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1HZWRQV QG /DZQRWHERRN 0DUFK Homework uestions from sheet In both these questions the key is constant 8 which means EE which means lefties righties and apples downies Draw the force diagram The weight is Fog we will use F for the applied force Since is constant we know there must be equal forces up down and left fright The upward force is the normal force and the force opposing the motion Ifrikh I 4N 7 60ON 11 Since EF Itg 600N F F 60 on f MN you can't N IF 600N re forget this µ fizzixis fun 60th GOON 0 too there are no units tou it is a ratio of 2 forces

Itg€™s_2nd_law.pdfm nyg a 1.96Mls's 70.2Ig 1HZWRQ V QG /DZ QRWHERRN 0DUFK P NJ) 1 :KDW LV WKH DFFHOHUDWLRQ RI WKH REMHFW" Now add friction c Draw in the forces 2 WriteNewton'sLawsequations

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Page 1: Itg€™s_2nd_law.pdfm nyg a 1.96Mls's 70.2Ig 1HZWRQ V QG /DZ QRWHERRN 0DUFK P NJ) 1 :KDW LV WKH DFFHOHUDWLRQ RI WKH REMHFW" Now add friction c Draw in the forces 2 WriteNewton'sLawsequations

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Homework uestions from sheetIn both thesequestionsthe key is constant8 whichmeans EE whichmeans lefties righties andapples downiesDraw the forcediagram The weight isFogwe will use F for the applied force Since isconstant we know there must be equalforcesup down and leftfright Theupward forceis thenormal force and the force opposingthe motionIfrikhI 4N

7 60ON

11

Since EFItg 600N

F F 60 on f MN youcan't

N IF 600N reforget this

µ fizzixisfun60thGOON 0 too

there are nounits tou itis a ratio of2 forces

Page 2: Itg€™s_2nd_law.pdfm nyg a 1.96Mls's 70.2Ig 1HZWRQ V QG /DZ QRWHERRN 0DUFK P NJ) 1 :KDW LV WKH DFFHOHUDWLRQ RI WKH REMHFW" Now add friction c Draw in the forces 2 WriteNewton'sLawsequations

2 gqy.E.gs5

500NThe first thing is we must realizepullingup at an angle is likepulling in 2 directionsat the same time we break F intohorizontal and vertical components Fxand Fy using trig below Since u asconstant again EFLefties righties uppies dowries

f Fx Faso NtFy Fg50.0N cos30 Ntfs in 30_Fg

a 43 3N Nt 50sin38 500Nc N 425N

d f MNµ 44

34 0.0912g

Page 3: Itg€™s_2nd_law.pdfm nyg a 1.96Mls's 70.2Ig 1HZWRQ V QG /DZ QRWHERRN 0DUFK P NJ) 1 :KDW LV WKH DFFHOHUDWLRQ RI WKH REMHFW" Now add friction c Draw in the forces 2 WriteNewton'sLawsequations

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So what if EF This iswhere Newton made a real breakthroughNewton's 2nd Law

K math terms this is a EIm

A direct proportion Ad EFmeans if we plot a vs IF it looks

like this

Whereas an inverse proportion A Atmlooks like this

I

Page 4: Itg€™s_2nd_law.pdfm nyg a 1.96Mls's 70.2Ig 1HZWRQ V QG /DZ QRWHERRN 0DUFK P NJ) 1 :KDW LV WKH DFFHOHUDWLRQ RI WKH REMHFW" Now add friction c Draw in the forces 2 WriteNewton'sLawsequations

In reality Newton recognized that ferag.ve obgeet

th EFa

constantwhich is how we define inertialmasy

How hard it is to change what an objectisdoing How hard EF change a

We usually remember this equationin linear form

EF a

Newton's 2nd Law

This one is worthy of an equationsheet

Page 5: Itg€™s_2nd_law.pdfm nyg a 1.96Mls's 70.2Ig 1HZWRQ V QG /DZ QRWHERRN 0DUFK P NJ) 1 :KDW LV WKH DFFHOHUDWLRQ RI WKH REMHFW" Now add friction c Draw in the forces 2 WriteNewton'sLawsequations

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In the example below µ so thereis no friction Because the object ison the floor there is a normal forceObviously the object is accelerating to theright

TN Ef y

f x

In the y dir In X derEF Cno a iny dir EF F ma

N Fg COON 20N ma

But what is mWe know from our discussion of mass and weightthat Fg omg 20 10.2kgCa

m nyg a 1.96Mls's

Ig70.2

Page 6: Itg€™s_2nd_law.pdfm nyg a 1.96Mls's 70.2Ig 1HZWRQ V QG /DZ QRWHERRN 0DUFK P NJ) 1 :KDW LV WKH DFFHOHUDWLRQ RI WKH REMHFW" Now add friction c Draw in the forces 2 WriteNewton'sLawsequations

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Now add frictionc Draw in the forces2 WriteNewton'sLawsequations in the

x and y directions3 Use f µN to find f4 Solve for a

f TN

LtgKir Ca x dir CatN Fg EE F f ma

mg106807

50

syr yaa

f MN to 29198N0kg

245N 2.5545c

Page 7: Itg€™s_2nd_law.pdfm nyg a 1.96Mls's 70.2Ig 1HZWRQ V QG /DZ QRWHERRN 0DUFK P NJ) 1 :KDW LV WKH DFFHOHUDWLRQ RI WKH REMHFW" Now add friction c Draw in the forces 2 WriteNewton'sLawsequations

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