# Ch Vectors Forces in teachersites Websites for Teachers

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```					http://www.saburchill.com/physics/chapters/0017.htm
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Forces in Two Dimensions
http://www.physics-net.com/force/sf000.htm

1
Objectives re Vectors
• Evaluate the sum of two or more vectors
both mathematically and graphically
• Determine the components of vectors
• Utilize basic trigonometry: sine, cosine,
tangent, Law of Cosines, Law of Sines

2
What is a vector?
• A quantity with both magnitude (number
with units) and direction. Examples are:
• Displacement        340 km W
• Force               24.0 n S
• Velocity            125.6 m/s E

3
Formulae to use
• Trig functions:
–   Sine q = opp / hyp
–   Cos q = adj / hyp
–   Tan q = opp / adj
–   Law of Cosines: R2 = A2 + B2 - 2AB(cosq)
–   Law of Sines: R = A          =    B
Sin q   Sin a    Sin b

• The first three only work with 90o between
vectors
• (Get handout if you have not had trigonometry)
4
When vectors are at right angles to
one another:
• Draw out vectors (use protractor with
metric rule) set to some scale (i.e. 1cm =
25km) using heel-to-toe
• Draw the resultant (measure direction and
magnitude)
• Find the angle between the resultant and
one of the given vectors
• Find the magnitude of the resultant
• So… let’s try one. Problem 1 in Ch5
• Protractor & Calculator needed!            5
Car driven 125.0 km W then 65.0
km S . Find mag & dir of resultant.
• Let scale be 1cm:10km (125xcm/10=12.5cm)
and 65xcm/10=6.5cm)
125.0 km W
• Draw to scale

65.0 km S
q
• Measure
– Magnitude of resultant(convert cm back to
km)
– Direction of resultant

6
be close to what you measured.
• This time we are going to find direction
first.
q

Opp 65
• Tan q = opp = 65

q = tan-1 65 = 27.5o S of E
•         125

7
Find the magnitude of the resultant
•   Let’s use sin relationship
•   sin q = opp/hyp = 65/R
•   R = 65/sin q = 65/sin 27.5o
•   R = 140.77 = 1.41x102 km
•   You could also use the Law of cosines and
get the same magnitude.

8
Another problem
• Find the displacement of two shoppers
who walk 250.0 m then turn 90o to the
right and walk 60.0 m. (Using Pythagorean
theorem…)
250.0 m
• R = \ (A2 + B2)                             60.0 m

• R = \ (250.0m)2+60.0m2)
• R = 257 m
• (Law of Cosines would work, too)
9
Force & Motion in Two Dimensions
• In this section you will
• Determine the force that produces
equilibrium when three forces act on an
object
• Determine the components of vectors
• Analyze the motion of an object on an
inclined plane with and without friction.

10
vectors to analyze situations in which the
forces acting on an object are at angles
other than 90°.
• Recall that when the net force on an object
is zero, the object is in equilibrium.
• According to Newton’s laws, the object will
not accelerate because there is no net
force acting on it; an object in equilibrium
is motionless or moves with constant
velocity.
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• Suppose that two forces are exerted on an
object and the sum is not zero.
• How could you find a third force that, when
zero, and therefore cause the object to be
in equilibrium?
• To find this force, first find the sum of the
two forces already being exerted on the
object.

12
• This single force that produces the same
effect as the two individual forces added
together, is called the resultant force.
• The force that you need to find is one with
the same magnitude as the resultant force,
but in the opposite direction.
• A force that puts an object in equilibrium is
called an equilibrant.

13
Finding the equilibrant of A and B

Asgn: 5/

14
Components of vectors
• Now we are going to go backwards and
resolve a vector into the vectors that
caused it (vector resolution)
+y

Ry
q       +x

Rx

15
To find the x component, Rx: cos q = adj/hyp = Rx / 5.0 km
cos q = Rx/5.0km  Rx = 5.0km cos q = 5.0km (cos 40)
Rx = 3.8 km
To find the y component, Ry: sin q = opp/hyp = Ry / 5.0 km
sin q = Ry/5.0km  Ry = 5.0km sin q = 5.0 km (sin 40)
Ry = 3.2 km
+y

opp = Ry
Ry
q = 40. o       +x

Rx

16
Section 5.1 assignment
• 3, 4, 7, 12-14 & 80 (angle for both E and F
is 45o)

• Section check frames 32-40

17
Review
What is the normal force, FN?
The perpendicular contact force exerted by
a surface on another object.
FN

Fw

18
Friction
• For objects at rest to move, they must
overcome inertia. To move a couch, you
must apply force to get it going then less
force to keep it going.
• Kinetic Friction Force: Ff kinetic = mk F N
• Static friction force: Ff static ≤ ms FN
• Sketch and analyze the problem.

19
#17. A girl exerts a 36 n horizontal
force while pulling a 52 n sled
across a sidewalk at constant
speed. Find the coefficient of
kinetic friction between the
sidewalk and the sled.
• Fn = mg = 52n
• Friction force equals girl’s pulling force.
• Ff kinetic = mkFN  so mk = Ff/Fn=36n/52n
mk = 0.69

20
#19. Corey is dragging a box full
of books (combined weight of
134n) from his house to the car.
How hard must Corey pull on
the box in order to start it
moving if the coefficient of static
friction between the pavement
and the box is 0.12?
21
Solution
• F Corey on Box = F friction
• F Corey = msFN = msmg
• F Corey = 0.12(134n) = 16.1n

22
#5/22. A 1.4 kg block slides
across a rough surface in such
a way that it slows down with an
acceleration of 1.25 m/s   2. Find

the coefficient of kinetic friction
between the block and the
surface.
23
FN   a = 1.25 m/s2

Fforward
Ffriction

Fw = mg

Is there a net force?

24
Fnet = mk FN
ma = mk mg
a = mkg
mk = a/g = 1.25 m/s2 / 9.80 m/s2 = 0.128

25
• New problem: 5/23
bookcase in the living room. You push with
a force of 65n and the bookcase
accelerates at 0.12 m/s2. What is the
coefficient of kinetic friction between the
bookcase and the carpet?

26
FN

Ff = mkFN             F
Fw

Fnet = F - mkFN = F - mkmg = ma
mk = (F - ma)/mg           working equation!
= [65n - (41kg)(0.12m/s2)] / 41kg(9.80m/s2)
= 0.15                                        27
• Your assignment for 5.2 is 18&20, 24,25

• Section check frames 67-74 of Publisher’s
PPt.

28
The drawing below shows the vectors acting on a skier going down a hill.
Acceleration is parallel to the slope, on one axis, usually the x-axis
The y-axis is perpendicular to the x-axis and perpendicular to the surface of
the slope.
There are two forces—normal and frictional forces. These forces are in the
direction of the coordinate axes. However, the weight is not.
When an object is placed on an inclined plane, the magnitude of the normal
force between the object and the plane will usually not be equal to the
object’s weight.

29
Apply Newton’s laws once in the x-direction and
once in the y-direction.
The weight does not point in either of these
directions, you will need to break this vector into
its x- and y-components before you can sum
your forces in these two directions

30
Let’s look at Ex. 6 on page 134
• A 62-kg person on skis is going down a
hill sloped at 37o. The coefficient of kinetic
friction between the skis and the snow is
0.15. How fast is the skier going 5.0 s after
starting from rest?

31
33. An ant climbs at a steady speed up the side
of its anthill, which is inclined 30.0° from
the vertical. Sketch a free-body diagram for
the ant.

32
• Assignment for 5.3 is 36 and 95, 97-99

• Section check 85-90

33
Multi Vector Problems

34
1
5

4           2

3

35
36
Magnitude

23.3 N
25 N

•   Sin q = a / h  h = O / Sin q
•   H = 23.3 N / Sin 43o
•   34.2 N
•   Add 180o and you get the equilibrant!

37

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