2.5 effects of a force

8/8/2019 2.5 Effects of a Force

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2.5Understanding

the effects of aForce



Learning OutcomesLearning OutcomesLearning OutcomesLearning Outcomes

By the end of this subtopic, you will be able toBy the end of this subtopic, you will be able to

Describe the effects of balanced forces acting onDescribe the effects of balanced forces acting on

an object,an object,Describe the effects of unbalanced forces acting onDescribe the effects of unbalanced forces acting onan object,an object,

Determine the relationship between force, massDetermine the relationship between force, massand acceleration , F = ma,and acceleration , F = ma,Solve problems using F =maSolve problems using F =ma



LE SSON 1LE SSON 1



INTR ODUC TI O NINTR ODUC TI O N

INTR ODUC TI O NINTR ODUC TI O N

A ll these daily activities need a force. Can you A ll these daily activities need a force. Can youdescribe what can a force do?describe what can a force do?

Take anobject

move anobject

lift anobject

Stop us from

falling down

Make thing

moves faster

Change our

action



1. Force is a vector quantity .1. Force is a vector quantity .

2. SI unit of force is kg m s2. SI unit of force is kg m s -- 22 or newton, Nor newton, N

3 . Effects of forces are :3 . Effects of forces are :(a) Moves a stationary object .(a) Moves a stationary object .(b) Slows down a moving object .(b) Slows down a moving object .(c) Accelerates an object .(c) Accelerates an object .

(d) Stop a moving object .(d) Stop a moving object .(e) Change the direction of motion .(e) Change the direction of motion .(f) Change the shape of an object .(f) Change the shape of an object .

(g) Change the size of an object .(g) Change the size of an object .

Change the speedof an object



EFFE C TS O F EFFE C TS O F BALA N CE D F O R CE BALA N CE D F O R CE

A N D A N DUN BALA N CE DUN BALA N CE D

F O R CEF O R CE

EFFE C TS O F EFFE C TS O F BALA N CE D F O R CE BALA N CE D F O R CE

A N D A N DUN BALA N CE DUN BALA N CE D

F O R CEF O R CE



BALA N CE D F O R CE S(1)BALA N CE D F O R CE S(1)BALA N CE D F O R CE S(1)BALA N CE D F O R CE S(1)

F1F2

No movement . The net force is zero .

F1 and F 2 are two equal and oppositeforces .




At rest . The net force is zero .

The weight is balanced by the fourreactions from the ground .

F 1

F 2

F 3F 4

Weight, W




Flying at the same height with constantvelocity .

The net force is zero .

Lift = weight & Thrust = drag .

Thrust

Weight

Drag

Lift



UN BALA N CE D F O R CE S(1)UN BALA N CE D F O R CE S(1)UN BALA N CE D F O R CE S(1)UN BALA N CE D F O R CE S(1)

A golf ball is hit by a force . No other forces to balance it .

The ball starts to move forward .



UN BALA N CE D F O R CE S(2)UN BALA N CE D F O R CE S(2)UN BALA N CE D F O R CE S(2)UN BALA N CE D F O R CE S(2)

The player hit a volleyball which movestowards him .

No other forces to balance it .

The ball bounces off and change

direction .



CO N CLU SIO NCO N CLU SIO NCO N CLU SIO NCO N CLU SIO N

1. When an object is at rest or moves withconstant velocity , it means that :

(a) No force is acting on the body OR

(b) All the forces acting on the bodyare balanced . The net force iszero .

2. When an object changes speed ordirection of motion, it means that :

there must be an unbalanced forceacting on the body . The netforce is called resultant force .



R E LA TI O NSHIP R E LA TI O NSHIP BE TW EE N BE TW EE N

F O R CE, MA SS A N DF O R CE, MA SS A N D

ACC E LE R A TI O N ACC E LE R A TI O N

R E LA TI O NSHIP R E LA TI O NSHIP BE TW EE N BE TW EE N

F O R CE, MA SS A N DF O R CE, MA SS A N D

ACC E LE R A TI O N ACC E LE R A TI O N



Carry out Exp 2.2 pg 29 andCarry out Exp 2.2 pg 29 andExp 2.3 pg31Exp 2.3 pg31

G roup A, B and CG roup A, B and C Exp 2.2 pg 29Exp 2.2 pg 29Aim : To investigate the relationshipAim : To investigate the relationship

betweenbetween aa andand F F

(constant mass)(constant mass)

G roup D, E and FG roup D, E and F Exp 2.2 pg 29Exp 2.2 pg 29Aim : To investigate the relationshipAim : To investigate the relationshipbetweenbetween aa andand mm

(Constant Force)(Constant Force)



Exp 2.2 pg 29Exp 2.2 pg 29

P reparation before experiment :P reparation before experiment :1.1. P repare a frictionP repare a friction- -compensated inclinedcompensated inclined

runway .runway .

Adjust the runway so that the trolley goes downAdjust the runway so that the trolley goes downthe runway with a constant speed when given athe runway with a constant speed when given aslight push .slight push .

2.2. Increasing the pulling force on the trolleyIncreasing the pulling force on the trolleyAlways stretch the elastic cord until the end of Always stretch the elastic cord until the end of the trolley .the trolley .

P ractice by pulling one cord, two cords andP ractice by pulling one cord, two cords and

three cords .three cords .

( constant mass)



Exp 2.3 pg 31Exp 2.3 pg 31

P reparation before experiment :P reparation before experiment :1.1. P repare a frictionP repare a friction- -compensated inclinedcompensated inclined

runway .runway .

Adjust the runway so that the trolley goes downAdjust the runway so that the trolley goes downthe runway with a constant speed when given athe runway with a constant speed when given aslight push .slight push .

2.2. P ull the trolley with a constant force .P ull the trolley with a constant force .Always stretch the elastic cord until the end of Always stretch the elastic cord until the end of the trolley .the trolley .

P ractice by pulling one trolley, two trolleys andP ractice by pulling one trolley, two trolleys and

three trolleys .three trolleys .

( constant force)



LE SSON 2LE SSON 2



Carry out theCarry out the

exp eriment for 30 exp eriment for 30 minutesminutes



Acceleration, a E Net force, F

Acceleration, aE 1

,Mass m

a E F

a E1

m

a E

F

m

F E ma

F =k ma

By taking k = 1 , we have F = ma

We define 1 N as the force needed to make amass 1 kg to produce an acceleration of 1 m s - 2 .

N ewton¶sSecond Law

of Motion



N ewton·s Second Law of MotionN ewton·s Second Law of Motion

states that thestates that the accelerationacceleration producedproduced

by a net force on an object isby a net force on an object is directlydirectlyproportional to theproportional to the magnitude of themagnitude of thenet forcenet force applied and isapplied and is inverselyinverselyproportional to the massproportional to the mass of theof theobject . The direction of theobject . The direction of theacceleration is the same as that of acceleration is the same as that of the net force .the net force .

F = maF = ma

a E F

m



What force is required to move a 2 kgWhat force is required to move a 2 kgobject with an acceleration of 3 m sobject with an acceleration of 3 m s -- 22 , if , if (a) the object is on a smooth surface ?(a) the object is on a smooth surface ?(b) the object is on a surface where the(b) the object is on a surface where the

average force of friction acting onaverage force of friction acting on

the object is 2 N ?the object is 2 N ?

Example 1

solution

(a)Net force, F = ma

= 2 (3)= 6 N

Ex ploring pg 93

(b)Net force, F = ma

F ± 2 = 2 (3)= 6 N

F = 8 N

F F2 N



Azhari applies a force of 50 N to move aAzhari applies a force of 50 N to move a12 kg carton at a constant velocity . What12 kg carton at a constant velocity . Whatis the frictional force?is the frictional force?

Example 2

50 N

Fsolution

50 ± F = 12 (0)= 0

Net force, F = ma

Ex ploring pg 93

F = 50 N



A car of mass 12 00 kg travelling at 1 5 m sA car of mass 12 00 kg travelling at 1 5 m s -- 11

is brought to rest over a distance of 30 m . is brought to rest over a distance of 30 m . FindFind

(a) the average retardation, and(a) the average retardation, and(b) the average braking force(b) the average braking force

Example 3

solution

(a)u= 1 5, v = 0, s=30Using v 2 =u 2 +2 as

0 = 1 52

+2 a(30)60a = - 1 5 2

a = -3 .7 5 m s - 2

Retardation=3 .7 5 m s- 2

Ex ploring pg 93

( b) F = maF = 12 00(-3 .7 5)

= - 4500 NBraking force= 4500 N



T he figure above shows a car of mass 1200 kg T he figure above shows a car of mass 1200 kg towing a caravan of mass 800 kg and both have antowing a caravan of mass 800 kg and both have anacceleration of 1.5 m sacceleration of 1.5 m s --2.2. CalculateCalculate

(a) the pulling force, F of the car,(a) the pulling force, F of the car,(b) the tension, T , in the coupling between the car(b) the tension, T , in the coupling between the car

and the caravan.and the caravan.

Example 4 T ex t book pg 41



solution

(a) Combine the car and the caravan,

Net force, F = maF =( 12 00 + 800)( 1. 5)

= 3000 N



solution

(b) Consider only the

caravan

Net force, F = ma

T =(800)( 1. 5)

= 12 00 N

OR

(b) Consider only the

car

Net force, F = ma

3000- T=( 12 00)( 1. 5)= 1 800

T =3000- 1 800= 12 00 N

T T3000N



SUMMA RY SUMMA RY SUMMA RY SUMMA RY



1. Force is a vector quantity .1. Force is a vector quantity .

2. SI unit of force is kg m s2. SI unit of force is kg m s -- 22 orornewton, Nnewton, N

3 . If an object is at rest or moving3 . If an object is at rest or movingwith constant velocity ,with constant velocity ,resultant force F = 0 .resultant force F = 0 .

4 . If an object is moving with an4 . If an object is moving with anacceleration,acceleration,resultant force F = maresultant force F = ma



Mastery P ractice 2.5 pg 4 2Mastery P ractice 2.5 pg 4 2

Question 1 , 2 , 3Question 1 , 2 , 3



Ex tra picturesEx tra pictures

2.5 effects of a force

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