 56.7160180tan

22211121 0 egeg VVTVVTU

45o

71.5645=26.56o Fromconservationofenergy:

mx 219.006.056.26cos1897.022

2219.02156.26sin1897.081.92218.081.922 k

DeformationofspringinCase2:

k=155.7N/m

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

2.025.0120216.0

212.05.025.0120

215.081.96.0

2

222

222

222111

12121221

Newton’s Second Law (Normal&Tangential Coordinates)

NN.

......NvmmgFNmaF

m.

dxyd

dxdy

dxyd,x

dxdyxy

k.k.kxy:guideparabolicofEquation

springnn

/

/

x

84250925608196020250120

250401

1

4042

25050

22

23

2

2

232

2

2

0

2

22

FBD of slider (at point B)

+t

+n

Fspring mg

N

10. The system shown is in equilibrium when = 0°. Initially when block C is in a state of

rest at = 90°, it is given a slight push. Determine the velocity of the block as it passes

from the position where =37o. Neglect the mass of the light rod.