Chapter 17
S e c ti o n 4
Pa scal’s Prin ciple a n d A r c h i m e d e s ’ Prin ciple
Ch 17 - Sec 4 - Pro 19 Page 1 / 2 A hyd raulic p r e s s c o n t a in i n g liquid g l y c e ri n e h a s a p i s to n o f sma ll c r o s s s e c ti o n a l a r e a a u s e d to e x e r t a f o rc e f o n t h e e n c l o se d l i q u id . A c o n n e c t in g p i p e l e a d s to a l a rg e r p i s to n o f c r o s s s e c ti o n a l a r e a A ( f i g u r e 17-4-1 9).
(A)
W h a t f o rc e F will t h e l a rg e r p i s to n s u s ta i n ? Figure 17-4-19
S o l u t io n : U s e t h e s t a n d a r d varia ble list.
(3)
(B)
I f t h e s m a l le r p i s to n h a s a d i a m e t er o f
p i s to n o n e o f
a n d t h e l a rg e r
, w h a t w e i g h t f o n t h e s m a l le r p i s to n will
s u p p o r t F = 2 t o n s o n t h e l a rg e p i s to n ? S o l u t io n : U s e t h e s t a n d a r d varia ble l i s t a n d t h e f o ll o w i n g l o g i ca l sequence of even ts.
Ch. 17 Pg. 251
Ch 17 - Sec 4 - Pro 19 Page 2 / 2
(6)
(9)
Ch 17 - Sec 4 - Pro 20 Parts a 6c Page 1 / 3 A c u b i c a l o b j e c t (c o p p e r , z i n c , g a l l iu m e t c . . . ) o f d i m e n s io n s L = 2 f t o n a s i d e ; W = 1 0 0 0 l b ( i n a v a c u u m ) is s u s p e n d e d b y a r o p e in a n o p e n t a n k o f w a t e r o f den sity
(mass density) as
in f i g u r e 1 7 - 4 - 2 0 a.
(A)
F i n d t h e d o w n w a r d f o rc e e x e r te d b y t h e w a t e r a n d t h e a t m o s p h e r e o n
the top of the object of area . S o l u t io n : U s e t h e s t a n d a r d varia ble s e t a n d t h e f o ll o w i n g t e r m s .
Pressure on the top face at a depth d = ½L due to water.
(5)
Ch. 17 Pg. 252
Ch 17 - Sec 4 - Pro 20 Parts a Page 2 / 3
Figure 17-4-20a
ü( 9 )
(B)
F i n d t h e f o rc e o n t h e b o t t o m o f t h e obje ct.
S o l u t io n : H e r e
.
Pressure on bottom face due to water.
ü( 1 6 )
(C)
F i n d t h e t e n s io n T in t h e r o p e . S o l u t io n : A n ana lysis o f t h e f o rc e o f t h e u n a c c e l er a te d c u b e y i e ld s ( s e e f i g u r e 1 7 - 4 - 2 0 a)
Ch. 17 Pg. 253
Ch 17 - Sec 4 - Pro 20 Part c Page 3 / 3
Figure 17-4-20
(Upward)
Ch 17 - Sec 4 - Pro 21 Parts a 6b Page 1 / 2
(A)
W h a t is t h e m i n i m u m a r e a o f a b l o c k o f i c e o f dep th d = 1 f t = 0 . 3 m t h i ck f lo a t in g o n w a t e r t h a t will hold u p a n auto mo bile o f m a s s m = 1 1 0 0 k g ( w e i g h i n g W = 2 5 0 0 l b )? S o l u t io n : I n this c a s e t h e i c e is bare ly com plete ly s u b m e r g e d . S o that, Figure 17-4-21a
V o lu m e o f i c e
;
v o l um e o f w a t e r d i s p l a ce d
;
car weight
Ch. 17 Pg. 254
Ch 17 - Sec 4 - Pro 21 Part a Page 2 / 2
ice weight
;
B = buoyant force
D u e to d i s p la c e d w a t e r .
ü( 7 )
(B)
D o e s it m a t t e r w h e r e t h e c a r is p l a c e d o n t h e i c e ?
S o l u t io n : Y e s . Its c e n t e r o f m a s s sho uld b e p l a c e d o v e r t h e c e n t e r o f m a s s o f t h e i c e . E l s e , t h e i c e will tilt a n d t h e c a r will slip off o f it. E v e n t h e t o t al s u r fa c e a r e a will c h a n g e . P r i m a r y b o o k s o l u t io n ( H R ) : ... a f te r t h e i c e tilts t h e r e will b e l e s s i c e s u b m e r g e d a n d t h e r e f o r e a n e w b u o y a n t f o rc e r e s u l t s . I t s e em s t h a t t h e H R s o l u ti o n v i o l at e s e q u a t io n s 2 – 5 .
Ch. 17 Pg. 255
Ch 17 - Sec 4 - Pro 22 Page 1 / 2 Three chemical engineers each weighing W = 80 lb (mass = 36kg) m a k e a r e c ta n g u l a r l o g raft b y l a s h in g t o g e th e r l o g s o f d i a m e t er d = 1 f t a n d leng th l = 6 f t . T a k e t h e r e la t iv e den sity o f w o o d to b e D = 0 . 8 0 . H o w m a n y l o g s will b e n e e d e d to k e e p t h e m afloat? S o l u t io n : S e e f i g u r e 1 7 - 4 - 2 2 a n d u s e t h e f o ll o w i n g varia ble set. Figure 17-4-22
V = V o lu m e o f w o o d = v o l um e o f w a t e r d i s p l a ce d .
(5)
T h e w e i g h t den sity o f w a t e r is =
(8)
Ch. 17 Pg. 256
Ch 17 - Sec 4 - Pro 22 Page 2 / 2
T h u s , t h e t o t al v o lu m e o f t h e w o o d is (9) T h e v o lu m e p e r l o g is
# of Logs =
Ch 17 - Sec 4 - Pro 23 Parts a 6b Page 1 / 2 A block of wood
floats
in w a t e r
with
s u b m e r g e d . I n oil it h a s 0 . 9 0 o f its v o lu m e
2/3
of
its v o lu m e
submerged.
(A)
F i n d t h e den sity o f t h e w o o d . S o l u t io n : S e e f i g u r e 1 7 - 4 - 2 3 . Figure 17-4-23 V o l um e o f w a t e r d i s p la c e d is
Block weight =
T h u s , w o o d den sity
(5)
Ch. 17 Pg. 257
Ch 17 - Sec 4 - Pro 23 Part a Page 2 / 2
ü( 7 )
(B)
F i n d t h e D den sity o f t h e oil. S o l u t io n : V = v o l um e o f o i l d i s p l a ce d , D = o i l d e n s i ty . Den sity of wood block.
(10)
Ch 17 - Sec 4 - Pro 24 Parts a 6b Page 1 / 6 A b l o c k o f w o o d h a s a m a s s o f m = 3 . 6 7 k g a n d a r e la t iv e den sity o f . I t is to b e l o a d e d with l e a d ( P b ) s o t h a t it will f lo a t in w a t e r with P b is
o f its v o lu m e
( b l o ck v o l u m e ) i m m e r s ed .
T h e den sity o f
.
(A)
W h a t w e i g h t o f P b is n e e d e d if t h e l e a d is o n t o p o f t h e w o o d ? S o l u t io n : S e e f i g u r e 1 7 - 4 - 2 4 a a n d u s e t h e s t a n d a r d varia ble set. Pb weight Buoyant force
; ;
block weight Total weight (2)
Ch. 17 Pg. 258
Ch 17 - Sec 4 - Pro 24 Part a Page 2 / 6
Figure 17-4-24a
(4)
= v o lu m e o f w a t e r d i s p la c e d .
T h e r e fo r e , (8) So,
(10)
(B)
W h a t is t h e w e i g h t o f t h e l e a d if it is a t ta c h e d b e l o w t h e w o o d ? S o l u t io n : S e e f i g u r e 1 7 - 4 - 2 4 b . Figure 17-4-24b
Block buoyant force Buoyant force of Pb Total buoyant force
ü( 2 )
Ch. 17 Pg. 259
Ch 17 - Sec 4 - Pro 24 Part b Page 3 / 6
ü( 5 )
d i s p la c e d b y P b = u n kn o w n
.
(10)
Re call t h a t
b e c a u s e only 9 0 % o f b l o c k is i m m e r s ed . T h e r e fo r e , ( 1 1 ) b e c o m e s
(16)
Ch. 17 Pg. 260
Ch 17 - Sec 4 - Pro 24 Part b Page 4 / 6
(C)
F o r this s e c ti o n s e e g r a p h 1 7 - 4 - 2 4 f o r a d e e p e r ana lysis o f o f e q n ( 1 4 ) . E v e n if P b w a s a m a t e r ia l o t h e r t h a n l e a d , s a y
wood or tin, then (18) Case 1:
Pb on top of the Block.
H e r e P b is s o d e n s e a n d c o m p a c t it d i s p la c e s v e r y little w a t e r a n d c o n t ri b u t es n o b u o y a n c y ,
.
Case 2:
(22)
The refore (31) becomes (25) Case 3:
b e c o m e s l a rg e o r m o r e P b is n e e d e d to pull t h e b l o c k d o w n . O n t h e f o ll o w i n g g r a p h o f ( 1 8 ) l e t weight of s om e ma terial.
Ch. 17 Pg. 261
Ch 17 - Sec 4 - Pro 24 Part c Page 5 / 6
Den sity of s om e ma terial. will s t a r t a t 1 . 0 5 o r
. Mg . 36N.
ü( 3 3 )
S e e G r a p h 1 7 - 4 - 2 4 b e lo w .
Ch. 17 Pg. 262
Ch 17 - Sec 4 - Pro 24 Part c Page 6 / 6
Graph 17-4-24
` Tab le 1 7 - 4 - 2 4
Ch. 17 Pg. 263
Ch 17 - Sec 4 - Pro 25 Page 1 / 2 A s su m e t h e den sity o f b r a s s we ights to b e air to b e
and that of
. W h a t p e r c e n t e r ro r a r is e s f ro m n e g l e c ti n g t h e
b u o y a n c y o f air in w e i g h i n g a n o b j e c t o f m a s s m a n d den sity D o n a balance beam? S o l u t io n : S e e f i g u r e 1 7 - 4 - 2 5 a. I n a v a c u u m w e h a v e , Figure 17-4-25a
When balanced
.
No Buoya ncy. I n a i r : S e e f i g u r e 1 7 - 4 - 2 5 b . T h e only d i ff e re n c e s a n d t h e c a u s e o f t h e e r ro r is t h e d i ff e re n c e in b u o y a n c i es o f t h e mass and the brass.
Figure 17-4-25b
(5)
T h e e r ro r p e r c e n t is d e f in e d b y ( a ft e r t r ia l and error)
(8)
S i n c e in a v a c u u m t h e b a l a n c e g i v e s
ü( 1 1 )
Ch. 17 Pg. 264
Ch 17 - Sec 4 - Pro 25 Page 2 / 2
T h e r e fo r e ,
(14)
Ch 17 - Sec 4 - Pro 26 Page 1 / 2 A h o l l o w s p h e r ic a l F e ( i ro n ) shell floats a l m o s t com plete ly s u b m e r g e d in w a t e r . I f t h e o u t e r d i a m e t er is r e la t iv e den sity o f F e is
, f in d t h e i n n e r d i a m e t er
S o l u t io n : U s e t h e s t a n d a r d varia ble set. W =
= weight of sphere in air Buoyant force v o l um e o f F e = u n k n o w n v o l um e o f w a t e r d i s p l a ce d . = outer diam eter. Ch. 17 Pg. 265
and the .
Ch 17 - Sec 4 - Pro 26 Page 2 / 2
(3)
N o w w e c a n f in d t h e i n n e r d i a m e t e r .
ü( 8 )
I n n e r d i a m e t er in m e t e r s . As a check:
(13)
T h u s , t h e c h e c k is o k . Ch. 17 Pg. 266
Ch 17 - Sec 4 - Pro 27 Page 1 / 2 A n i r o n (F e ) c a s ti n g c o n t a in i n g a n u m b e r o f c a v i ti e s w e i g h s a n d h a s a mass, respectively, of
in air a n d in w a t e r . W h a t is t h e v o lu m e o f t h e
in t h e c a s ti n g ? I f t h e den sity o f w a t e r is D den sity o f F e to b e
t h e n a s su m e t h a t t h e
c a v i ti e s r e la t iv e
.
S o l u t io n : L e t ’ s f i r s t d e a l with t h e F e in air a n d t h e n in w a t e r . In air: (1) In water:
d o to t h e b u o y a n t f o rc e B . (4) V = v o l um e o f w a t e r d i s p l a ce d b y F e . T h e r e fo r e , (5)
v o lu m e o f t h e c a v i ti e s = Thu s,
(8)
O r in o t h e r u n i t s ,
Ch. 17 Pg. 267
Ch 17 - Sec 4 - Pro 27 Page 2 / 2
Note:
(11)
S o c o n t in u i n g f ro m ( 1 0 ) , t h e v o lu m e o f t h e c a v i ti e s is
Ch 17 - Sec 4 - Pro 28 Parts a 6b Page 1 / 3 A solid c u b e o f c a r b o n g r a p h i t e is f lo a t in g o n H g a n d h a s o n e q u a r t e r o f its v o lu m e s u b m e r g e d . I f e n o u g h w a t e r is a d d e d to c o v e r t h e cube then
(A)
w h a t f ra c t io n f will rem ain i m m e r s ed in t h e m e r c u r y ? T h e r e la t iv e
den sity o f m e r c u r y with r e s p ec t to w a t e r is
.
S o l u t io n : S e e f i g u r e 1 7 - 4 - 2 8 a, t h e c a s e w i t h o u t w a t e r . Figure 17-4-28a
V o lu m e o f c u b e = Displaced Hg Buoyancy on cube Carbon grap hite cube
ü( 6 ) L a t e r w e will u s e this to f in d
.
S o l e t ’ s c o n t in u e . Ch. 17 Pg. 268
Ch 17 - Sec 4 - Pro 28 Page 2 / 3
(8)
N o w l e t ’ s a d d t h e w a t e r into t h e s ys te m .
Figure 17-4-28b
Total buoyant force on cube Buoyancy due to water
(16)
So now,
(21)
N o w simp ly r e a rr a n g e ( 2 3 ) to g e t t h e v o lu m e o f H g d i s p la c e d .
Ch. 17 Pg. 269
Ch 17 - Sec 4 - Pro 28 Part a Page 3 / 3
(24) Let
(28)
Th is is t h e v o lu m e o f t h e c u b e in m e r c u r y o r a b o u t 20% or
(B)
.
Does the answer depend on the shape of the body?
S o l u t io n : N o . On ly t h e w e i g h t a n d v o lu m e o f t h e c a r b o n g r a p h i t e c u b e i n fl u e n c e t h e o u tc o m e o f t h e e q u a t i o n s .
Ch. 17 Pg. 270