Download Chapter 5 Forces in Two Dimensions

Chapter 5
Forces in Two Dimensions
Vectors Again!
How do we find the resultant in multiple dimensions?
1. Pythagorean Theorem- if the two vectors are at
right angles
R 2= A 2 + B 2
2. At an angle other than 90°
a. Law of Cosines
R2 = A2 +B2 –2AB cos 
b. Law of Sines
R= A = B
Examples
1. A car is driven 125.0 km due west,
then 65.0 km due south. What is the
magnitude of its displacement?
2. Find the magnitude of the sum of two
forces, one 20.0 N and the other 7.0
N, when the angle between them is
30.0 °.
Does a vector have
components?
A = Ax + Ay
A
Ay
Ax
Vector Resolution
The process of breaking a vector into its
components
How can we determine the vector
components?
By using Trigonometry
Trig. Functions
1. SOH
sin  = opposite/hyp.
2. CAH
cos  = adj./ hyp.
3. TOA
tan  = opp./adj.
Example
As the 60-Newton tension force acts
upward and rightward on Fido at an
angle of 40 degrees, the components of
this force can be determined using
trigonometric functions.
Adding 2/more Vectors
By resolving each vector into its x and y
components then add the x components to
form the x component of the resultant then
add the y components to form the y
component of the resultant
Rx= Ax + Bx + Cx
Ry= Ay + By + Cy
Then
R2 = Rx2 + Ry2
How do we find the angle of
the resultant?
Use the Angle of Resultant Vector
= tan-1 (Ry/Rx)
Example: A hiker travels 4.0 m South
then 7.3 m Northwest. Find the
displacement and angle of the hiker.
What is Friction?
A force opposing motion
2 Types of Friction
1. Kinetic Friction(Ffk)- friction created
between moving surfaces
2. Static Friction(Ffs)-force between 2
nonmoving surfaces
What does Frictional Force
depend upon?
1. Surface materials- depends on the
nature of the surfacesCoefficient of
Friction- value describing the nature of
the surfaces in contact
2. Normal force- perpendicular contact
force exerted by a surface on an
object
How is it Determined?
1. Kinetic Friction(Ffk)
Ffk= kFN
2. Static Friction(Ffs)
Ffs sFN
Example Problem 3 p.128
You push a 25.0 kg wooden box across a
wooden floor at a constant speed of 1.0
m/s. How much force do you exert on
the box?
What do we know?
M= 25.0 kg
Fapp.=?
V= 1.0 m/s
a= 0.0m/s/s
= .20(Table 5-1)
Practice Problem p.130, #22
A 1.4 kg block slides across a rough
surface that it slows down with an
acceleration of 1.25 m/s/s. What is the
coefficient of friction between the block
and the surface?
Motion Along an Inclined
Plane
A crate weighing 562 N is resting on a
plane inclined 30.0° above the
horizontal. Find the components of the
weight forces that are parallel and
perpendicular to the plane.
Example Problem #6, p.134
A 62 kg person on skis is going down a
hill sloped at 37°. The coefficient of
kinetic friction between the skis and the
snow is 0.15. How fast is the skier
going after 5.0 s after starting from
rest?