1. The member OA rotates about a horizontal axis through O with a

constant angular velocity ω = 3 rad/s. As it passes the position θ = 0,

a small block of mass m is placed on it at a radial distance r = 450 mm.

If the block is observed to slip at θ = 50°, determine the coefficient

of static friction µs between the block and the member. (3/54)

ω = 3 rad/s (cst), at θ = 0, a small block of mass m placed at radial distance r = 450 mm, block is observed to slip at θ = 50°, determine µs.

θ=50°

+θ +rW=mg

Ff=µsNN

ω =3 rad/s

θ

( )

( ) ( )

549.050cos81.9

)3(45.050tan

50cos50tan

50cos

50sin

50sin50cos50sin50cos

50cos

050sin,

50cos

050cos,02

222

22

22

=−=−=−

=

−=/−=/−/

==

−=−−==

=

=−=+==

∑

∑

g

rw

g

rwg

rwggrwmgmgm

mgNF

rwmmgFrrmmaF

mgN

mgNrrmmaF

s

ss

ssf

frr

µ

µµ

µµ

θ

θθθθ

&&&

&&&&

2. A 1500-kg car enters an S-curve at A with a speed of 96 km/h

with brakes applied to reduce the speed to 72 km/h at a uniform rate

in a distance of 90 m measured along the curve from A to B. The

radius of curvature of the path of the car at B is 180 m. Calculate the

total friction force exerted by the road on the tires at B. The road

at B lies in a horizontal plane. (3/64)

m = 1500-kg, vA = 96 km/h, vB = 72 km/h, s=90 m, ρΒ = 180 m, calculate the total friction force exerted by the road on the tires at B.

n

Ft-t

Fn

smh

kmv

sms

h

km

m

h

kmv

B

A

/2072

/67.266.3

96

3600

1

1

100096

=

=

==

=

=

at

an

+t

NFFF

NFmaF

NFmaF

smv

a

smaassavv

ntfrictiontotal

nnn

ttt

B

n

ttAB

422033302590

3330)22.2(1500,

2590)728.1(1500,

/22.2180

20

/728.1,)90(267.2620,)(2

2222

222

222

0

22

=+=+=

===

===

===

=−=−−=

∑∑

ρ

3. The two 0.2 kg sliders A and B move without friction in the

horizontal-plane circular slot. Determine the acceleration of each

slider and the normal reaction force exerted on each when the

system starts from rest in the position shown and is acted upon by

the 4-N force P. Also find the tension in the inextensible connecting

cord AB. (3/87)

mA = mB = 0.2 kg, determine the acceleration of each slider and the normal reaction force, system starts from rest, also find T

)0(

045sin

2.045cos

0 =

=+−=

=−=

∑∑

vrestfromstartssystem

TNmaF

aTPmaF

Ann

ttt

+t

n

n NA

T

A

�

�

B

2.045cos aTmaF ==∑ �

45°

+t

NB

45°

T

T

)(

045sin

2.045cos

BandAslidersforsametheisa

TNmaF

aTmaF

t

Bnn

ttt

=+−=

==

∑∑

NTNNTN

sma

aaaaT

aTaT

BA

t

tttt

tt

245sin,245sin

/10

44.0,2.042.0,2.045cos

2.0445cos,2.045cos4

2

====

=

=−==

−==−

�

�

4. The pin A is forced to move in the fixed parabolic slot by the moving vertical guide. The centerline of the slot is described by x = 10y2 [m]. When x = 0.1 m, the speed v of the vertical guide is 0.4 m/s and is not changing at this instant. Determine the magnitudes of the forces acting on

y v

magnitudes of the forces acting on the 250 g pin exerted by the vertical guide and the fixed plate. x = 10y2

A

x

x

x = 10y2 [m], x = 0.1 m, v = 0.4 m/s (cst), m = 250 g, determine the magnitudes of the forces acting on the pin exerted by the vertical guide and the fixed plate

x = 10y2

A

y v

vA

aA

vAx

vAy α = 26.57°[ ]m

dx

yd

dx

dy

mx

y

56.05.2

5.01,5.2

57.26,tan5.0

1.010

1.0

10

2/32

2

2

2/1

2/1

2/1

2/1

=−

+=−=

===

===

ρ

αα o

2

/2.0)2(104.0

10

smyyyx

yx

=⇒==

=

&&& x = 10y

x

x22

2

/4.0

20200

/2.0)2(104.0

smy

yay

yyyx

smyyyx

y −=−==

+==

=⇒==

&&&

&&&&&

&&&

NFNN

NN

NmaF

NFNFmaF

yy

xx

17.163.2

353.2894.057.26cos

)4.0(25.0)81.9(25.057.26cos

57.26sin,057.26sin0

1.0453.2

==

==

−=−=

==−==

∑∑

4342143421

+t

nW=mg

FN

26.57°

26.57°

x

y

5. The small object is placed on the inner surface of the conical dish at the radius shown. If the coefficient of static friction between the object and the conical surface is 0.30, for what range of angular velocities ω 30°

ω m

0.2 m

range of angular velocities ωabout the vertical axis will the block remain on the dish without slipping? Assume that speed changes are made slowly so that any angular acceleration may be neglected.

30°

µs = 0.30, for what range of angular velocities ω about the vertical axis will the block remain on the dish without slipping? Speed changes are made slowly angular acceleration may be neglected.

30°

ω m

0.2 m

+t

ω

r = 0.2 m

n

0.2 m

ω

W=mg

N

Ff

30°

y

n

30°30°

{( )

( )30sin30cos

,30sin30cos

030sin30cos0

µµ

µµ

±==±

==±−=∑

mgNmgN

NFFmgNF f

N

fy

{

( )30cos30sin

,30cos30sin

30cos30sin

22

2

µω

ωµ

ωµ

mm

m

mrNmrN

mrmaFNmaF n

N

fnn

==

===∑

µs = 0.30, for what range of angular velocities ω about the vertical axis will the block remain on the dish without slipping? Speed changes are made slowly angular acceleration may be neglected.

30°

ω m

0.2 m

30sin30cos µ±=

mgN

2

µωmr

=

0.2 m

ω

W=mg

N

Ff

30°

y

n

30°30°

+t

ω

r = 0.2 m

n

30sin30cos µ±=N

30cos30sin µm=

sradsrad

sradsrad

r

g

rmgm

/21.7/41.3

/21.7/41.3

30sin30cos

30cos30sin

30cos30sin30sin30cos

21

2

2

<<

==

±=

/=±/

ω

ωω

µµ

ω

µω

µ

m

m

6. The 2 kg collar is forced to move on the parabolic guide given

by y = 4 – x2, by an attached spring with an unstretched length of

1.5 m and a stiffness of k = 20 N/m.

It is known that when

the collar passes the

position x = 1 m, its Dimensions in “m”

velocity is v = 2 m/s.

Determine the reaction

force acting on the

collar from the guide

and the total

acceleration of the

collar.

Dimensions in “m”

m = 2 kg, y = 4 – x2, lo = 1.5 m, k = 20 N/m, when x = 1m, v = 2 m/s. Determine the reaction force acting on the collar from the guide and the total acceleration of the collar.

Dimensions in “m”

+t

n

W=mg

N

Fspring

θ

θ

2.8 m

3 m

4.104 m

α

γ

β

o02.43

93.03

8.2tan

=

==

α

α

2

43.63

tan2

314

4

2

2

2

−=

=

=−=

=−=

−=

dx

yd

dx

dy

m

xy

oθ

θ

o57.2690 =−= θβ 2.8 mdx

o

o

45.1657.2602.43

57.2690

=−=−=

=−=

βαγ

θβ

[ ] s

n smv

am /716.0,59.52

)2(1 22/32

===−−+

=ρ

ρ

( )

NNN

N

maFmgNmaF

NxxkxkF

nspringnn

spring

403.7,431.1747.14776.8

)716.0(245.16sin08.5243.63cos)81.9(2

sincos

08.52)5.1104.4(20

62.19

12

−==−+−

=−+−

=−+−=

=−=−=∆=

∑321

γθ

m = 2 kg, y = 4 – x2, lo = 1.5 m, k = 20 N/m, when x = 1m, v = 2 m/s. Determine the reaction force acting on the collar from the guide and the total acceleration of the collar.

Dimensions in “m”

+t

n

W=mg

N

Fspring

θ

θ

2.8 m

3 m

4.104 m

α

γ

β

o02.43

93.03

8.2tan

=

==

α

αo43.63=θ

o

o

45.16

57.26

=

=

γ

β

2.8 m

22222

2

62.19

/76.3375.33716.0

/75.33,255.1795.49

243.63sin)81.9(245.16cos08.52

sincos

smaaa

smaa

a

mamgFmaF

tn

tt

t

tspringtt

=+=+=

==+

=−+

=−+=∑321

θγ

7. The slotted arm is rotating

at a rate of =5 rad/s and

=2 rad/s2 when θ = 90o.

Determine the normal force

the slotted arm must exert on

the 0.5 kg particle if the

θ&

θ&&

the 0.5 kg particle if the

particle is confined to move

along the path defined by the

horizontal hyperbolic spiral

rθ = 0.2 m.

=5 rad/s, =2 rad/s2, θ = 90° , m = 0.5 kg, rθ = 0.2 m, determinethe normal force the slotted arm must exert on the particleθ& θ&&

2/2,/5,2

90 sradsradrad ==== θθπ

θ &&&o

/405.0

0)5(1273.02

,0

1273.0,2.02

,2.0

smr

rrr

mrrr

−=

=+

=+

==

=

&&&&

&

&&&π

θθ

πθ

+t

+r

+θ

α

R

Nvvr

vθα

637.0

405.0arctan

=α

2/418.2

0)2)(1273.0()5)(405.0(22

0

smr

r

rrrr

=

=+−+

=+++

&&

&&

&&&&&&&&

π

θθθθ

n

α

θ& θ&&,

smeev

smrvsmrvevevv

r

rrr

/637.0405.0

/637.0)5(1273.0,/405.0,

θ

θθθ θvvv

&&vvv

+−=

===−==+=

o49.32=

α

=5 rad/s, =2 rad/s2, θ = 90° , m = 0.5 kg, rθ = 0.2 m, determinethe normal force the slotted arm must exert on the particleθ& θ&&

2/2,/5,2

90 sradsradrad ==== θθπ

θ &&&o

2/418.2

/405.0,1273.0

smr

smrmr

=

−==

&&

&

+t

+r

+θ

α

R

Nvvr

vθ

α

2

22

/765.0

)5)(1273.0(418.2

sma

rra

eaeaa

r

r

rr

−=

−=−=

+= θθ

θ&&&

vvv

637.0

405.0arctan

=α

n

α

θ& θ&&,2/798.3

)2)(1273.0()5)(405.0(22

sma

rra

r

−=

+−=+=

θ

θ θθ &&&&o49.32=

α

NR

NRmaF

NN

NmaF rr

655.1

)798.3(5.049.32sin

453.0

)765.0(5.049.32cos

−=

−=−=

−=

−==

∑

∑

θθ

8. Cylinder C which has a mass of 2 kg, can move freely within the

slot of arm OA as it rolls along the horizontal surface. Friction is

neglected. As arm OA rotates in the vertical plane with a

counterclockwise angular velocity of ω = 5 rad/s, the magnitude of

this velocity decreases 2.5 rad/s per second. Determine the forces

acting on cylinder C for the instant represented. Treat cylinder C as

a particle.a particle.

O

0.45 m

8

15

C

θθ&

A

m = 2 kg, ω = 5 rad/s, α = -2.5 rad/s2 . Determine the forces acting on cylinder C.

O

θ

0.45 m

8

15

C

θθ&