A great triumph of Galileo's investigations was his success at discerning the rules of nature in the absence of friction. The devices and world available to him were anything but frictionless, yet Galileo was able to penetrate this ugly reality and to envision the ideal. Thus he could see that in the absence of friction a body would continue to travel forever.

Newton extended this to say that as long as the forces on a body are balanced (that is, there is no NET force), it will stand or move along at a constant velocity. Further, if an object moves at a constant velocity, the forces acting on it MUST be balanced. Conceptually, that is a much harder thing to accept, as you will see.

Think a little more about what forces act on objects with which you are familiar.

1. Draw the forces acting on each object:

a) an object on a table	b) an object falling freely without air drag
c) an object falling freely with air drag, but not yet at terminal velocity	d) A ball at the peak of its flight, but without significant air drag.

a)

b)

c)

d)

The diagrams you've drawn are called free-body diagrams. They get increasingly helpful as problems become difficult.

2. A horizontal force acts on a block as shown. Draw the horizontal forces on the block when (a) there is no friction, (b) the block accelerates despite some friction, and (c) the block is held to a constant speed by the friction. Be sure the relative lengths of the forces are meaningful.

a)

b)

c)

The first time you think about it, it is usually difficult to understand that when the block slides at a constant velocity the forward and backward forces are precisely equal. If there is no net force on the block, what makes it go forward? This is a perplexing question because it is the wrong question. Once the block begins to move (and we may well ask what gets it moving), we expect it to continue forward at a constant velocity. It's doing what a block should do. If it were to slow down, then we would have a puzzle, because slowing down is the unnatural thing to do.

3. As a 2.6 pound rutabaga is pushed by a 0.8 lb force, it generates a 0.3 lb frictional force. What is its acceleration?

F = ma .8 lb - .3 lb = (2.6 lb/32 slug/lb)a a ~= 6.15 ft/sec2

So it's time to look more deeply into friction, and we will consider only the sliding friction between two flat (but not always level) solid surfaces. Its origins are complex. No surface is perfectly flat, and as the micro-bumps ride over one another they generate drag. Further, the molecules of the materials may bond and have to be ripped away from their parent material. Dirt and grease further alter the significance of these forces. So friction is too complex to deal with from a strictly theoretical perspective, yet in the real world its gross behavior is quite simple.

It turns out that contact area between the surfaces, the speed with which they're sliding, the color of the objects, the time of day, and a million other things, are NOT important. What IS important is the force pressing the surfaces together. We call this the NORMAL FORCE.

To understand friction, perform the following thought experiment: place a block on a level surface. Add weights to the top, and with each additional weight, measure the force required to slide the block slowly and steadily along the surface. As one might expect, when the weight pressing on the surface is doubled or tripled, the force required to push it is doubled or tripled. Here is the way we graph it:

Mathematically this is similar to the spring. This time we can say the force of friction is proportional to the normal force, or Ff N. As before, we can restate this as Ff = µN where µ is a constant. Note that since µ compares two forces, it can have no units. It is called the ";coefficient of friction"; and is the Greek letter pronounced ";mu";. The numerical value of µ is determined by experiment and depends on the slipperiness of the surfaces involved. It typically has values from zero to 0.5, but it often goes higher.

Try a few to get the hang of it. Remember, to calculate the frictional force, use F_f = µN.

4. Calculate the unknowns for each accelerated block.

a)	b)	c)
a) F = ma F - .2(18 * 9.8) = 18 * 3.7 F ~= 102 nt	b) F = ma (8)(5) - .2(6 * 9.8) = 6a a ~= 4.71 m/sec2	c) F = ma 41 - .3(m * 32) = m * 5.6 41 = m[5.6 + (.3 * 32)] m = 2.70 slug

5. And a few more:

a)	b)
a) F = ma 8200 - µ(37 * 980) = 37(155) µ ~= .0680	b) F = ma 7 * 5 - .3(60) = (60/32)a a ~= 9.07 ft/sec2

6. Marion Haste weighs 130 pounds and carries a fox terrier. If her coefficient of friction is 0.3 and it takes a force of 105 pounds to push her along with a 4 ft/sec² acceleration, what is the weight of her dog?

F = ma 105 - .3(130 + w) = (130 + w)/32 * 4 105 - (130/32) * 4 - .3(130) = w * 4 + .3w w ~= 11.5 lb

Before we go on, we must look again at the ";Normal force";, N. Remember that it is the force pressing two surfaces together, so it is always the sum of the components of force perpendicular to the surface, not necessarily the vertical forces.

7. For starts, find the normal force in each of these.

a)	b)	c)
d)	e)	f)
g)	h)	i)
a) N = w	b) N = 0 Nothing is pressing w against the wall	c) N = wcos(Ø)
d) N = w1 + w2	e) N = Fsin(Ø) + mg	f) N = w + F1sin(Ø) + kxsin(Ø2)
g) N = wcos(Ø) + kx	h) N = -kxsin(Ø1) + mgcos(Ø2)	i) N = Fsin(Ø)

8. Now write F = ma equations for #7e - 7i, assuming that each is accelerating in the direction of the applied force, and that each surface has a coefficient of friction, µ.

e) F = ma Fcos(Ø) - µ(mg + Fsin(Ø) = ma	f) F = ma F1cos(Ø1) - kxcos(Ø2) - µ[F1(Ø1) + kxsin(Ø2) + w] = (w/g)a	g) F = ma wsin(Ø) - µ(wcos(Ø) + kx) = (w/g)a
h) F = ma mgsin(Ø2) + kxcos(Ø1) - µ[mgcos(Ø2) - kxsin((Ø1)] = ma	i) F = ma Fcos(Ø) - w - µ[Fsin(Ø)] = (w/g)a

These free-body diagrams are quite important, so try a few more.

9. Draw a free body diagram for a block sliding down a frictionless incline. Determine its acceleration.

Perpendicular to the plane, N = mgcos(Ø) so that there is no net force in the perpendicular. Parallel to the plane the unbalanced force is mgsin(Ø). So: F = ma mgsin(Ø) = ma a = gsin(Ø)

10. A Hostess Ding Dong rests on a frictional incline. Draw its free body diagram.

Here the frictional force, f, is probably not equal to µN, unless the ding dong is about to slip: Note that the forces must add to zero because acceleration is zero.

11. A banana placed on a frictional incline accelerates as it slides. Draw its free body diagram.

N = mgcos(Ø) Ff = µN Ff = µmgcos(Ø) because block slips F = mgsin(Ø) - µmgcos(Ø) = net force

12. The banana in problem 11 is now skidded up the same incline so that it quickly comes to rest. Draw the forces acting on it.

N = mgcos(Ø) Ff = µN Ff = µmgcos(Ø) F = mgsin(Ø) + µmgcos(Ø) = net force

13. A train of 3 masses is pulled along as shown on a frictionless surface. Draw FBD's for each mass and calculate the tensions in the strings.

First identify the tensions (F = ma applied to each block): 30 - T1 = 8a T1 - T2 = 5a T2 = 13a Acceleration can be found by considering the whole string as one unit: 30 = (8 + 5 + 13)a a ~= 1.154 m/s2 To find the tensions in each string, simply plug in the acceleration into each force equation: T1 = 30 - 8(1.154) T1 ~= 20.8 nt T2 = T1 - 5a T2 = 20.8 - 5(1.154) T2 ~= 15.0 nt

14. Express the tension, T, in terms of the other variables.

a)	b)
a) F = ma T - µmg = ma T = µmg + ma T = m(µg + a)	b) F = ma T - mg = ma T - p(lwh)g = plwha T = plwh( g + a)

15. A wrecking ball is photographed in the midst of a swing. If, at that instant, the cable is at angle q, what does its free body diagram look like?

Tension pulls wrecking ball up while mass pulls wrecking ball down

I will frequently use less-than-complete free body diagrams when forces in one plane balance out. Thus the forces on a mass sliding down an incline are:

	and they can become.
A mass pulled along a frictional floor is changed from the complete:
	to the abbreviated

16. A 6 kg mass is pulled down a 42° incline by a 12 nt force. If the incline is of µ = .4, what is the acceleration?

F = ma 12 + mgsin(Ø) - µmgcos(Ø) = ma 12 + (6 * 9.8)sin(42) - .4[(6 * 9.8)cos(42)] = 6a a ~= 5.64 m/sec2

17. Two masses tethered together accelerate down an incline. Find the following. Include simplified FBD's for each mass.
a) the acceleration
b) the tension in the string

a and b) F = ma 16 + mgsin(Ø) - T - µmgcos(Ø) = ma 16 + (11 * 9.8)sin(35) - T - .2[(11 * 9.8)cos(35)] = 11a 60.17 - T = 11a F = ma T + mgsin(Ø) - µmgcos(Ø) = ma T + (19 * 9.8)sin(35) - .6[(19 * 9.8)cos(35)] = 19a T + 15.28 = 19a Add the two simplified equations. Notice that the T's cancel: 60.17 + 15.28 = (11 + 19)a a ~= 2.52 m/sec2 Plug in the acceleration into either equation to find T: T = 19a - 15.28 T = 32.5 nt

18. Three weights are accelerated upward as shown. Find the tensions in each string.

For the entire unit, F = ma. So we can write one large equation: F = ma 116 - (20 + 38 + 17) = [(20 + 38 + 17)/32]a a ~= 17.49 ft/sec2 Plug in the acceleration into an equation for the forces acting on a chosen weight to find the tension: 116 - 20 - T1 = (20/32)(17.49) T1 ~= 85.1 lb T1 - 38 - T2 = (38/32)(17.49) 85.1 - 38 - T2 = (38/32)(17.49) T2 ~= 26.3 lb

19. Find the volume of the block.

Note: weight = mg = pVg = 4.3 * V * 32 = 137.6V F - Fg- Ff = ma 57 - 137.6Vsin(33) - .3[(137.6V)cos(33)] = 4.3V(3.9) 57 = V[137.6sin(33) + .3[137.6cos(33)] + 2.3 * 3.9] V ~= .451 ft3

20. What force will accelerate the mass at 12 ft/sec²?

F + Fg- Ff = ma F + (5 * 32)sin(27) - .4[(5 * 32)cos(27)] = 5 * 12 F ~= 44.4 lb

21. If a 2 kg mass slips at 1.3 m/sec² down a 32° incline , what is the coefficient of friction?

Fg- Ff = ma (2 * 9.8)sin(32) - µ[(2 * 9.8)cos(32)] = 2 * 1.3 µ = [(2 * 9.8)sin(32) - (2 * 1.3)]/[(2 * 9.8)cos(32)] µ ~= .468

22. What is the acceleration of the mass?

F - mgsin(Ø) - µmgcos(Ø) = ma a = [F - mgsin(Ø) - µmgcos(Ø)]/m

23. Find the acceleration. Remember that forces applied at angles change the normal force.

First calculate the frictional force: Ff = N Ff = .2[6 * 9.8 + 22sin(33)] Ff ~= 14.2 nt F = ma 22cos(33) - 14.2 = 6a a ~= .708 m/sec2

24. Here are a couple more.

a)	b)
a) F = ma F - µN = ma 11cos(25) - µ[47 - 11sin(25)] = (47/9.8)(1.3) µ ~= .0882	b) F = ma F - µN = ma Fcos(37) - .2[80 + Fsin(37)] = (80/9.8)(2.6) F[cos(37) - .2sin(37)] - .2(80) = (80/9.8)(2.6) F ~= 54.9 nt

25. Write an equation relating the variables.

a)	b)
a) F = ma Fcos(Ø) - µ[mg + Fsin(Ø)] = ma	b) F = ma F - Ff - Fg = ma Fcos(Ø2) - µ[mgcos(Ø1) - Fsin(Ø2)] - mgsin(Ø1) = ma

When friction is introduced into the problem it does not fundamentally change the way it is solved. So try some two-mass systems using the same basic method you did in Forces in Equilibrium.

26. Find the acceleration of the system.

Assign a tension T to the string and write the two force equations: equation #1 F = ma T - µmg = ma T - .3(11 * 9.8) = 11a equation #2 F = ma mg - T = ma 5 * 9.8 - T = 5a Add the two equation together: T - .3(11 * 9.8) = 11a 5 * 9.8 - T = 5a --------------------------- 5 * 9.8 - .3(11 * 9.8) = 16a a ~= 1.04 m/sec2

The introduction of a second mass makes it important to be consistent with the signs of the accelerations. While the choice of positive and negative directions is discretionary, I strongly recommend the direction of the velocity be chosen as the positive direction. This eliminates all kinds of confusion that seems to develop otherwise.

27. Again, find the acceleration.

Assign a tension T to the string and write the two force equations: equation #1 F = ma T - µmgcos(Ø) - mgsin(Ø) = ma T - (2 * 32)sin(28) - .2[(2 * 32)cos(28)] = 2a equation #2 F = ma mg - T = ma 1 * 32 - T = 1a Add the two equation together: T - (2 * 32)sin(28) - .2[(2 * 32)cos(28)] = 2a 1 * 32 - T = 1a ------------------------------- -(2 * 32)sin(28) - ..2[(2 * 32)cos(28)] + 1 * 32 = 3a a ~= -3.12 ft/sec2 We have a problem of interpretation. The block has been assumed to be sliding up the incline. Thus we can conclude it is sliding up but slowly down. In other words, the block has a positive velocity but its acceleration is negative.

28. You just can't practice enough. By now they should be coming fairly easily.

a)	b)
a) equation #1 T - .3(5 * 9.8) = 5a equation #2 (4 * .5)32 - T = (4 * .5)a Add equations: 2 * 32 - .3(5 * 9.8) = 7a a ~= 2.29 ft/sec2	b) equation #1 T - µ(112) = (112/9.8)(2.3) equation #2 50 - T = (50/9.8)(2.3) Add equations: 50 - µ(112) = [(112 + 50)/9.8](2.3) µ ~= .107

29. A 8 kg mass is pulled up a 40° incline by a 6 kg mass hanging over the side of a ledge. If the coefficient of friction is .2, what is the acceleration?

equation #1 T - .2[(8 * 9.8)cos(40)] - (8 * 9.8)sin(40) = 8a equation #2 6 * 9.8 - T = 6a Add: 6 * 9.8 - .2[(8 * 9.8)cos(40)] - (8 * 9.8)sin(40) = 14a a ~= -.258 m/sec2 Thus, the block is sliding up, but slowing to a stop

30. Denton Fender, massing 4.7 slugs, with coefficient of friction 0.4, was sitting on a 37° slope until he was struck in the back by a 7.5 slug block with coefficient of friction 0.1. Now they both accelerate down to the slope to an uncertain future. What is their acceleration, and what is the force does the block exert on Denton?

For Denton Fender: F = ma (4.7 * 32)sin(37) + f - .4[(4.7 * 32)cos(37)] = 4.7a For the block: F = ma (7.5 * 32)sin(37) - f - .1[(7.5 * 32)cos(37)] = 7.5a Adding the equations to find the acceleration: (4.7 + 7.5)* 32sin(37) - [(.4 * 4.7) + (.1 * 7.5)] * 32cos(37) = (4.7 + 7.5)a a ~= 13.7 ft/sec2 Plug in the acceleration into one of the first equations to find f: (4.7 * 32)sin(37) + f - .4[(4.7 * 32)cos(37)] = 4.7(13.7) f ~= 22.2 lb

The addition of a third mass makes the problem complex, but not really any more difficult. You must remember, though, the tensions in the two strings will not be equal.

31. Find the acceleration.

F = ma equation #1 T1 - m1g = m1a equation #2 T2 - T1 - µm2g = m2a equation #3 m3g - T2 = m3a Add: m3g - µm2g - m1g = (m1 + m2 + m3)a a = (m3 - µm2 - m1)g/(m1 + m2 + m3)

32. Again, find the acceleration.

F = ma equation #1 T1 - m1g = m1a equation #2 T2 + m2gsin(Ø) - T1 - µm2gcos(Ø) = m2a equation #3 m3g - T2 = m3a Add: m3g - m1g + m2gsin(Ø) - µm2gcos(Ø) = (m1 + m2 + m3)a a = [m3 - m1 + m2sin(Ø) - µm2cos(Ø)]g/ (m1 + m2 + m3)