Rectilinear MotionF=ma

1. Determine the steady-state angle a if the constant force

P is applied to the cart of mass M. The pendulum bob has

mass m and the rigid bar of length L has negligible mass.

Ignore all friction. Evaluate your expression for P = 0.

x

y

30o

N

mg (M+m)g

Txxx mamgTmaF a sinsin

0coscos0 a mgTFy

1

2

xxx amMgmMPmaF sin 3

mM

gmMPax

sin

a

costan

gmM

Pa

y

x

a

Rectilinear MotionF=ma

2. The spring of constant k=200 N/m is attached toboth the support and the 2-kg cylinder, which slidesfreely on the horizontal guide. If a constant 10-N forceis applied to the cylinder at time t=0 when the spring isundeformed and the system is at rest, determine thevelocity of the cylinder when x=40 mm. Also determinethe maximum displacement of the cylinder.

N

mg

xa

axmaF

x

x

F

xx

spring

1005

220010

040.0

0

22

040.0

00

5052

1

1005

xxv

dxxvdvdxavdv

v

x

Fspring

smv /49.0

mxorxxx 1.000505 2

the maximum displacement of the cylinder

(initialcondition)

222

50525052

xxvxxv

3. A heavy chain with a mass r per unit length is pulled bythe constant force P along a horizontal surface consistingof a smooth section and a rough section. The chain isinitially at rest on the rough surface with x=0. If thecoefficient of kinetic friction between the chain and therough surface is mk, determine the velocity v of the chainwhen x=L.The force P is greater than mkrgL in order toinitiate motion.

Ffrictionmkrg(Lx)

W1=rg(Lx)

L

xgg

L

Pa

LaxLgP

maFPmaF

kkx

xk

xfrictionxx

mmr

rrm

(Acceleration is function of displacement.)

gLP

vL

LggLL

L

P

L

xggxx

L

Pv

dxL

xgg

L

Pvdvdxavdv

kkk

L

kk

L

kk

v

vx

mr

mmr

mmr

mmr

2

222

1 2

0

22

000

Lx

W2=rgx

N2=rgxN1=rg(Lx)

Rectilinear MotionF=ma

Curvilinear MotionF=ma

4. The 2 kg collar is forced to move on the parabolic guide given by y=4x2, by anattached spring with an unstretched length of 1.5 m and a stiffness of k=20 N/m. It isknown that when the collar passes the position x=1 m, its velocity is v=2 m/s.Determine the reaction force acting on the collar from the guide and the totalacceleration of the collar.

Dimensions in “m”

k=20 N/m

y=4 – x2

m=2 kg

o

o

x

xdx

dy

975.468.2

3tan

435.6322tan1

m

dx

yd

dx

dy

59.52

211

2/32

2

2

2/32

r

NlkFsp 07.525.138.220 22

mg

+ t

+ n

N

Fsp

NNammgN

smv

amaFmgNmaF

n

kg

nnspnn

411.7975.46435.63sin07.52435.63cos

/7156.059.5

2sincos

2

222

r

1

N

2

2/744.33cossin

cossin

smam

Fga

vamaFmgmaF

t

sp

t

ttsptt

Total acceleration:222 /752.33 smaaa nn

5. The slotted arm AB drives particle C with mass 5 kg through the spiral groove described by the

equation 𝑟 =1.5

𝜃m, where is in radians. If the arm is driven at an angular velocity of ሶ𝜃 = 4

𝑟𝑎𝑑

𝑠and

angular acceleration ሷ𝜃 = 4𝑟𝑎𝑑

𝑠2, determine the forces acting on particle C by the slotted arm and spiral

groove when =120o. (Motion is in horizontal plane.)

Curvilinear MotionF=ma

𝒓𝜽 = 𝟏. 𝟓

eev

smrv

smr

r

rr

mr

rr

ererv

r

rad

dt

d

r

864.2367.1

/864.24716.0

/367.1

04716.0180

120

0

716.0

5.1180

1205.1

09.2

𝒓𝜽 = 𝟏. 𝟓

eev r

864.2367.1 +r

+

+r

+v

rv

v

+t

a

+n

oa 48.64367.1

864.2tan

a

+n

o52.25

o52.25

NR

2

22

2

/072.82

/594.7

/862.3

0

05.1

smrra

smrra

smr

rrrr

rrr

r

NRmaRmaF 23.2252.25sin075.42

NNmaNmaF rrr 075.4252.25cos N

Work&Energy6. The 0.2 kg slider moves freely along the fixed curvedrod from A to B in the vertical plane under the action ofthe constant 5 N tension in the cord. If the slider isreleased from rest at A, calculate its velocity v as itreaches B.

2

12121221 eegg VVVVTTU

1

Reference line

s/m.v...v..

mghmv..

48425081920202

152

2

11506505

2

2

2

2

2

2

length of cord

m...l 65025060 22

1

m.l 1502

1

2

Work&Energy7. The spring has an unstretched length of 625 mm. If thesystem is released from rest in the position shown,determine the speed v of the ball (a) when it has droppeda vertical distance of 250 mm and (b) when the rod hasrotated 35°.

2

12121221 eegg VVVVTTU

1

Ref.00 46762290622

600

250...tana

aa

Length of spring: m...lm.sin.l 919065065072102

6502 22

21

s/m.v......v

kxkxmghmv

9230625072102102

162509190210

2

125081944

2

10

2

1

2

1

2

10

2

222

2

2

1

2

21

2

2

1 2

l1

l2

Work&Energy

3

13131331 eegg VVVVTTU

1

Ref.0622.

Length of spring:

ml

l

ml

013.1

4.102cos65.065.0265.065.0

721.02

sin65.02

3

4.1290

22

3

1

s/m.v

.......sin..v

kxkxmghmghmv

4830

625072102102

162500131210

2

1250819441265081944

2

10

2

1

2

1

2

10

3

222

3

2

1

2

313

2

3

0412.

l1

l3

when the rod has rotated

35°

Impulse-Momentum8. The stationary 20-kg block is subjectedto the time-varying horizontal forcewhose magnitude P is shown in the plot.Note that the force is zero for all timesgreater than 3 s. Determine the time ts atwhich the block comes to rest.

mgN mgN Fy 00 x

y

P

W=mg

NFf=mN

t

mvFdtmv

sttP

021

3050

s.t

.t....

s

s

463

0352819204035232

352501500

The start time is s.t..tNP s 352819206050 m

s.352

35250 .

tP 50

Impulse-Momentum

9. Car B is initially stationary and is struck by car Amoving with initial speed v1 = 30 km/h. The carsmove together with speed v′ after the collision. If thetime duration of the collision is 0.1 s, determine thecommon final speed v′, (b) the magnitude R of theaverage force exerted by each car on the other carduring the impact. brakes are released during thecollision.

x

y

R

W=mBg

NB

h/kms/m.v

v.

vmmvm BAAA

205555

900180063

301800

B

NR

..RvmFdtvmt

BBB

50000

55559001000

(a)

(b)

9.

smvv

vmmvGGdtFx

/91.1901.0314.060014.0 22

2112

0

121212 eegg VVVVTTE

kJE

JTTE

18.17

74.1718160014.02

191.19044.0

2

1 22

12

Impulse-Momentum

11. The collar slides down the rod. The coefficient

of friction is mk = 0.35. The tension in the cable

attached to the collar is constant as 80 N and the

angle between the cable and the horizontal

remains the same as a throughout the motion. The

weight of the collar is 60 N. If the collar stops after

10 seconds it starts sliding down the rod with an

initial velocity of 5 m/s, what should a be?

Rectilinear MotionF=ma

+x

y

N

FfmkN

W=mg

a

T

m

a

cos28849.14

cos80426.4235.0

cos80426.42

cos8045cos60

045cos45cos

0

f

kf

y

F

NF

N

N

TWN

F

sin08.13cos578.4509.4

81.9

6045sin60sin80cos28849.14

45sinsin

x

x

xfxx

a

a

mamgTFmaF

dtdv

dtadvdt

dva

fv

v

xxx

x

10

0

0

5

sin08.13cos578.4509.4

0

(Acceleration is constant.)

cos

.sin.cos.sin.cos..

00095081357841008135784509450

o..

.

.

.tan 488545

891

4840

0330

85360

a

22914.0tan611.2sec

914.0tan611.2sec0sec009.5tan08.13578.4

0164.0tan773.4tan817.5

tan1sec836.0tan773.4tan817.6sec

2

2222

Curvilinear MotionF=ma

12. The horizontal platform is rotating with

a constant angular velocity of w=5 rad/s.

Two identical blocks, each with a mass of 2

kg are held in the smooth slot within the

platform by a pair of springs each with a

stiffness of k=250 N/m. If the unstretched

length of each spring is lo= 0.18 m, what

should angle be for steady state position?

Springs are attached to the platform at

point A.

(Consider only one block for the solution of

the problem.)

w= 5 rad/s (cst), m = 2 kg

2 rrmsinFcosN

maF

spring

rr

circular motion, steady state position

2

mrsinxkcosN

xkFspring

sin

cosxkN

cosxkcosFsinN

rrmcosFsinN

maF

spring

spring

2

w

r

0.3 m

x

y+r

N

Fspring

x2

sin2 rx

xy

N

Fspring

W

R

1

2

Putting into

2

mrsinxkcos

sin

cosxk

87.36

75.03.0

225.0

cos

sintan

225.0sin

45sin200,sin5045sin250

sin5218.0250sin250

sin18.0sin

18.0sin18.0

2

2

212

r

r

r

rrr

rr

mrrk

rxlengthinitiallengthfinalxxx

sinmrsincosxk

sinmrsinxkcosxk

sinmrsinxkcossin

cosxk

2

1

22

222

2

w

r

0.3 m

x

y

N

Fspring

x2

sin2 rx

Multiplying by sin