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Ch05: Force and Motion - INewton’s First LawForceMassNewton’s Second LawSome particular forces; Gravitational Force, Weight, Normal force, Friction, and TensionNewton’s Third LawApplications to Newton's laws

5.1: what is physics

5.2: Newtonian mechanicsThe study of the relation between a force and acceleration,

first studied by Isaac Newton 1642-1727, is called Newtonian mechanics. It is applied to relatively slow objects (much less than the speed of light) and large objects (much more than atom size)

In previous chapters, we have studied the change in velocity as a result of acceleration,

What cause acceleration and hence velocity change? It is the Force

A force is a push or pull upon an object resulting from the object's interaction with another object. When the interaction ceases, the objects no longer experience the force

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5.3: Newton’s First LawIn the absence of an external net force …

an object in motion remains in motion with constant velocity in a straight line.an object at rest remains at rest

This means if net force Fnet= 0 acceleration is zero ( a = 0)Object is said to be at equilibrium

Newton’s First Law: If no net force acts on a body, the body’s velocity cannot change; that is, the body cannot accelerate.

5.4: ForceOnly a force can cause a change in velocity acceleration

Ex: Push, pull, throw an object; gravity; magnetic attraction

The force exerted on a standard mass of 1 kg to produce an acceleration of 1 m/s2 has a magnitude of 1 Newton (N)

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Contact forces act through physical contact

Field forces act through empty space

5.4: Force: types

Force is a vector quantityForce is measured as the elongation of the spring.The total (net) force is the vector sum of all forces acting on the spring.

5.4: Force: measuring force

5.5: MassMass is an intrinsic characteristic of a body; a characteristic

that automatically comes with the existence of the body.mass is a scalar quantity.Relates the force on the body to the resulting acceleration

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5.6: Newton’s Second LawThe acceleration of an object is directly proportional to the net force acting upon it and inversely proportional to its mass.

amFF netrrr

==∑xx maF =∑ yy maF =∑ zz maF =∑

Newton’s Second Law: The net force on a body is equal to the product of the body’s mass and its acceleration.

5.6: Newton’s Second LawThe SI unit of force is Newton (N) 1 N =(1 kg)(1 m/s2) = 1 kgm/s2.

Some force units in other systems of units are given in Table shown

When solving with problems with Newton’s laws, we usually draw Free-body diagram (FBD); Draw the forces vectors act on each mass and its acceleration vector.

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²/293.07.87.8

7.847.4

60cos20cos 2121

smaNmaF

NF

FFFFF

xxx

x

xxx

==⇒==

=+=

°+°=+=

∑∑∑

²/4.173.0

22.522.5

22.593.671.1

60sin20sin 2121

smaNmaF

NF

FFFFF

yyy

y

yyy

==⇒==

=+−=

°+°−=+=

∑∑∑

5.6: Newton’s Second Law: ExampleTwo hokey players hits a puck as shown. What is the acceleration of the puck?

0.3 kg

°==

=+=+=+=⇒

− 3119

4.17tan

²/8.33²4.17²29 ˆ4.17ˆ29ˆˆ

1θ

smajijaiaa yxrr

5.6: Newton’s Second Law: ExampleIn the Fig. the acceleration, , shown is caused by the three forces

. Find the magnitude and direction of the not shown force

(x-components)

(y-components)

and

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gmFamF gnetrrrr

=⇒=

Gravity is the attractive force every object feels towards Earth.

mgFmgF gg =⇒−=−

Gravitational Force

Magnitude of gravitational force

Weight can change from planet to planet depending on gravity acceleration but mass is constant.

5.7: Some Particular Forces: The Gravitational force and Weight

The magnitude of the gravitational force is the called the weight (W); which is minimum force needed to lift a mass m

mgW =

Ex: Block of mass m at rest on a table

5.7: Some Particular Forces: The Normal Force The Normal Force

When a body presses on a surface, the surface pushes on the body with a normal force that is perpendicular to the surface.

mgFmgF

FF

maFF

N

N

gN

yynet

=⇒=−=

=−=

==∑

0

0

∑ ==⇒

=

0

)0rest at (mass

amF

arr

r

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5.7: Some Particular Forces: Tension

When a cord (or a rope, cable) is attached to a body and pulled taut, the cord pulls on the body with a tension forcedirected away from the body and along the cord

Note that, when two objects connected by a cord in moving system, at a given instant, both objects will have same acceleration and velocity

In the previous example, if we pull the object on table by a cord with tension force T, but it still at rest on table (a=0)

N

N

Ny

FmgTTmgF

maa mgTFF

−=−=⇒

=⇒==−+=∑

andsmallerget force Normal

rest)at object - 00(0

5.7: Some Particular Forces: Tension

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If mass is lifted by the cord and accelerated (T>mg), the block is no more pressing on table Normal force disappear

mg maT

mamgTF yy

+=⇒

=−=∑

For example, if m=12kg and a=0.2 m/s²

N..g)m(amgmaT

ma mgTFy

120)8920(12 =+=+=+=⇒

=−=⇒∑

To have acceleration, T must be larger the weight

5.7: Some Particular Forces: Tension

An elevator moving up and comes to rest (slowing down)

weight than theless is tension the

direction -yin forcesonly

gF

⇒−=⇒−=−=⇒

=−=

=

∑

MaMgTMaMgT

MaMTamF

yy

netrr

5.7: Some Particular Forces: Tension

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If two objects interact, the force F12 exerted by object 2 on object 1 is equal in magnitude and opposite in direction to the force F21 exerted by object 1 on object 2.

5.8: Newton’s Third Law

Sketch the situationCategorize the problem (moving or at rest)Isolate an object and identify the forces acting upon it (draw free body diagram-FBD).Establish a convenient coordinate system (x-y) so that one of the axis is parallel to the accelerationApply Newton’s second law to each diagramSolve for desired quantitiesCheck your answer

5.9: Applying Newton’s Laws: Hints for solving problems using Newton’s laws

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Box pulled by a force F at an angle θ. The box has a horizontal acceleration a. find

a) The acceleration b) The normal force

Solution:

1- draw the situation as shown

2- draw free body diagram (FBD)

FBD3- add coordinate system (x-y)

5.9: Applying Newton’s Laws: Example

Solution:

4- analyse forces in directions of x and y

Fx=Fcosθ, Fy=Fsinθ

5- apply newtons laws

FBDaF rrm=∑

mFamaF

maF xx

θθ coscos =⇒=

=∑

θ

θ

sin

0)(a 0sin y

Fmgn

mgFn

maF yy

−=⇒

==−+

=∑For x-direction

For y-direction

Fx

Fy

5.9: Applying Newton’s Laws: Example –continued from previous slide

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Atwood machine

Find a) a = ? b) T = ?

FBD

hanged on Light cord and massless frictionless pulley, find

5.9: Applying Newton’s Laws: Example –Atwood machine

1).........(.......... 11 amgmT

maF yy

=−

=∑

2).........(.......... 22 amgmT

maF yy

−=−

=∑

).....(2' 22 amgmT −=

For m1

For m2

From (2)

Sub. (2’) in (1)

21

12

2112

1122

mmgmgma

amamgmgmamgmamgm

+−

=⇒

+=−⇒=−−

Sub. a in (1)

gmmmmT

21

212+

=

5.9: Applying Newton’s Laws: Example –continued from previous example

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Traffic weight 122N, findT1, T2, and T3

mg=122N

FBD

5.9: Applying Newton’s Laws: Example –traffic at rest

For the traffic light

For the cables joint points

From (1)

5.9: Applying Newton’s Laws: Example –continued from previous slide

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Two blocks A and B are placed in contact with eachother on a frictionless, horizontal surface, as shown. A horizontal force Fapp is applied to A as shown. Find (a) the magnitude of the acceleration of the system (b) the force exerted by A on B.Fapp = 20N, mA = 4kg, and mB = 6kg

Solution:

(a)We can consider the two masses as a single mass mA+mB.

Apply Newtons law

5.9: Applying Newton’s Laws: Example –continued from previous slide

ar

(b) the force exerted by A on B due to action reaction between the two blocks (Newton’s 3rd law) shown in the figure

5.9: Applying Newton’s Laws: Example –continued from previous slide

Hence due to Newton’s third law

For block B, we have force from block A

But they are opposite in direction

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Assume object of mass m on an incline plane of incline angle θ. We can calculate the acceleration and normal force as follows

1) Good to place our x-axis parallel to the acceleration parallel to the incline

2) The gravity force components are

parallel to the plane = mg sinθ

perpendicular to the plane = mg cosθ

3) Object will slide with acceleration along the plane (x-axis) by the effect of mgsinθ

θθ

sinsin

gamamg

maF

x

x

xx

=⇒=

=∑

θθ cos 0cos

0)(a 0 y

mgnmgn

maF yy

=⇒=−⇒

===∑

5.9: Applying Newton’s Laws: motion on an incline plane - Example

a cord pulls on a box up along a frictionless plane inclined at θ=30°.The box has mass m = 5 kg, and the force from the cord has magnitude T = 25 N. What is the box’s acceleration component a along the inclined plane?

5.9: Applying Newton’s Laws: motion on an incline plane - Example

m/s².m

θmgTa

mamgTmaF xx

10sinsin

=−

=⇒

=−

=∑θ

mg sinθ

mg cosθ

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Example: Two blocks connected as shown. Find (a) the acceleration of block S, (b) the acceleration of block H, and (c) the tension in the cord.

5.9: Applying Newton’s Laws: two blocks connected by a cord

FBD

For block M

For block m

(1)

(2)

and

Sub. (1) in (2)

5.9: Applying Newton’s Laws: two blocks connected by a cord

gmM

ma

amMTTmg

+=⇒

+=+− )(Note that direction of motion is taken to be +ve

or

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Example: Two objects of masses m1 and m2are attached by a lightweight cord thatpasses over a frictionless pulley of negligiblemass as shown. Find the a) magnitude of the acceleration of the twoobjects andb) the tension in the cord.

From m1

(1)

5.9: Applying Newton’s Laws: two blocks connected by a cord

Ex: Two Objects Connected by a CordFrom m2

(2)

From (1) Sub in (2)

Sub. In (2)