Forces of Friction -...
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Forces of Friction
l When an object is in motion on a surface or through a viscous medium, there will be a resistance to the motion l This is due to the interactions between the object
and its environment l This resistance is called the force of friction
Forces of Friction, cont. l Friction is proportional to the normal force
l ƒs ≤ µs n and ƒk= µk n l μ is the coefficient of friction
l These equations relate the magnitudes of the forces, they are not vector equations
l For static friction, the equals sign is valid only at impeding motion, the surfaces are on the verge of slipping
l Use the inequality if the surfaces are not on the verge of slipping
Forces of Friction, final l The coefficient of friction depends on the
surfaces in contact l The force of static friction is generally greater
than the force of kinetic friction l The direction of the frictional force is opposite
the direction of motion and parallel to the surfaces in contact
l The coefficients of friction are nearly independent of the area of contact
Static Friction l Static friction acts to keep the
object from moving l If increases, so does l If decreases, so does l ƒs ≤ µs n
l Remember, the equality holds when the surfaces are on the verge of slipping
Fr
Fr
ƒsr
ƒsr
Kinetic Friction l The force of kinetic
friction acts when the object is in motion
l Although µk can vary with speed, we shall neglect any such variations
l ƒk = µk n
Explore Forces of Friction
l Vary the applied force l Note the value of the
frictional force l Compare the values
l Note what happens when the can starts to move
Friction in Newton’s Laws Problems
l Friction is a force, so it simply is included in the in Newton’s Laws
l The rules of friction allow you to determine the direction and magnitude of the force of friction
!F∑
Friction Example, 1 l The block is sliding down the
plane, so friction acts up the plane
l This setup can be used to experimentally determine the coefficient of friction
l µ = tan θ l For µs, use the angle where
the block just slips l For µk, use the angle where
the block slides down at a constant speed
Friction, Example 2 l Draw the free-body
diagram, including the force of kinetic friction l Opposes the motion l Is parallel to the surfaces in
contact l Continue with the solution
as with any Newton’s Law problem
l This example gives information about the motion which can be used to find the acceleration to use in Newton’s Laws
Friction, Example 3
l Friction acts only on the object in contact with another surface l Draw the free-body diagrams l Apply Newton’s Laws as in any other multiple object system
problem
Analysis Model Summary l Particle under a net force
l If a particle experiences a non-zero net force, its acceleration is related to the force by Newton’s Second Law
l May also include using a particle under constant acceleration model to relate force and kinematic information
l Particle in equilibrium l If a particle maintains a constant velocity (including a value
of zero), the forces on the particle balance and Newton’s Second Law becomes
!F∑ = 0
Uniform Circular Motion, Acceleration
l A particle moves with a constant speed in a circular path of radius r with an acceleration:
l The centripetal acceleration, is directed toward the center of the circle
l The centripetal acceleration is always perpendicular to the velocity
2
cvar
=
!ac
Uniform Circular Motion, Force l A force, , is
associated with the centripetal acceleration
l The force is also directed toward the center of the circle
l Applying Newton’s Second Law along the radial direction gives
2
cvF ma mr
= =∑
!Fr
Uniform Circular Motion, cont
l A force causing a centripetal acceleration acts toward the center of the circle
l It causes a change in the direction of the velocity vector
l If the force vanishes, the object would move in a straight-line path tangent to the circle l See various release points in
the active figure
Conical Pendulum l The object is in
equilibrium in the vertical direction and undergoes uniform circular motion in the horizontal direction l ∑Fy = 0 → T cos θ = mg l ∑Fx = T sin θ = m ac
l v is independent of m
sin tanv Lg θ θ=(divide first 2 equations and use tan=sin/cos)
Motion in a Horizontal Circle
l The speed at which the object moves depends on the mass of the object and the tension in the cord
l The centripetal force is supplied by the tension
Trvm
=
Horizontal (Flat) Curve l The force of static friction
supplies the centripetal force
l The maximum speed at which the car can negotiate the curve is
l Note, this does not depend on the mass of the car
sv grµ=
Banked Curve
l These are designed with friction equaling zero
l There is a component of the normal force that supplies the centripetal force
tan vrg
θ =2
Banked Curve, 2
l The banking angle is independent of the mass of the vehicle
l If the car rounds the curve at less than the design speed, friction is necessary to keep it from sliding down the bank
l If the car rounds the curve at more than the design speed, friction is necessary to keep it from sliding up the bank
Loop-the-Loop
l This is an example of a vertical circle
l At the bottom of the loop (b), the upward force (the normal) experienced by the object is greater than its weight
F = nbot −mg =mv 2
r∑
nbot =mg 1+v 2
rg
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