Transcript Friction - mrwilterdink

Friction
Ffriction = μFNormal
Ff = μFN
• Ff = friction force (direction is always the
opposite of velocity causing it to slow objects
down)
Ff = μFN
• Ff = friction force (direction is always the
opposite of velocity causing it to slow objects
down)
• μ = coefficient of friction (experimentally
determined value based on the two surfaces
in contact with one another)
Ff = μFN
• Ff = friction force (direction is always the
opposite of velocity causing it to slow objects
down)
• μ = coefficient of friction (experimentally
determined value based on the two surfaces
in contact with one another)
• FN = normal force (force perpendicular to the
surface boundary)
Types of Friction
• static friction – friction that prevents an object
from moving.
Types of Friction
• static friction – friction that prevents an object
from moving.
• kinetic (sliding) friction – friction that slows
down an object in motion.
Types of Friction
• static friction – friction that prevents an object
from moving.
• kinetic (sliding) friction – friction that slows
down an object in motion.
• static friction is always greater than kinetic
friction for the same object.
Types of Friction
• static friction – friction that prevents an object
from moving.
• kinetic (sliding) friction – friction that slows
down an object in motion.
• static friction is always greater than kinetic
friction for the same object.
• This means that it takes more force to get an
object moving than it takes to keep it moving.
Friction on Horizontal Surfaces
• On a horizontal surface, the normal force will
always be equal to the weight of the object.
Friction on Horizontal Surfaces
• On a horizontal surface, the normal force will
always be equal to the weight of the object.
• Example: You push a 25.0 kg wooden box
across a wooden floor at a constant speed.
How much force do you exert on the box?
Example: You push a 25.0 kg wooden box
across a wooden floor at a constant speed.
How much force do you exert on the box?
• Start by drawing a free-body diagram.
Example: You push a 25.0 kg wooden box
across a wooden floor at a constant speed.
How much force do you exert on the box?
• Start by drawing a free-body diagram.
Example: You push a 25.0 kg wooden box
across a wooden floor at a constant speed.
How much force do you exert on the box?
• Start by drawing a free-body diagram.
• All of the forces must cancel out because the box is
moving at a constant speed.
• This means that the pushing force must be the
same magnitude as the friction force.
• This means that the pushing force must be the
same magnitude as the friction force.
• The coefficient of friction can be obtained
from the table on p.129. We want the value
for kinetic friction since the box is moving.
• This means that the pushing force must be the
same magnitude as the friction force.
• The coefficient of friction can be obtained
from the table on p.129. We want the value
for kinetic friction since the box is moving.
• The normal force will be the weight of the box
since the surface boundary is horizontal.
• This means that the pushing force must be the
same magnitude as the friction force.
• The coefficient of friction can be obtained
from the table on p.129. We want the value
for kinetic friction since the box is moving.
• The normal force will be the weight of the box
since the surface boundary is horizontal.
• Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N
Calculate the minimum force required to get the
box moving from rest in the previous example.
Calculate the minimum force required to get the
box moving from rest in the previous example.
• This time we must use the coefficient of static
friction from the table. The normal force
remains the same.
Calculate the minimum force required to get the
box moving from rest in the previous example.
• This time we must use the coefficient of static
friction from the table. The normal force
remains the same.
• Ff = (0.50)(25.0 kg x 9.80 m/s2) = 123 N
Calculate the minimum force required to get the
box moving from rest in the previous example.
• This time we must use the coefficient of static
friction from the table. The normal force
remains the same.
• Ff = (0.50)(25.0 kg x 9.80 m/s2) = 123 N
• It takes 123 N to get the box moving, but just
49 N to keep it moving.
Why do you suppose it is bad to lock
the brakes on a moving vehicle?
Why do you suppose it is bad to lock
the brakes on a moving vehicle?
• When the wheel is not spinning, the friction
produced is kinetic friction.
Why do you suppose it is bad to lock
the brakes on a moving vehicle?
• When the wheel is not spinning, the friction
produced is kinetic friction.
• When the wheel is spinning, static friction is
taking place.
Why do you suppose it is bad to lock
the brakes on a moving vehicle?
• When the wheel is not spinning, the friction
produced is kinetic friction.
• When the wheel is spinning, static friction is
taking place.
• Because static friction is always greater, there
is more friction between your tires and the
road when the wheels are spinning. This gives
you greater control over the vehicle.
What pushing force is needed to accelerate the
25.0 kg box at a rate of 1.0 m/s2?
What pushing force is needed to accelerate the
25.0 kg box at a rate of 1.0 m/s2?
• Draw another free body diagram.
What pushing force is needed to accelerate the
25.0 kg box at a rate of 1.0 m/s2?
• Draw another free body diagram.
What pushing force is needed to accelerate the
25.0 kg box at a rate of 1.0 m/s2?
• Draw another free body diagram.
• To get the box to accelerate, the pushing force
has to be greater than the friction force.
• The net force can be calculated using
Newton’s 2nd Law (Fnet = ma)
• The net force can be calculated using
Newton’s 2nd Law (Fnet = ma)
• Fnet = (25.0 kg)(1.0 m/s2) = 25 N
• The net force can be calculated using
Newton’s 2nd Law (Fnet = ma)
• Fnet = (25.0 kg)(1.0 m/s2) = 25 N
• This means that the pushing force is 25 N
greater than the friction force.
• The net force can be calculated using
Newton’s 2nd Law (Fnet = ma)
• Fnet = (25.0 kg)(1.0 m/s2) = 25 N
• This means that the pushing force is 25 N
greater than the friction force.
• The friction force is always the same for this
box on this surface
Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N
• The net force can be calculated using
Newton’s 2nd Law (Fnet = ma)
• Fnet = (25.0 kg)(1.0 m/s2) = 25 N
• This means that the pushing force is 25 N
greater than the friction force.
• The friction force is always the same for this
box on this surface
Ff = (0.20)(25.0 kg x 9.80 m/s2) = 49.0 N
• The pushing force must be 74.0 N (49 + 25).
Homework
• Read section 5.2 (p.126-130)
• Problems 17-26 beginning on p.128
Friction on an incline
Friction on an incline
• The normal force is no longer the weight of
the object.
Friction on an incline
• The normal force is no longer the weight of the
object.
• The weight vector can be broken down into 2
components, parallel to the incline and
perpendicular to the incline.
Friction on an incline
• The normal force is no longer the weight of
the object.
• The weight vector can be broken down into 2
components, parallel to the incline and
perpendicular to the incline.
Friction on an incline
• The normal force is no longer the weight of
the object.
• The weight vector can be broken down into 2
components, parallel to the incline and
perpendicular to the incline.
• The perpendicular
component is the normal force.
Friction on an incline
• The normal force is no longer the weight of
the object.
• The weight vector can be broken down into 2
components, parallel to the incline and
perpendicular to the incline.
• The perpendicular
component is the normal force.
• The parallel component
represents a pulling force.
Calculating normal force
• SOH CAH TOA can be used to determine both
the normal force and the pulling force.
Calculating normal force
• SOH CAH TOA can be used to determine both
the normal force and the pulling force.
• cos21o = FN/weight
(w = mg)
Calculating normal force
• SOH CAH TOA can be used to determine both
the normal force and the pulling force.
• cos21o = FN/weight (w = mg)
• sin21o = Fpull/weight (w = mg)
Determine the coefficient of static friction for a
metal box on a wooden ramp if the box has a
mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.
Determine the coefficient of static friction for a
metal box on a wooden ramp if the box has a
mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.
• Start by drawing a free-body diagram. It is
helpful to draw the weight force as 2
component forces (parallel and perpendicular)
Determine the coefficient of static friction for a
metal box on a wooden ramp if the box has a
mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.
• Start by drawing a free-body diagram. It is
helpful to draw the weight force as 2
component forces (parallel and perpendicular)
• The box will begin to slide when the friction
force and the parallel force are equal.
Determine the coefficient of static friction for a
metal box on a wooden ramp if the box has a
mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.
• Start by drawing a free-body diagram. It is
helpful to draw the weight force as 2
component forces (parallel and perpendicular)
• The box will begin to slide when the friction
force and the parallel force are equal.
• sin38o = (F||)/(132 N)
Determine the coefficient of static friction for a
metal box on a wooden ramp if the box has a
mass of 13.5 kg and begins to slide down the
ramp when the incline reaches 38o.
• Start by drawing a free-body diagram. It is
helpful to draw the weight force as 2
component forces (parallel and perpendicular)
• The box will begin to slide when the friction
force and the parallel force are equal.
• sin38o = (F||)/(132 N)
• F|| = 81.3 N
• F|| = 81.3 N
• Next, we need to determine the normal force
(perpendicular component) to use in the
friction equation.
• F|| = 81.3 N
• Next, we need to determine the normal force
(perpendicular component) to use in the
friction equation.
• cos38o = (FN)/(132 N)
• F|| = 81.3 N
• Next, we need to determine the normal force
(perpendicular component) to use in the
friction equation.
• cos38o = (FN)/(132 N)
• FN = 104 N
• F|| = 81.3 N
• Next, we need to determine the normal force
(perpendicular component) to use in the
friction equation.
• cos38o = (FN)/(132 N)
• FN = 104 N
• Now, we can solve for μ.
81.3 N = μ(104 N)
• F|| = 81.3 N
• Next, we need to determine the normal force
(perpendicular component) to use in the
friction equation.
• cos38o = (FN)/(132 N)
• FN = 104 N
• Now, we can solve for μ.
81.3 N = μ(104 N)
μ = 0.782