FrictionAffects Motion

Static

Friction is a force that opposes the

motion between two surfaces that are in contact

There are two types STATIC friction: opposes the start of motion

between two surfaces that are not in relative motion.

KINETIC friction: the force between two surfaces that are in relative motion. It always acts in the opposite direction to the motion of the object.

Friction

Static friction is ALWAYS

bigger than kinetic friction

Friction

Flat Surface FrictionEx: 1 A westward force of 50 N is applied to a 50 kg mass resting

on a surface that will exert a frictional force of 16 N. If the force is applied for 5 seconds, what is the resulting velocity?

Fnet = ma34 N [W] = (50kg)(a)a = 0.68 m/s2

a = vf – vi

tFnet = Fa + FfFnet = 50 N [W] + 16

N [E]Fnet = 34 N [W]

3.4 m/s [W]

Flat Surface FrictionEx 2: A 30 kg mass is dragged over a surface with a frictional

force of 12 N by pulling on a rope attached to the mass. The rope makes an angle of 30o to the horizontal and the applied force is 60 N. What is the resulting acceleration?Vertical No acceleration

HorizontalFnet = Ff + Fa(h)

Fnet = -12 N + (cos30 • 60)Fnet = 39.96 N (in the positive

direction)

Fnet = ma39.96 = (30)(a)a = 1.3 m/s2

Flat Surface FrictionEx 3: The magnitude of the applied

force in Figure 3.71 is 165 N and 30.0o. If the desk remains stationary, calculate the force of static friction acting on the desk.

Desk is stationary so …Fnet = 0 in both x and y

HORIZONTALFnet = Fa + Ff0 = (165)(cos 30.0) + FfFf = - 143 N [horizontal]

Calculating

Friction Friction is directly proportional to

the weight of an object (FN)

Nf FF

f NF F

fF

= force due to friction (N) = coefficient of friction

NF

= normal force (N)

= the Greek letter “mu”

“mew”

The coefficient of friction is a number (no units) that depends on the two surfaces in contact. The larger the number is, the more “sticky” the situation. So, a very slippery surface (ice) would have a low coefficient of friction.

0.06 – 0.20

0.8 – 1.2

1.0 – 1.6

Calculating

FrictionEx 4: A 12 kg piece of wood is placed on top of another piece of

wood. There is 35 N of maximum static friction measured between them. Determine the coefficient of friction between the two pieces of wood.

FN = Fg = mgFN = (12 kg)(9.84m/s2)FN = 1.2x 102 N

Ff = μ FN

35 N = (μ)(1.2 x 102 N)μ = 0.30

0.30

Calculating

FrictionEx 5: What frictional force must be overcome to start a 50 kg

object sliding across a surface with a static coefficient of friction equal to 0.35?

FN = Fg = mgFN = (50 kg)(9.84m/s2)FN = 490.5 N

Ff = μ FN

Ff = (0.35)(490.5 N)Ff = 171.675 N

1.7 x 102 N

Calculating

FrictionEx 6: A 10 kg box is dragged over a horizontal surface by a force

of 40 N. If the box moves with a constant speed of 0.5 m/s, what is the coefficient of kinetic friction for the surface?

0.41

Calculating

FrictionEx 7: A 10 kg box is dragged across a level floor with a force of

60 N. The force is applied at an angle of 30o above the horizontal. If the coefficient of kinetic friction is 0.20, what is the acceleration of the box?

3.2 m/s2