EGR 232 Dynamics: Homework Set 10 Fall 2013
Problem 12:42
As part of an outdoor display, a 12 lb model C of the earth is attached to wires AC and BC and revolves at constant speed v in the horizontal circle shown. Determine the range of the allowable values of v if both wires are to remain taut and if the tension in either of the wires is not to exceed
26 lb.
___________________________________________________________________________
Solution:
The maximum tension in either wire should be less than 26 lb.
To remain taut, the min tension in either wire should be greater than 0. The weight is 12 lb.
Examine two cases....Case 1) Max of 26 lbs.
Assume that wire AC has the max tension.
 
0
50
75 o o
T
T
2
1
W a
T
1 sin 50 o
T
1 sin 50 o

T
2

T
2 sin 75 o sin 75 o


W
12
 lb
0
Let T
1
= 26 lbs then
T
2

T
1 sin 50 o 
12

26sin 50 o 
12 sin 75 o sin 75 o

8.20
lb
Since T
2
is less than 26 lbs, we made the correct assumption. To find the velocity
  x ma n
T
1 cos 50 o 
T
2 cos 75 o m

2

 v

 m

T
1 cos 50 o 
T
2 cos 75 o


3 ft
12 lb
32.2
/
2
 lb o 
(8.2) cos 75 o v

12.3
/
 for the minimum speed case let T
2
= 0 then:
T
1

T
2 sin 75 o 
12 sin 50 o lb

0sin 75 o 
12 sin 50 o lb so then
 v
 m

T
1 cos 50 o 
T
2 cos 75 o


3 ft

15.6
lb f
12 lb
32.2
/
2

(15.6) cos 50 o 
(0) cos 75 o


8.98
ft s therefore the range of velocities is 8.98 ft/s to 12.3 ft/s

EGR 232 Dynamics: Homework Set 10 Fall 2013
Problem 12:58
A semicircular slot of 10 in radius is cut in a flat plate which rotates about the vertical
AD at a constant rate of 14 rad/s. A small, 0.8 lb block E is designed to slide in the slot as the plate rotates. Knowing that the coefficients of friction are μ s
= 0.35 and μ k
= 0.25, determine whether the block will slide in the slot if it is released in the position corresponding to a)
θ
= 80 degrees, b)
θ
= 40 degrees. Also determine the magnitude and the direction of the friction force exerted on the block immediately after it is released.
------------------------------------------------------------------------------------------------------------
Solution: a) at theta = 80 deg. assume that the block is in contact with the outer surface of the slot and has a tendency to slide down
W
N
F f m a
FBD AD a
N where:
 
 v
2


14 /
 
26 10sin
 so for 80 deg: a
N
 
2    
Newton’s 2 nd Law: Assume F f
can be whatever

3168 in s 2 value it needs to keep the block from sliding and it is at point of impending motion. W =0.8 lb

F y
 ma

N cos
 y

F f sin
 
W

0

N sin


F
N
N
N
 cos sin ma


N



F f
F f
F f sin
 
0.8
cos


 ma
N
 lb in s
2 in s
2 cos (0.8 / 386.4 / )(3168 / )
N sin
 
F f cos
 
6.56
lb
ρ
θ

Solve for N and F f
F f

0.8

N cos
 sin
 and
F f

6.56

N sin
 cos

0.8

N sin
 cos


6.56

N cos
 sin
0.8cos
 
N cos
2
 

6.56sin
N cos
N
2

  
 
N sin

N sin
6.56sin
2



6.56sin
0.8cos
 
0.8cos
2

 for 80 deg:
N

6.56sin 80 o

0.8cos80
o

6.32
lb and
F f

 o

1.927
lb sin 80 o
In order to not slide then the needed coef of friction would be.
 
F
N f 
1.927

0.305
6.32
Since the available coef of static friction is 0.35, the block should stay in place and the actual friction force will be the 1.927 lb.
------------------------------------------------------------------------------------------------------ b) Now try the same model when the angle drops down to 40 degrees.
 

14 /

26 10sin
 so for 80 deg: a
N
 
2     
3836 in s 2 then the F=ma equations become:

F y
 ma y


F
N
 ma
N


N
N cos sin




F
F f
N
N

N sin cos sin






F f
F f
F f f sin sin cos






W


0.8
7.942
0 lb cos cos


 ma
N lb in s 2 in s 2 )

F f

0.8

N cos
 sin
 and
F f

7.942

N sin
 cos

0.8

N sin
 cos


7.942
 cos
N
 sin
0.8cos
 
N cos 2
 

7.942sin
N cos
N
2

  
 
N sin

N sin 2
7.942sin
 

7.942sin
0.8cos
 
0.8cos
2

 for 80 deg:
N

7.942sin 40 o 
0.8cos 40 o 
4.492
lb and
F f

 o sin 40 o

6.598
lb
In order to not slide then the needed coef of friction would be.
 
F f 
N
6.598

1.469
4.492
Since the available coef of static friction is 0.35, the block sill slide in this position, therefore the actual friction force will be given by
F f
  k
N

0.25(4.492)

1.123
lb

EGR 232 Dynamics: Homework Set 10 Fall 2013
Problem 12:68
Rod OA rotates about O in a horizontal plane. The motion of the 5 lb collar B is defined by the relations r
=
10
and
 
 

, where r is expressed in feet, t in t+4 seconds and
θ
in radians. Determine the radial and transverse components of the force exerted on the collar when a) t = 1 s, b) t = 6 s.
___________________________________________________________________________
Solution:
F r
Given: r ma r
F
 ma r
 
  d 

  dt
 
  ma


F

 ma
 d 
  dt
    and
F
θ r
= r 
10 dt t+4 dr
= -20(t+4) -2 
-20
(t+4)
2 r  dr dt
= 60(t+4)
-3 
60
(t+4)
3
θ
F r a
θ m at t = 1 s
 
  d 

   dt d 
  dt
 
   2
     and r
=
10
1+4 r r

-20
(1+4)
2
60

(1+4)
3
 2 ft
 

0.8 /
0.48
/ 2 a r at t = 6 s



 




 d dt d  dt

 

     and r
= r 
10
6+4
-20
(6+4)
2
 1 ft
  0.2 / r 
60
(6+4)
3
 0.06
/ 2

then (at 1 second) a
  a e  (   2 ) e r
 ( r  a

  7.42
 
ˆ r

2 e r

3.2
ˆ

/ 2
 
 then
F r

5 lbf
(
32.2
/
2
 ft s
2
)
 


ˆ
 e

1.15
lb f
F


5 lbf
(
32.2
/
2
)(3.2
ft s
2
)

0.50
lb f then (at 6 second) a a

  a e
  3.94
ˆ r

 (   2 ) e r
 ( r 
2 e r

0.8
ˆ

/ 2
  then
  ˆ

ˆ
 e
F r

5 lbf
(
32.2
/
2
 ft s
2
)
 
0.612
lb f
F


5 lbf
(
32.2
/
2
 ft s
2
)
 
0.124
lb f

EGR 232 Dynamics: Homework Set 10 Fall 2013
Problem 12:71
A 100 g pin B slide along the slot in the rotating arm OC and along the slot DE which is cut in a fixed horizontal plate. Neglecting friction and knowing that rod OC rotates at the constant rate of

.
0
 12
, determine for any given value of
θ
a) the radial and transverse components of the resultant force F exerted on pin B, b) the forces P and Q exerted on pin B by rod OC and the wall of slot DE, respectively.
____________________________________________________________________________
Solution:
Since system is horizontal, the weight of the pin
75 o acts into the page. a
P
Q
50 o
T
1

EGR 232 Dynamics: Homework Set 10 Fall 2012
Problem 12:71
A 10 g pin B slide along the slot in the rotating arm OC and along the slot DE which is cut in a fixed horizontal plate. Neglecting friction and knowing that rod OC rotates at the constant rate of
  12
, determine for any given value of
θ
a) the radial and transverse components of the resultant force F exerted on pin B, b) the forces P and Q exerted on pin B by rod OC and the wall of slot DE, respectively.
____________________________________________________________________________
Solution: m = 0.01 kg
  12 rad s then
  0 r
θ
0.2 from trig: cos r 
 
0.2
cos
0.2 / r

 then r r and


 
    2 d  dt r 
  1
    2      2  0.4
 2  r 
     2  cos 3 
  2  therefore: r r







 2 
2.4sin
cos 2
0.2(12)cos

2

  cos 3 
2 
 acceleration is given as
  a e  (   2 ) e r
 ( r    ˆ

 
2.4 cos 2   28.8 sin 2 cos 3 


a


2.4cos
2   cos 3
28.8sin
2



0.2
cos 
(12) 2

 e r



0.2
cos 


2.4sin
 cos 2 


(12)


 e
 a 


2.4 cos 2   cos 3
28.8 sin 2



28.8
cos 


 e r




57.6 sin cos 2 



 e

Applying 2 nd Law:
  r ma r
F r
 ma r
F
 ma

F

 ma
 then
F

 ma


(.01)


57.6 sin cos
2

 

F
θ
θ

0.576
sin
 cos
2

F r a
θ m a r
F r
 ma r

(0.01)


2.4 cos
2
  cos 3
28.8sin
2



28.8
cos

  r ma r
Q cos
F
 ma

  ma r so: sin
  ma

P
0.024 cos
2
 
0.288sin
2 cos 3


θ
Q
 ma r cos


Q

0.024
1 cos
2



0.024 cos
2
  cos 4

0.288
sin 2 cos
4


0.288sin
2




0.288
1 cos
2

0.288
cos 2




0.288
sin 2 cos
4


Q

0.288
cos


 a
θ m

0.264
1 cos
2

P

Q sin
  ma


0.024 cos 2
 sin
 
0.288sin
3
 cos 4


0.288sin
 cos 2

 sin

0.576
cos 2

P

0.024
sin
 cos
2

 sin
3
0.288
cos
4



0.288
sin
 cos
2


0.576
sin
 cos
2

P

0.288
sin
3 cos
4



0.312
sin
 cos
2
 a r