Post on 25-Feb-2018
7/25/2019 Ch6 Solved Problems
1/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
1
7/25/2019 Ch6 Solved Problems
2/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
2
7/25/2019 Ch6 Solved Problems
3/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
3
7/25/2019 Ch6 Solved Problems
4/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
4
7/25/2019 Ch6 Solved Problems
5/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
5
7/25/2019 Ch6 Solved Problems
6/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
6
7/25/2019 Ch6 Solved Problems
7/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
7
7/25/2019 Ch6 Solved Problems
8/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
8
7/25/2019 Ch6 Solved Problems
9/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
9
7/25/2019 Ch6 Solved Problems
10/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
10
7/25/2019 Ch6 Solved Problems
11/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
11
at the highest point on the path. ote that the gravitational force has the same magnitude
and direction at each poi nt on the circular path.The tension force varies in magnitude at
different points and isalways directed toward the center of the path.
(c) Atthetop ofthecircle,= mv2/r = T + Fg,or
T=m---F=m---mg=m -- g)
r g r r
20
= (0.275 kg)[(5. mfs? - 9.80 m/s2]= 16.05 Nl0.850 m
(d) At the bottom of the circle,Fe = mv2/r = T - Fg = T - mg, and solving for the
speed gives
v2 = ;,(T-mg)= r(- g) andIf thestring is at thebreaking point at thebottom of the circle, then T=22.5N, and the
speed of the object at thispoint must be
v = (0.850 m)22.5 N
( 0.275 kg- 9.80 m/s2
) = 17.82 m/s l
48. In a home laundry dryer, a cylindrica l tub containing wet
clothes is rotated steadily about a horizontalaxis as shown
in Figure P6.48.So that the clothes wiH dry uniformly, they
are made to tumble. T he rate of rotation of the smooth-
FigureP6.48
walled tub is chosen o that a small piece of cloth will lose
contact with the mb when the cloth is at an angle of() =68.0 above the horizontal. If the radius of the tub is r =
0.330 m, what rate of revolution is needed?
7/25/2019 Ch6 Solved Problems
12/12
Chapter 6 Solution of selected problems Dr. Khaled Abdelsabour Elsayed
12
P6.48 When the cloth is at a lower angle 9,the radial component ofLF = ma reads. mv
2
n+mgsmO=-r
p
mg
pC:U in68
mgcos68
ANS FIG.P6.48
v2
At8 = 68.0, the normal force drops to zero and g sin 68 =-.r
The rate of revolution is
angu lar speed = (1.73 m/s)(1rev)( t1Cr ))=I 0.835 rev/s I = 50.1 rev/tnin
2JCr 2JC 0.33 m