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“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.