## Chapter 17 Section 4

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 ...
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