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Net Force. Physics 11 Adv. Comprehension Check. An object has a weight of 12.2N; what is its mass? If you were to take the weight of a 1.0kg object on the moon, what would it be? A 5.00kg wooden crate is slid across a wooden floor; what frictional force is experienced by the crate? - PowerPoint PPT Presentation
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Net ForceNet Force
Physics 11 AdvPhysics 11 Adv
Comprehension CheckComprehension Check
1.1. An object has a weight of 12.2N; what is An object has a weight of 12.2N; what is its mass?its mass?
2.2. If you were to take the weight of a 1.0kg If you were to take the weight of a 1.0kg object on the moon, what would it be?object on the moon, what would it be?
3.3. A 5.00kg wooden crate is slid across a A 5.00kg wooden crate is slid across a wooden floor; what frictional force is wooden floor; what frictional force is experienced by the crate?experienced by the crate?
4.4. If a 150N force is applied to a 25kg If a 150N force is applied to a 25kg object, what would its acceleration be?object, what would its acceleration be?
Comprehension CheckComprehension Check
1.1. 1.24kg1.24kg
2.2. 1.64N1.64N
3.3. 9.8N9.8N
4.4. 6.0m/s6.0m/s22
Putting it All TogetherPutting it All Together
Now that we have considered Now that we have considered Newton’s Second Law, you can use Newton’s Second Law, you can use that to analyze kinematics problems that to analyze kinematics problems with less information than we have with less information than we have used previouslyused previously
We can either use dynamics We can either use dynamics information to then apply to a information to then apply to a kinematic situation or vice versakinematic situation or vice versa
Free Body DiagramsFree Body Diagrams
A free body diagram A free body diagram will be used in most will be used in most dynamics problems in dynamics problems in order to simplify the order to simplify the situationsituation
In a FBD, the object is In a FBD, the object is reduced to a point and reduced to a point and forces are drawn forces are drawn starting from the pointstarting from the point
Fg
FN
Fa
Ff
The Net ForceThe Net Force
In most situations, there is more than In most situations, there is more than one force acting on an object at any one force acting on an object at any given timegiven time
When we draw the FBD we should When we draw the FBD we should label all forces that are acting on an label all forces that are acting on an object and also determine which object and also determine which would cancel each other outwould cancel each other out
Ones that do not completely cancel Ones that do not completely cancel out will be used to determine the net out will be used to determine the net forceforce
The Net ForceThe Net Force
The net force is a vector sum which The net force is a vector sum which means that both the magnitude and means that both the magnitude and direction of the forces must be direction of the forces must be consideredconsidered
In most situations we consider if In most situations we consider if Physics 11, the forces we consider Physics 11, the forces we consider will be parallel or anti-parallelwill be parallel or anti-parallel
An ExampleAn Example
A 25kg crate is slid from rest across A 25kg crate is slid from rest across a floor with an applied force 72N a floor with an applied force 72N applied force. If the coefficient of applied force. If the coefficient of kinetic friction is .27, determine:kinetic friction is .27, determine:• The acceleration of the crate?The acceleration of the crate?• The time it would take to slide the crate The time it would take to slide the crate
5.0m across the floor.5.0m across the floor.
FBDFBD
Fg=-250N
FN=250N
Fa=72NFf=?
Use the frictional force equation to Use the frictional force equation to determine the magnitude of the determine the magnitude of the
frictional forcefrictional force
NF
NF
NF
FF
f
f
f
Nf
66
66
)250)(27(.
The net force is the sum of the The net force is the sum of the forces (acting parallel or anti-forces (acting parallel or anti-
parallel)parallel)
NF
NNF
FFF
FF
net
net
afnet
inet
8.5
7266
Use Newton’s Second Law to solve Use Newton’s Second Law to solve for the accelerationfor the acceleration
2/23.0
)25(8.5
sma
akgN
amFnet
Use kinematics to solve for the time Use kinematics to solve for the time taken to cross the floortaken to cross the floor
stsm
mt
tsmm
dtvta
td
6.6/23.0
)0.5(2
2
/23.00.5
2)(
2
22
00
2
Pushing or Pulling a Sled – Pushing or Pulling a Sled – Which is Better?Which is Better?
Pushing a SledPushing a Sled
A child of 35kg is sitting on a sled A child of 35kg is sitting on a sled that has a coefficient of kinetic that has a coefficient of kinetic friction of 0.10 with the snow. If the friction of 0.10 with the snow. If the parent who is pushing the sled parent who is pushing the sled pushes down with an angle of 25pushes down with an angle of 25°, °, with what force do they need to push with what force do they need to push to keep the child and sled sliding to keep the child and sled sliding with a constant velocity?with a constant velocity?
FBDFBD
Fg
Ff
FN
Fa
Break Applied Force into Break Applied Force into ComponentsComponents
Fay
Fax
Fa
aay
aay
aay
aax
aax
aax
FF
FF
FF
FF
FF
FF
)423.(
)25sin(
sin
)906(.
)25cos(
cos
Consider x and y independentlyConsider x and y independently
NxF
NF
FNF
FFF
NFF
NFF
FFFF
a
a
aa
axfnetx
aN
aN
gayNnety
1100.4
3.34864.
)906(.)343)423)((.10(.0
343)423(.
343)423.(0
Pulling a SledPulling a Sled
A child of 35kg is sitting on a sled A child of 35kg is sitting on a sled that has a coefficient of kinetic that has a coefficient of kinetic friction of 0.10 with the snow. If the friction of 0.10 with the snow. If the parent who is pulling the sled pulls parent who is pulling the sled pulls up with an angle of 25up with an angle of 25°, with what °, with what force do they need to pull to keep force do they need to pull to keep the child and sled sliding with a the child and sled sliding with a constant velocity?constant velocity?
FBDFBD
Fg
Ff
FN
Fa
Break Applied Force into Break Applied Force into ComponentsComponents
Fay
Fax
Fa
aay
aay
aay
aax
aax
aax
FF
FF
FF
FF
FF
FF
)423(.
)25sin(
sin
)906(.
)25cos(
cos
Consider x and y Consider x and y independentlyindependently
NF
NF
FNF
FFF
FF
NFF
NFF
FFFF
a
a
aa
axfnetx
aax
aN
aN
gayNnety
36
3.34948.
)906(.)343)423.)((10(.0
)906(.
343)423(.
343)423(.0
Practice ProblemsPractice Problems
Page 168Page 168• Questions 4-8Questions 4-8
Page 170Page 170• 9 - 139 - 13
Physics 20 Physics 20 • Page 108Page 108
1-10 and 171-10 and 17
