presentation day1

“WATER WORLD”

Essential Question Buoyant Force

Problem Solving Activity Conclusions & Assignment

Mystery Revealed

What makes one stand out?

http://www.youtube.com/watch?v=s-afOwI0NQU

TITANICLength over-all: 882.5 ft

Gross tonnage: 46, 329 tons

Net tonnage: 24, 900 tons

Depth: 59.5 ft

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

FA= FB

Horizontal Forces

Thus, Σ Fx =FA+ (-FB) =0

Vertical Forces

P1 ⟩ P2

Then, F1⟩ F2

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

Thu s, Σ Fy =F1+ (-F2) =0

F1+ (-F2) =FB

FB = F1 - F2

OR

BUOYANT FORCE

“ The Net Upward Force ex erted by a fluid on a submerged or immersed object.”

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

FB = P1A - P2A = ( P1 – P2 ) A

FB = P1A - P2A

FB = F1 - F2

FB = P1A - P2A

OR

FB = ( P1 – P2 ) A

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

Note : The difference in pressure at two different elevations in a fluid is

P1 - P2 = ρg ( y 2 – y 1 )

FB = ( P1 – P2 )A

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

By substitution

FB = (P1 – P2)A

FB = ρg(y2 – y1)AV block= (y2 – y1)A = Vfluid displaced

Thus, FB = ρg(y2 – y1)A

FB = ρfluidgVfluid

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

FB = ρfluidgVfluid

FB = mfluidg

displaced fluidFB = W

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

F2

PA = FA /A PB = FB /A

P1= F1 /A

FA

F1

P2 = F2 /A

FB

FB = Wdisplaced fluid

ARCHIMEDES’ PRINCIPLE

“ An immersed object is buoyed up by a force equal to the weight of the fluid it displaces.”

ARCHIMEDES’ PRINCIPLE

FB = W in air – W in water

PROBLEM SOLVING ACTIVITY

MATERIALS:

clay graduated cylinder Electronic weighing scale beakerCalculator basin

PROBLEM SOLVING ACTIVITY

TASK 1: Determine the buoyant force on

ball of clay submerged in water.

TASK 2: Determine the buoyant force on

basin-like clay immersed in water.

BALL OF CLAY BASIN-LIKE CLAY

*Weight of ball of clay:

W = mg

*Weight of the

displaced fluid

m=ρV

W = mg

*Weight of basin-like clay:

W = mg

*Weight of the

displaced fluid

m=ρV

W = mg

BUOYANT FORCE = Weight of displaced fluid

CONCLUSIONS

If the weight of an object is greater than the weight of the displaced fluid or BUOYANT FORCE, object sinks in the fluid.

If the weight of an object is less than the weight of the displaced fluid or BUOYANT FORCE, object floats in the fluid.

ASSIGNMENT

1. Read more about buoyant force.

2. Research how Archimedes’ Principle is applied in swimming and in sea transport.

3. Research on good internet sites about buoyancy and

Archimedes’ Principle.

Navaza, Delia C. & Valdes, Bienvenido J. (2004). Physics. Quezon

Avenue, Quezon City: Phoenix Publishing House Inc.

Giancoli, Douglas C. (1998). Physics: Fifth edition. 23 First

Lok Yang Road, Singapore: Pearson Education South Asia

Pte. Ltd.

REFERENCE LIST

Hewitt, Paul G. (2002). Conceptual Physics: Ninth edition. 23 First

Lok Yang Road, Singapore: Pearson Education South Asia

Pte. Ltd.

Cameron, James (Producer and Director). (June 23, 1999).

Titanic [Motion picture]. Hollywood: Columbia Pictures.

REFERENCE LIST

Movie Animation:http://www.movieanimation.org/

REFERENCE LIST

Van Heuvelen, Allan (1986). Physics:A general Introduction.

Second edition. United States of America: Little, Brown and

Company Ltd.