Transcript Forces in Two Dimensions Power Point

Forces in Two Dimensions - Objectives
1.
2.
3.
4.
5.
Addition of Forces
Resolution of Forces
Equilibrium and Static
Net Force Problems Revisited
Inclined Planes
Addition of Forces
• Newton’s 2nd Law: ∑F = ma (∑F is the net force)
Two ways to add vectors
1.Graphical Method
• Head and tail
• parallelogram
2.Mathematical method: Using vector
resolution and Pythagorean Theorem
to determine magnitude and tangent
function to determine direction
Equilibrium and Static
• When all the forces that act upon an object are
balanced, then the object is said to be in a state of
equilibrium.
∑F = 0, a = 0
• An object at equilibrium is either ...
– at rest and staying at rest - "static equilibrium.”
– in motion and continuing in motion with the same
speed and direction.
example
• Forces A, B, C acting on a point. Determine if they
produce equilibrium on the point.
A
B
C
Force A
Force B
Force C
Magnitude
3.4 N
9.2 N
9.8 N
Direction
161 deg.
70 deg.
270 deg
Head and tail method to determine
resultant
Parallelogram method to determine
resultant
Mathematical Method
1. Resolve vectors
2. Use Pythagorean Theorem to
determine magnitude
3. Use Tangent function to
determine direction
Forces
Fx
Fy
A
(3.4 N)(cos161o) = -3.2 N
(3.4N)(sin161o) = 1.11 N
B
(9.2N)(cos70o) = 3.2 N
(9.2N)(sin70o) = 8.65 N
C
(9.8N)(cos270o) = 0 N
(9.8 N)(sin270o) = -9.8 N
R
0N
0N
• The data in the table above show that the forces nearly balance.
We could say it's "close enough."
Example
• A frame is shown with the given tension. Determine the
weight of the frame.
B
A
(50N)sin30 = ½ (50 N)
(50N)cos120 = -43 N
(50N)cos30 = 43 N
Fg = 50 N
C
Cx = 0
Cy = - 50 N
The weight is 50 N
downward
Forces
Fx
Fy
A
(50 N)(cos150o) = -43 N
(50 N)(sin150o) = 25 N
B
(50 N)(cos30o) = 43 N
(50 N)(sin30o) = 25 N
C
Cx= ?
Cy = ?
R
0N
0N
Example
• A sign is shown with the given mass of 5 kg.
Determine the tension of each cable.
Tsin40o + Tsin40o
A=T
B=T
40o
Tcos140o
40o
Tcos40o
Fg = mg = 49 N
Forces
Fx
Fy
A
T∙cos140o
T∙sin140o
Tsin40o + Tsin140o = 49 N
B
T∙cos40o
T∙sin40o
T = 38.1 N
C
0N
Cy = mg = 49 N
R
0N
0N
An important principle
Fg = 10 N
• As the angle with the horizontal increases, the amount of
tensional force required to hold the sign at equilibrium
decreases.
In conclusion
• Equilibrium is the state of an object in which all the forces
acting upon it are balanced. In such cases, the net force is 0
Newton. Knowing the forces acting upon an object,
trigonometric functions can be utilized to determine the
horizontal and vertical components of each force. At
equilibrium, all the vertical components must balance and all
the horizontal components must balance.
Net Force Problems Revisited
• A force directed an angle can be resolved into two
components - a horizontal and a vertical component.
•
•
To determine Net Force, add all the forces by components,
then use Pythagorean Theorem to solve for the magnitude
and tangent function to determine direction
The acceleration of an object can be determined by using
Newton's second law.
Example - Determine the net force and
acceleration
Object moves in horizontal direction
Fnet = 69.9 N, right
m = (Fgrav / g) = 20 kg
a = (69.9 N) / (20 kg) =3.50 m/s/s, right
Example - Determine the net force and
acceleration
Object moves in horizontal direction
Fnet = 30.7 N, right
a = 1.23 m/s/s, right.
Procedures for adding vectors at an angle
with horizontal
1.
2.
3.
4.
5.
6.
7.
Resolve the vectors at an angle into x and y
components.
Add all the x components together
Add all the y components together
Use Pythagorean Theorem to find the resultant
(hypotenuse)
Resultant2 = x2 + y2
Use trigonometric function to determine the
direction: tanθ = opp / adj
Use Newton’s 2nd Law to determine acceleration.
Practice 1
A block of 10 kg mass is pushed
along a frictionless, horizontal
surface with a force of 100 N at an
angle of 30° above horizontal.
FN
FAY
FA
30˚
This applied force (FA)
o) = 87 N
FAxcan
= 100cos(30
be broken
into
F =COMPONENTS
100sin(30o) = 50 N
Ay
X verticalYforce must
The total
be 0, so
FAY
RyFAX
= FN + FAY
–Fg = 0
FN = Fg F–g FAY
FN
FAX
Fg
Total =R
FAX
= Rx Total
= Fax= 0
Acceleration depends only on
FAX
Practice 2
• A man pulls a 40 kilogram crate across a
smooth, frictionless floor with a force of 20 N
that is 45˚ above horizontal.
What is the net force on the sled?
How could the
Fnetacceleration
= FA cos θ be increased?
Fnet = (20
N)(cos
45°) F greater and
Pushing at a smaller
angle
will make
net
Fnetincrease
= 14.14acceleration.
N
therefore
What is the crate’s acceleration?
a = Fnet / m
a = (14.14 N) / (40 kg)
a = 0.35 m/s2
Pushing on an Angle
A block is pushed along a
frictionless, horizontal surface with a
force of 100 newtons at an angle of
30° below horizontal.
FAX
FAY
The total verticalg force must
N
be 0, Fso
= FgTotal
+ FAY
Total F
=N
FAX
=0
-30˚
FA
Fg
FAX
F
FN
FAY
This applied force (FA)
canXbe broken
Y into
COMPONENTS
Acceleration depends only on
FAX
Practice 3
• A girl pushes a 30 kilogram lawnmower
with a force of 15 N at an angle of 60˚
below horizontal.
Assuming there is no friction, what is the
acceleration of the lawnmower?
Fnet = FA cos θ
Fnet = (15 N)(cos 60°)
Fnet = 7.5 N
a = Fnet / m
a = (7.5 N) / (30 kg)
a = 0.25 m/s2
What could she do to reduce her acceleration?
Push at an greater angle
Practice 4
Practice 5
• A student moves a box of books by attaching a rope to the box
and pulling with a force of 90.0 N at an angle of 30.0 degrees.
The box of books has a mass of 20.0 kg, and the coefficient of
kinetic friction between the bottom of the box and the
sidewalk is 0.50. find the acceleration of the box.
determine the net force and acceleration
Fnet2 = (∑Fx)2 + (∑Fy)2
Fnet = 39 N
θ = tan-1(-23/32) = -36o
θ = 324o CCW
a = Fnet / m = 39 N / 5 kg = 8.0 m/s2,
same direction as the force
Inclined Planes
• Objects accelerate down inclined planes because of an
unbalanced force. The unbalanced force is caused by gravity.
• Note: the normal force is not directed in
the direction that we are accustomed to.
The normal forces are always directed
perpendicular to the surface that the
object is on.
Fg on Inclined Plane
An important idea
• The process of analyzing the forces acting upon objects on
inclined planes will involve resolving the weight vector (Fgrav)
into two perpendicular components.
• The perpendicular component of the force of gravity is directed
opposite the normal force and as such balances the normal force.
• The parallel component of the force of gravity is part of the net
force that is responsible to object’s motion along the incline.
Forces on an Incline Calculations
• Consider forces:
– Perpendicular
• F┴ = Fg cos θ
• Cancel out Normal (FN )
θ
– Parallel
• F// = Fg sin θ
• All the parallel components
(including the friction force)
add together to yield the net
force. Which should directed
along the incline.
Tilt you head method
θ
Essential Knowledge
• What happens to the component of weight that is
perpendicular to the plane as the angle is increased?
Decreases – Fg perpendicular
• What happens to the component of weight that points
ALONG the plane as the angle is increased?
Increases – Fg parallel
• What happens to the normal force as the angle is
increased?
Decreases – depends on Fg perpendicular
• What happens to the friction force as the angle is
increased?
Decreases – depends on normal force
Example
Fg = 50N
30°
• What is the magnitude of the normal force?
FN = Fg perpendicular = Fg cos θ = 43.3 N
• If the box is sliding with a constant velocity,
what is the magnitude of the friction force?
Ff = Fg parallel = Fg sin θ = 25 N
Example 1
• The free-body diagram shows the forces acting upon a 100-kg
crate that is sliding down an inclined plane. The plane is inclined
at an angle of 30 degrees. The coefficient of friction between the
crate and the incline is 0.3. Determine the net force and
acceleration of the crate.
F┴ = Fgrav∙cos30o = 850 N
F// = Fgrav∙sin30o = 500 N
In perpendicular direction:
Fnorm = F┴ = 850 N
In parallel direction:
Fnet = F// - Ff
Fnet = 500 N - µFnorm
Fnet = 235 N
a = Fnet / m = 2.35 m/s2
Example
practice
1.
An 8.0-newton block is accelerating down a frictionless ramp inclined at
15° to the horizontal, as shown in the diagram below. What is the
magnitude of the net force causing the block’s acceleration?
2.
A child pulls a wagon at a constant velocity along a level sidewalk. The
child does this by applying a 22-newton force to the wagon handle, which
is inclined at 35° to the sidewalk as shown below. What is the magnitude
of the force of friction on the wagon?
3.
The diagram below shows a 1.0 × 105-newton truck at rest on a hill that
makes an angle of 8.0° with the horizontal. What is the component of the
truck’s weight parallel to the hill?
4.
A block weighing 10.0 newtons is on a ramp inclined at 30.0° to the
horizontal. A 3.0-newton force of friction, Ff , acts on the block as it is
pulled up the ramp at constant velocity with force F, which is parallel to
the ramp, as shown in the diagram below. What is the magnitude of force
F?
5.
a.
b.
A force of 60. newtons is applied to a rope to pull a sled across a horizontal
surface at a constant velocity. The rope is at an angle of 30. degrees above the
horizontal.
Determine the magnitude of the frictional force acting on the sled.
Calculate the magnitude of the component of the 60.-newton force that is
parallel to the horizontal surface.
6. The diagram below represents a block at rest on an incline. Draw arrows to
indicate the directions of friction force (Ff), normal force (FN), and gravity (Fg)
7. The diagram shows a sled and rider sliding down a snowcovered hill that makes an angle of 30.° with the horizontal.
Draw an arrow from the center of the rider to indicate the
direction of normal force.
8. A book weighing 20. newtons slides at a constant velocity down a ramp
inclined at 30.° to the horizontal as shown in the diagram below. What
is the force of friction between the book and the ramp?