Forces due to FrictionFriction: A force between the contacted surfaces of two objects that resists motion.

If an object is not moving, that does not mean that there are no friction forces at work:

mg

NIn this case, there is no motion, nor any additional external force that may produce motion, so there is not friction present!

No Motion F = fstatic

F fs Accelerated Motion

F fk

a

and then:F > fkinetic

Uniform Motion

F fk

F = fk

If Fmax > fs-max then acc.

Friction (N)

time (s)

fs

max fs

fk

Static friction builds to a maximum value at which point the object “breaks free” -- it accelerates and kinetic friction is present!

No Acceleration Static Friction

The ratio of the maximum force of static friction to the magnitude of the normal force is called the coefficient of static friction: s = fs

N

The ratio of the magnitude of the force of kinetic friction to the magnitude of the normal force is called the coefficient of kinetic friction:

k = fk

N

FN

.5mg

fs ∑Fx = 0 = F – N

∑Fy = 0 = fs – .5mg

µs = fs

NN = F = fs

µs= .5(75)(9.80)

.41

= .5mg

µs

= 900 N

Examine the right side:

m1

m2Ø

Ø = 34˚m1 = 9.5 kg

m2 = 2.6 kg

Will the blocks accelerate, and if so, what will that acceleration be?

µs = .24

µk = .15

m1g

m1gcosøø

m1gsinø

T1

N

fs

m2g

T2

T1 = T2 = T

If the block will stay, then:∑Fx1 = 0 = T - (fs + m1gsinø)∑Fy1 = 0 = N - m1gcosø

N = m1gcosø

∑Fy2 = 0 = m2g - T2

T = fs + m1gsinøm2g = µs(m1gcosø) + m1gsinøm2 = m1(µscosø + sinø)

2.6 < 8.5The tension in the string is not strong enough!OR: ∑Fx = 0 = T + fs - m1gsinøT = m1gsinø - fs

m2g = m1gsinø - µsm1gcosø

2.6 < 7.2 again, T is not enough to hold

therefore: T < fs + m1gsinø

To find the acceleration:∑Fx = m1a1 = T + fk - m1gsinø

∑Fy = m2a2 = T - m2g

∑Fy = 0 = N - m1gcosø

in terms of the magnitude of a: a1 = a2 = a

T = m1a + m1gsinø - fk

T = m2a + m2g

m1a + m1gsinø - fk = m2a + m2g

m1a + m1gsinø - µkm1gcosø = m2a + m2g

a(m1 + m2) = m2g + m1g(µkcosø - sinø)

a = [m2g + m1g(µkcosø - sinø)]

(m1 + m2)

a = 25 + 93[(.15)(.83) - (.56)]

12.1

= - 1.3 m/s2

Note well what the negative sign means:• for block 1, acceleration is in the negative x direction- opposite of up (+) the incline.• for block 2, opposite of g (which we used as +)

The dynamics of Uniform Circular Motion

Remember that for any object traveling in uniform circular motion, the magnitude of the net (centripetal) force acting upon the object is:

∑F = ma = mv2/r

For the purposes of discussing motion of an object in uniform circular motion, we will not use x and y axes!Instead, vector will be described in the z direction (perpendicular to the plane of motion) and r direction (along the radius of the path)!

We will consider 3 specific examples of forces that act centripetally:

The Conical Pendulum

L

m

ø

z

r

mg

T

ø

R

∑Fr = ma = mv2/R = Tsinø

∑Fz = 0 = Tcosø - mg

Setting the components in ratio form:

Tsinø = mv2/R

Tcosø mg

Solving for v: v = √ Rgtanø

This equation gives the constant speed needed for the pendulum to maintain its circle.

The constant speed of a body would be:

v = 2πR

TSubstituting and solving for time:

T = 2π (Lcosø)/g

This would give the period of the motion, which is independent of the mass of the bob!

The Rotor

mg

fsN

The Rotor

An object is kept in uniform circular motion by frictional forces!

∑Fz = 0 = fs - mg∑Fr = mar = N

fs = mgN = mv2/R

Solving these equations together we can derive an equation for speed needed to prevent slipping:

v = √ gR/µs

The Banked Curve

The Banked Curve

ø

mg

N

∑Fz = 0 = Ncosø - mg

∑Fr = mar = Nsinø

manipulating these equations will give:

tanø = v2/Rg

this is the angle-speed combination where friction

is not present!

tanø = v2/Rg NOTE:

• the angle of the bank is independent of the mass of the vehicle or even friction!• when a banked curve is constructed, the road is banked according to an average calculated speed.

• as with the conical pendulum, the constant speed needed to maintain a given banked curve (neglecting any frictional forces):

v = Rgtanø

Drag Forces and ProjectilesDue to air resistance, objects will only accelerate until reaching a final /maximum speed.

That max. speed is called terminal velocity.

Terminal velocity depends on the properties of the falling object (size and shape and density) as well as the properties of the fluid (esp. density).

One particular characteristic of drag forces is that they depend on velocity.

Air Resistance (Aerodynamic Drag)

Unlike kinetic friction, air resistance (Drag) is not constant but increases as speed increases.

The amount of Drag will depend on the shape of the object (drag coefficient [C]), the density of the air (ρ), the cross sectional area (A) and the square of the velocity of the object:

D = .5CρAv2

Terminal Velocity occurs when the object stops accelerating and the Drag equals weight!

∑Fy = 0 = mg - D

mg = .5CρAv2

D = bv

For a relatively tiny particle traveling through a thick fluid, Drag is proportional to only velocity:

b is a constant value dependent upon the characteristics of the object and the fluid

bvT = mg

Terminal velocity in the fluid would still occur when the Drag equals the weight force: