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Unformatted text preview: \ NEWTON’S LAWS (POINT OBJECTS) FIRST OBJECTS DO NOT CHANGE THEIR STATE OF MOTION (vel. 12 For NOW) SPONTANEOUSLY
DEFINES INERTIA
Examples: SEAT BELTS, STUMBLE, GRASS MOWING SECOND a), EVERY OBJECT HAS AN INTRINSIC PROPERTY CALLED INERTIAL MASS (M) _
b) AN OBJECT OF MASS M CAN HAVE A NONZERO ACCELERATION IF
AND ONLY IF THERE IS A FORCE E PRESENT SUCH THAT COROLLARIES: (i) IF AN OBJECT IS IN EQUILIBRIUM (a m) (g = O ), THE VECTOR SUM OF ALL THE FORCES ACTING ON IT MUST BE ZERO
E+F2+F3...=O p ' Object does not have to be at rest, it must not change 3. (ii) IF g at 0 .at a SPACE POINT AT A TIME t, THERE MUST BE A FORCE
ACTING AT THE 2: AT THAT TIME. THIRD WHEN TWO OBJECTS INTERACT THE FORCES ACTING ON THEM
FORM ACTION—REACTION PAIRS ( F21 acts on object 1, E2 on object 2) ﬁz—FIZ * 42
FORCES ~72“ W»
In order to use Newton’s Laws we need the Forces that occur in various physical systems. For our discussion in 121 we deal with Mechanical Forces only. Also, we do not discuss in detail the
origin of the force in Every Case. Near Earth Every unsupported object has an acceleration
g = —9.8m/s2)”/ = —g;f/ So it must experience a force 313 = —9.8M)9 2 —Mg)“/ where M is its mass. This force is weight and is a vector perpendicular to the
Earth’ssurface directed toward the center of the Earth. More precisely, Since the
Earth is a sphere the force should be written as y = —9.8Mz9 Vt I WEIGHT where r9 is a unit vector along the radius. It is a manifestation of Newton’s
universal law of Gravitation (discussed in detail later) _G A
FG = ——MfM r
—> RE where M E = Mass of Earth
R E = Radius of Earth 2
G=6.7><10“” N (kg)
By Newton’s 3rd law it follows that the reaction force to 313 acts at the center of the Earth So Earth pulls on M, M pulls on Earth with an Equal and opposite force, 11 CONTACT FORCE OR NORMAL FORCE
Comes into play when an object is in contact with the surface of a solid. It acts perpendicular to surface of the solid: hence Normal Force (N R ). It comes about because the atoms/molecules of a solid oppose the attempt by any foreign object to enter the solid. For example, put the massM
of the above discussion on the Earth. Now M is in E In so the sum of the forces acting on it must be zero. 7')
g = n}? 143 = —Mg3”2 'v _ LA
> g + w = 0 fwdﬂ
)1 WM . l . Ex 2 M l and M 2 are lying on a smooth horizontal surface. Apply a force E = F)? to M1 as
shown; Both M l and M 2 acquire an acceleration Question: Whlch force causes M 2 to accelerate? Answer: Contact force between M I and M 2.
Let us draw all the forces acting on each mass (Free Body diagrams) y) l A
1 " \ «ma é
042.:
WE r? Inna
\ ﬂ, :2.
V
M .13
By the 3rd law 1112 + n21 = O ("12 + "21)”& : 0
To calculate g we must use force acting at that mass at that time so
M19; 2 [email protected] = Fat—n21)?
M 29 = 7g = 11129?
Add
(M1+M2)9, : Fﬁ+(n21 _n12)5&
= F)”: ‘F A
a: — x
“1 {M,+M2] EX 3 To weigh yourself you stand on a weighing machine. You have two forces acting on
you w = —ngz and g = n)"; the normal force which the machine exerts on you. You are in a In so n = Mg . You push down on machine with g 2 —nj» so machine records It and
hence w. EX 4 If the surface is not horizontal y will have to adjust so that three is a m perpendicular
(normal) to the surface while there is acceleration g sin (*9 down the ramp. The force picture is M 2"
QWKBMX/AC 2% W3 _,,____.___.
. l' M
l to surface there is a m so 3., n—MgcosG =0; 71 =Mgcos® H to surface there is acceleration caused by Mg sin® Not surprisingly Q is maximum when (9 = 0 (horizontal surface) and goes to zero when G) = Z 2
(surface is vertical) ' EX 5 Another way to change 72 is to apply a force 5 at an angle 6) above the xaxis. Now for —=— m along y we have
n+Fsin®~Mg =0
so 1: n=Mg—Fsin® p
while along x there is acceleration given by
M g = F cos @fc ’l’l III TENSION IN A MASSLESS INEXTENSIBLE STRING 6”?” F”; 4; "" <6“ I; You are holding one end of a light string. Your friend catches hold of the other end. Suppose
she pulls on it with a force E = —F)”c , toward the left. In order to keep it in E 171 you have to pull on the right with E = +1755. How come? Well, when she applied ~ Fat at A and the string wants to be in E m it must develop + F)? at A, gain a to keep 5 m everywhere inside, it needs 5 ’s at every point balancing each other out until point B is reached where string pulls to the left. So for
E m at B you must pull with+ F22. A force E applied at one end of the string causes a tension T = F to appear in the string such that at the ends Z: acts toward the middle and at the middleZ
is directed toward the ends. ’ IV SPRING FORCE (HOOKE’S LAW) We already used this in establishing a measure for Force because we used a SPRING
BALANCE. This force appears if you stretch a spring or squeeze it. The spring resists the
change in its length so this force always opposed the stretch (or squeeze). For small changes in
length the force is proportional to the change in length hence we write FSP : ‘MAxﬁe _p where Ax = change in length
'k = spring constant Minusisign ensures that FSP is opposite to Ax .= Axfc. So if k : 104 N / m , it will cost you ION of force to change its length by 1mm. V FRICTION . , I . This force arises because surfaces of solids are never totally smooth so when two surfaces are
made to slide past one another they resist it by developing the force of friction. Indeed, as we
showed in class if the applied force is less than a certain value no motion occurs and we talk of static friction ( fs ). Note: friction always opposes motion M ,I
F7 M ﬁg M5 Recall the experiment we did in class. We slowly increased Fa and since no motion occurred pp
we said fs = —Fﬂpp. That means .that as long as there is no motion of f vs. Fm, forms a straight line of slope 1. Finally, sliding starts because )2 has a maximum value. That is A S(#SNR) where ,us is called coefﬁcient of static friction. ,us is determined by the properties of the We surfaces. If f W > ,uSNR , sliding begins but frictional force in NOT zero. It is given by f k = 1“ka
M is called coefﬁcient of kinetic ﬁ‘iction. Note: f iNR , f always opposed Famed —_.__, ...
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 Spring '10
 Shawhan
 Physics, Force, Normal Force

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