Transcript FORCES at WORK

FORCES at WORK
Today is Component forces and friction.
Friction results from
the mutual contact of
irregularities in the
surfaces. (The atoms
cling together at many
points of contact. As
sliding occurs, the
atoms snap apart or
are torn from one
surface to the other.)
• Static Friction the resistance force that
must overcome to start an object in
motion.
• Kinetic or Sliding Friction the resistance
force between two surfaces already in
motion
• Rolling Friction the resistance force
between a surface and a rolling object.
• Fluid Friction the resistance force of a
gas or liquid as an object passes through.
• The direction of friction is ALWAYS in a
direction opposing motion. An object
sliding to the right, experiences friction
toward the left. An object sliding down an
inclined plane experiences friction up the
inclined plane.
• Sliding friction is always less than starting
friction.
Coefficient of friction, μ,
μ = Friction force
Normal force
Tables of coefficients of friction are printed
in references and textbooks.
Friction isn’t always bad
• Traction
• Brakes
If a force of 50 N is applied to a block of
mass 10 kg and it does not move, what is
the coefficient of friction for the block and
table.
After his car stalls, George pushes slowly on
his 2000 kg car at a constant speed, with a
force of 300 N, what is the coefficient of
sliding friction?
Force can be divided into COMPONENTS
We will use our
knowledge of
trig. Functions
to calculate
these values.
Fy = F sin Θ
Fx = F cos Θ
Okay You Try…
You are walking
your dog Fido,
and he is pulling
on the chain at
with a force 60 N
at an angle of 40º
What are the component forces of the pull?
If the velocity is constant, and Fido has a
weight of 70 N, what is the coefficient of
friction between Fido and the ground?
What is the net force and acceleration
of the box?
Fill in the blanks….
STATICS
FORCES In EQUILIBRIUM
When forces are in
EQUILIBRIUM, all the
forces acting on a body are
balanced and the body is
NOT accelerating.
Because the object is not accelerating we know
The sum of the forces in horizontal is zero
and the sum of the forces in vertical is zero
OR
Σ Fx = 0 and Σ Fy = 0
MHS
A 50 N sign hangs by a
single wire as shown.
What is the tension in
the wire?
1) Construct one of the triangles.
2) Calculate the part of the weight it supports
3) Use trig to find the missing side
Tension in the
wire is 38.9 N
Suppose the tension in each wire is
found to be 50 N, what is the weight
of the picture?
1. Draw the triangle
2. Solve for fy
Since two Fy pulling up, the weight is 50 N
A picture has a weight of 10 N, what is the
tension in the wire at the following angles?
@ 60º Tension = 5.8 N
@ 45º Tension = 7.1 N
@ 15º Tension = 19.3 N
Consider the tow truck at the right. If the tensional
force in the cable is 1000 N and if the cable makes a
60-degree angle with the horizontal, then what is the
vertical component of force which contributes to
lifting the car off the ground?
The triangle
W
The weight side
Sin Θ = opp/hyp
Sin 60 = W/1000 N
W = 866 N
A 50 kg crate is hanging on a 4 m long rope which
sags 5.0 cm. What is the tension in the rope?
What is the angle?
tan Θ = opp/adj
tan Θ = .05m/2m
Θ = 1.4
What is the tension?
sin Θ = opp/hyp
sin 1.4 = 250N/T
T = 10,232 N