Ch. 4 Forces

Use coefficients of friction to calculate frictional force. 2. Describe air resistance as a form of friction. 3. Understand and explain terminal velocity and drag.

The Force of Friction The force that opposes motion between two surfaces that are touching Two types Static Friction Kinetic Friction

Friction Force Friction depends on the surfaces in contact Every surface has microwelds Microwelds are microscopic grooves in the surface of every object. The more points or grooves in contact, the more friction is present

Static Friction (Fs) The force that keeps an object from moving even if there is an applied force Keeps objects still Must be overcome to make objects move

Static Friction (Fs) Static friction reaches its maximum when the applied force is as great as it can be without moving the object

Kinetic Friction (Fk) The force that opposes the motion of two surfaces sliding past each other The frictional force of an object in motion To keep objects moving, a force must be continually applied to overcome kinetic friction

Kinetic Friction (Fk) Right before an object starts moving, Fk is less than Fs(max)

Coefficient of friction (μ) μ – the lowercase Greek letter mu The ratio of the force of friction to the normal force acting between two objects The normal force is the force that opposes gravity

Coefficient of Kinetic Friction Ratio of the force of kinetic friction to the normal friction Fk = μkFN

Coefficient of Static Friction Ratio of the force of static friction to the normal friction Fs = μsFN

Coefficients of Friction (don’t write this – you are getting a copy Coefficients of Friction (don’t write this – you are getting a copy. Make sure you keep it!)

Coefficients of Friction If you are not looking for the coefficient of friction in your problem, look up the value for it in the table Make sure you use the correct coefficient There are differences between the coefficients for kinetic friction and static friction

A 24 kg crate initially at rest on a horizontal floor requires a 75 N horizontal force to set it in motion. Find the coefficient of static friction between the crate and the floor.

Once the crate in the previous problem is in motion, a horizontal force of 53 N keeps the crate moving with a constant velocity. Find μk between the crate and the floor.

Drag and Terminal Velocity A fluid is anything that can flow, generally a gas or liquid. When a body moves through a fluid or a fluid moves around a body there is a drag force ( D ) that opposes motion and points in the direction in which the fluid flows relative to the body.

Air Resistance Although we often ignore it, air resistance, R, is usually significant in real life. R depends on: speed (approximately proportional to v 2 ) cross-sectional area air density other factors like shape R is not a constant; it changes as the speed changes R m mg

Volume & Cross-sectional Area 2z z Area y Area x 2y Volume = xyz Area = xy 2x Volume = 8 xyz Area = 4 xy If all dimensions of an object are doubled the cross-sectional area gets 4 times bigger, but the volume goes up by a factor of 8.

Falling in Air 4 R R m 8 m A mg 4 A 8 mg With all sides doubled, the area exposed to air is quadrupled, so the resistance force is 4 times greater. However, since the volume goes up by a factor of 8, the weight is 8 times greater (as long as we’re dealing with the same materials). Conclusion: when the only difference is size, bigger objects fall faster in air. 8 mg

D = ½ C ρ Av2 ρ is the density of air (mass per volume) A is the effective cross sectional area of the body (cross section taken perpendicular to the velocity) C is the drag coefficient (typically valued b/w 0.4 - 1.0) but varies based off velocity

When a blunt object falls from rest through the air, drag is directed upward, and its mag gradually increases from zero as the object speeds up. D opposes Fg therefore D – Fg = ma If the body falls long enough, D eventually equals Fg; therefore a = 0 and the body falls at constant speed called terminal velocity

0 = ½ C ρ Av2 – Fg Terminal velocity (vt) To find vt we set a = 0 in the drag eqtn’ 0 = ½ C ρ Av2 – Fg Which gives vt = √

The terminal speed of a sky diver is 160 km/h in a spread-eagle position and 310 km/h in the nosedive position. Assuming that the diver’s drag coefficient C does not change from one position to the other, find the ratio of the effective cross-sectional area A in the slower position to that in the faster position

Terminal Velocity Suppose a daredevil frog jumps out of a skyscraper window. At first v = 0, so R = 0 too, and a = -g. As the frog speeds up, R increases, and his acceleration diminishes. If he falls long enough his speed will be big enough to make R as big as mg. When this happens the net force is zero, so the acceleration must be zero too. R This means this frog’s velocity can’t change any more. He has reached his terminal velocity. Small objects, like raindrops and insects, reach terminal velocity more quickly than large objects. mg