Post on 11-Aug-2018
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&