Generally in the history of science there have been long stretches of time where little seemed to be happening. Individuals would gather bits of information, some related, some apparently not. In this process the stage would be set for a great synthesis a fusion of seemingly unrelated principles into one or a few universal laws.

By the middle of the 17th century such a moment was imminent. The work of Galileo on moving bodies, experimental work done on Earth, was united with the clockwork movements of the planets discovered by the mathematician Johannes Kepler. The final synthesis produced a single, unified world-view that stretched from Earth to the most distant reaches of space. It seemed to have the power to explain everything from an apple falling from a tree to a wave breaking on a beach, to a distant galaxy turning slowly in the emptiness of space.

The man who achieved this was Isaac Newton born on Christmas day in 1642, the year of Galileo's death. It would be difficult to overstate the importance of his work. Besides its impact on science, it touched philosophy, religion, even art. It carried our species from an imprecise view of nature, where events happen somewhat arbitrarily and unpredictably, to a clean mathematical cause-and-effect view of the universe. It was an enormous revolution in thinking. In problems of interplanetary navigation, Newton's laws are used today and require modification only in rare and extreme circumstances.

Newton stated three laws of motion. The first was taken from Galileo, that a body at rest would remain at rest, and a body in motion would continue in uniform motion, unless acted on by an outside force. You can observe this by watching a marble roll across a hard surface onto a carpet. It will keep rolling a long time on the hard surface until the outside force of the carpet's fibers brings it to rest. Or in shaking catsup out of a bottle -- you get the catsup and bottle moving and then quickly pull back the bottle, anticipating that the catsup will continue in a forward direction onto your hamburger. You can see that a body tends to stay at rest staying at rest by laying a banana on a piece of paper and then jerking the paper out from under it.

Watch around you for examples of the first law, but for now we will concentrate our attention on Newton's second law of motion. It can be observed from simple experiment that if we exert an unbalanced force on an object, it will not only move, but accelerate. The rate of acceleration can be shown to increase proportionally to the unbalanced force. Thus, if we plot our data, we get something like the graph to the right.

Mathematically this can be stated F = ka where k is some constant. The proportionality constant k turns out to be dependent on the amount of matter being accelerated, so the equation becomes F=ma Force equals mass times acceleration.

One must pause for a moment in quiet wonder at the simple profundity of this statement. It does more than say that force and acceleration are related by mass. It tells us that nothing else -- not density, composition, color, shape, speed, or anything -- has ANY effect on this interaction. With it we unite forces and motion and gain the ability to understand and predict the motion of objects from atoms to galaxies. Always remember F is the net or unbalanced force. If the forces acting on a mass are unbalanced, the mass will accelerate. If a mass is seen to accelerate, there must be unbalanced forces. It's that simple.

Using Newton's second law, we may now define the unit of force: Place a 1 kg mass on a frictionless table and apply just enough force to accelerate it at 1 m/sec². The force you apply will then be, by definition, 1 newton. Look at the units of this force:

F = ma = (kilograms)(meter/second²), so a newton is a kg-m/sec².

Before we get into dynamics, let's look at the concept of net force or unbalanced force. To do this we will study all the forces acting on the body.

First, to get you used to this important relationship, solve a few basic problems.

1. A 6.4 kg mass receives a force of 23 nt. What is its resulting acceleration?

F = ma 23 = 6.4a a ~= 3.59 m/sec2

2. What force is required to accelerate a 0.744 slug mass at 3.6 ft/sec²?

F = ma F = 0.744 slug * 3.6 ft/sec2 F~= 2.68 lb

3. Two forces of 3800 dynes and 1090 dynes act oppositely as shown on a 17 gm gumdrop. What acceleration does the gumdrop undergo?

F = ma 3800-1090 = 17a a ~= 159 cm/sec2

Let's take a moment to review units in the 3 systems used in this text, the SI (international system), the cgs (cm-gm-sec system), and the English system.

	length	mass	time	acceleration	force		force
SI	meter	kilogram	second	m/sec2	kg-m/sec2	=	newton
cgs	centimeter	gram	second	cm/sec2	gm-cm/sec2	=	dyne
English	foot	slug	second	ft/sec2	slug-ft/sec2	=	pound

You can state your weight quite properly as, say, 135 slug-ft/sec². Mass and force are in bold face because it is so important that you know which is which.

Note -- It takes 1 pound to accelerate 1 slug at 1 ft/sec²
Note -- 1 dyne accelerates 1 gm at 1 cm/sec².

4. Find the unknown as each object accelerates on a frictionless surface.

a)	b) >	c)
a) F = ma 38 - 13 = m*5 m = 5.00 slug	b) F = ma 32 - 21 = .075a a = 160 ft/sec2	c) F = ma 43 - Ff = (8.8)(3.1) Ff = 15.7 nt

5. If two or more forces act, the F in F=ma is the NET force. Find the unknown mass, acceleration, and force of friction.

a)	b)	c)
a) F = ma 38 - 13 = m*5 m = 5.00 slug	b) F = ma 32 - 21 = .075a a = 160 ft/sec2	c) F = ma 43 - Ff = (8.8)(3.1) Ff = 15.7 nt

6. A 4.0 kg chihuahua released on Earth accelerates downward at 9.8 m/sec². What force must be acting on it?

F = ma F = (4.0 kg)(9.8 m/sec2) F = 39.2 nt

7. A 150 lb frictionless crate of bananas is pushed along by a 30 lb force. What is its acceleration? (careful: it is F = Ma)

F = ma 30 = (150/32)a a ~= 6.40 ft/sec2

8. Rico Alcohol weighs 152 pounds and holds a small, 25 pound rocket as shown. While adjusting the Velcro on one of his roller blades, he inadvertently ignites the rocket which exerts a 96 lb thrust. What is his resulting acceleration?

F = ma 96 = [(152 + 25)/32]a a ~= 17.4 ft/sec2

Note how naturally W = mg follows from F = ma.

9. I have again mixed together weight and mass. Be careful to use them appropriately. Assume Earth's gravity.

a)	b)	c)
a) F = ma 35 = (370/9.8)a a ~= .927 m/sec2	b) F = ma 72 - 50 = (50/32)a a ~= 14.1 ft/sec2	c) F = ma F = [(8.5 * 105)/980] * 160 F ~= 1.39 * 105 dyne

10. Pierre O'Boots accelerates a 0.4 slug banana with a cord tied around its center by lifting on the cord with a force of 20 pounds. Draw the forces acting on the banana and calculate its upward acceleration.

F = ma 20 - (.4 * 32) = .4a a = 18.0 ft/sec2

Think like this:

11. Farah Flying wishes to raise a 13 kg case of Cheez-Whiz at 4 m/sec². What upward force must she exert?

F = ma F - 13 * 9.8 = (13)(4) a ~= 179 nt

Remember, Farah's upward force must exceed the downward force of gravity by an amount m * a.

12. All the objects below hang from strings in Earth's gravity.Find the unknown in each case:

a)	b)	c)	d)
a) F = ma 3 - (.5 * 9.8) = .5a a ~= -3.80 m/sec2 Down!	b) F = ma 340 - 8 * 32 = 8a a ~= 10.5 ft/sec2	c) F = ma 3.3 * 104 - 17 * 980 = 17a a ~= 961 cm/sec2	d) F = ma 40 - 2 * 32 = 2a a = -12.0 ft/sec2

When time-measurement devices were crude, accurate determinations of the acceleration of gravity were difficult to accomplish. A man named Atwood came up with a device to slow the process down to manageable speeds. It has become known as the Atwood Machine, and looks like this: To keep our calculations simple for the moment, we will assume the pulley to be massless and frictionless.

13. Consider an Atwood machine with masses of 5 kg and 7 kg. Write an equation of motion for each mass and calculate the acceleration.

step 1) Assign a tention T to the string step 2) Diagram the forces acting on each mass step 3) create equations of motion as before: F = ma T - 5 * 9.8 = 5a and 7 * 9.8 - T = 7a Add the two equations. The T's cancel: T - 5 * 9.8 = 5a 7 * 9.8 - T = 7a ----------------- 7 * 9.8 - 5 * 9.8 = 5a + 7a a ~= 1.63 m/sec2

Think like this:

14. Theresa Crowd, with a mass of 52 kg, notes that when she grabs on to a rope hanging in her barn, she falls at 4.4 m/sec². What is the counter-mass on the other end of the rope?

step 1) Assign a tention T to the string step 2) Diagram the forces acting on each mass step 3) create equations of motion as before: F = ma 52 * 9.8 - T = 52 * 4.4 and T - m * 9.8 = m(4.4) Add the two equations. The T's cancel: 52 * 9.8 - T = 52 * 4.4 T - m * 9.8 = m(4.4) ----------------- 52 * 9.8 - 9.8m = 4.4m + 52 * 4.4 m = 19.8 kg

15. What counter-mass will cause the 87 kg mass to fall 2.6 m in 1.7 sec?

First find the acceleration: S = 1/2at2 2.6 = (1/2)a(1.7)2 a = 1.80 m/sec2 Then motion equations: F = ma 87 * 9.8 - T = 87(1.80) and T - m * 9.8 = m(1.80) 696 = 11.6m m = 60.0 kg

16. Beth Wishes finds she can accelerate a tiny 290 gm car by attaching a 120 gm mass as shown. What is the resulting acceleration?

For the car: F = ma T = 290a For the mass: F = ma 120 * 980 - T = 120a Adding the two equations: T = 290a 120 * 980 - T = 120a a ~= 287 cm/sec2

Forces can come from a number of different sources. Some are easier than others to deal with, and we will now take a diversion to study the simple way in which springs exert forces. Imagine a spring hanging vertically next to a ruler. A 5 nt weight is placed on the end, and the spring stretches 30 cm. In the diagram to the right, indicate how far the spring will stretch as additional 5 nt weights are added to it. The interesting, and somewhat surprising, thing is that the simplest answer is correct. The spring will stretch an additional 30 cm for each additional 5 nt weight placed on it. This calls for an experimental test, and if you doit you will get results somewhat like this:

Careful: if you load so many weights on it that the graph ceases to be straight, you will have "exceeded the elastic limit" and the spring will be permanently deformed. You have probably seen a few of these in your experience.

17. Consider a spring that is stretched 8 cm when a 13 nt force is applied. How far will it be stretched when 26 nt is applied?

Since stretch is proportional to force: (8cm / 13 nt) = (x / 26 nt) x = 16 cm

Because of this proportionality, we can say F = kx where k is called the "stiffness constant" of the spring and is a property that differs from spring to spring. Try a few problems with it.

18. A 7.3 kg mass is placed on a spring with a stiffness constant of 34 nt/cm. How much does this stretch the spring?

F = kx (7.3 * 9.8)nt = (34 nt/cm) * x x ~= 2.10 cm

You may have noted with consternation that I have mixed newtons and centimeters, normally a no-no in physics. I can get by with it here as long as the stretch is measured in centimeters. Check it out. (nt/cm)(cm) = nt. Another way to view it is by converting the stiffness constant to 3400 nt/m, and the answer becomes .0210 m = 2.10 cm. The numbers are awkward to work with, so as long as you're careful, you can be a little loose.

19. Mel Adjusted, in a futile attempt to strengthen his pectoral muscles, stretches a spring exercise device 0.73 meters by exerting a force of 177 nt. What is the stiffness constant of the device?

F = kx 177 nt = k * .73 m k ~= 242 nt/m

20. The diagrams below show various objects accelerated by springs. Find the unknown mass, stiffness constant, stretch, or acceleration.

a)	b)	c)	d)
a) F = ma 26 * 4 = 8.2a a ~= 12.7 m/sec2	b) F = ma k * 6 = (22/9.8)(4.2) k ~= 1.57 nt/cm	c) F = ma m * 9.8 - (8400)(.05) = m * 2.9 m(9.8 - 2.9) = 8400 * .05 m ~= 60.9 kg	d) F = ma 6 * 3 - 12 = (12/32)a a = 16.0 ft/sec2

21. Find the stretch of each spring.

a)	b)
a) The tension in the spring is kx, or in this case, 6x. F = kx 2(6x) = 25 x ~= 2.08 cm	b) Watch the angle! F = kx 13xSin(39) = 2 * 32 x ~= 7.82 in

22. Find the stretch in each spring

a)	b)
a) Work your way through the problem, assigning values. If you need help with pulleys, see Forces in Equilibrium F = kx T = 6x 12x + 6x = 16 * 9.8 x ~= 8.71 cm	b) F = kx 5.5x + 11x = 5.5x - 8 + 40 x ~= 2.91 cm

23*. Two springs, one of constant 8 nt/cm and the other 4 nt/cm, support a weight as shown. They are positioned so the weight stretches each equally. How much stretch will a 40 nt weight produce? What is the effective combined stiffness constant of the two springs? Find the stretch in each spring.

Say each spring stetches a distance x, then we could say: 8x + 4x - 40 nt = 0 (up and down forces are equal to zero) x = [40 nt/ (12 nt/m)] x = 3.33 cm F = kx k = (40 nt/ 3.33 cm) k ~= 12 nt/cm

24*. Two springs, again of constants 8 nt/cm and 4 nt/cm support a 40 nt weight as shown.
a) How much will the 8 nt/cm spring stretch?
b) How much will the 4 nt/cm spring stretch?
c) What is their combined stiffness constant?

a) F = kx x1 = F1/k1 x1 = 40 nt / (8 nt/cm) x = 5.00 cm

b) x2 = F2 / k2 x2 = 40 nt / (4 nt/cm) x = 10.0 cm

c) xT = x1 + x2 xT = 5.00 cm + 10.0 cm xT = 15.0 cm kT = F / xT kT = 40 nt / 15 cm kT ~= 2.67 nt/cm

And now back to some dynamics.

25. A 0.17 kg Twinky hangs on a 2.40 nt/m spring. Rolland Butter pulls the Twinky down an additional 11 cm and releases it. What is its initial acceleration?

The spring is initially stretched by the weight of the Twinky. Additional stretch will cause additional force to be exerted, which will go into accelerating the Twinky: F = kx F = (2.40 nt/m)(.11 m) F = .264 nt F = ma .264 nt = (.17 kg)a a ~= 1.55 m/sec2

A short time after release of the mass, the spring will be stretched less, will exert less force, and the Twinky will continue to accelerate upward, but at a lesser rate. It will begin to slow down only after it passes the neutral position it started at.

26. A 43 nt force presses on a 15 kg block at a 26° relative to the horizontal. What is its acceleration?

Since the vertical components of force balance out, only the horizontal component, 43cos(26), accelerates the block: F = ma 43cos(26) = (15 kg)a a ~= 2.58 m/sec2

27. Find the unknown in each situation.

a)	b)	c)	d)
a) F = ma Fcos(35) = (4.6 slug)(2.9 ft/sec2) F ~= 16.3 lb	b) F = ma 42 * 6cos(21) = (90/9.8)a a ~= 27.4 m/sec2	c) F = ma 20cos(27) + 5 * 5cos(57) = (44/32)a a ~= 22.9 ft/sec2/b>	d) F = ma 59cos(33) - 2.6 * 9.8 = 2.6a a ~= 9.23 m/sec2

28. A 9 kg mass is placed on a 38° incline and released. What is its acceleration?

The component of force down the slope is mgsin(Ø): F = ma 9 * 9.8sin(38) = 9a a ~= 6.03 m/sec2

Take a look at this component-of-force thing. When a block sits on an incline, we can look at the forces on it in two ways:

or

To calculate the magnitude of the two components, study this:
Having trouble swallowing the equality of the two angles? Check out these diagrams.
So now look at what this says about the magnitude of the forces:
Since the incline presses perpendicularly against the block just enough to cancel out the WcosØ, the only remaining force acting is WsinØ.

29. A 6 pound force pushes a 20 pound weight down a 19° frictionless slope. What is its acceleration?

F = ma 6 + 20sin(19) = (20/32)a a = 20.0 ft/sec2

30. Find the unknown in each case.

a)	b)	c)	d)
a) F = ma 11 * 6 - 80sin(27) = (80/9.8)a a ~= 3.64 m/sec2	b) F = ma 30 + m * 9.8sin(22) = 7m 30 = m[7 - 9.8sin(22)] m ~= 9.01 kg	c) F = ma k * 9 - 3 * 32sin(57) = 3 * 5 a ~= 10.6 ft/sec2	d) F = ma 17 + 9 * 9.8sin(18) - 6 = 9a a ~= 4.25 m/sec2

31. A 23 kg mass is pulled by a spring (spring constant 9 nt/cm) stretched 13 cm and pushed by a 43 nt force pointed at a 40° angle. If the incline is banked at a 28° angle, what is the acceleration?

F = ma 43cos(40) + (9 * 13) - 23 * 9.8sin(28) = 23a a ~= 1.92 m/sec2

32. And this...just write a true F = ma equation, but don't solve it for anything.

F = ma Fcos("theta 1") + mgsin("theta 2") + kx = ma

We have, up to now, dealt only slightly with the interaction of one mass with another, yet forces between masses are of fundamental importance in figuring out the real world. Try some more.

33. Daring space traveler Andy Jestion notes 48 and 35 kg masses being hauled behind a small space vehicle at 3.6 m/sec².

a) Draw the forces acting on each mass. b) Write the F = ma equation for each. c) Solve for the two tensions in the tethers.

a) Force diagrams:

b) F = ma T1 = 48(3.6) T2 - T1 = (35)(3.6)

c) Add the equations to cancel T1: T1 = 48(3.6) T2 - T1 = (35)(3.6) Solve for T2: T2 = 299 nt Plug T2 into the second equation and solve for T1: 299 - T1 = (35)(3.6) T1 = 173 nt

In this last problem, I introduced variables T₁ and T₂. In doing this I invoked the fundamental principle of algebra, which is rarely explained to students in their algebra classes. So, lest you wait any longer, HERE IT IS:

Thou Shall Replace Ignorance with a Letter and Proceed.

It is remarkable how powerful this idea is. Rather than stopping to puzzle over an unknown, just slap a symbolic name on it and continue to write true things.

Try that idea on a few more situations.

34. Three masses are tethered together as shown. The tension in the lead string is 44nt.

a) What is the acceleration of the system? b) What are the other tensions?

a) Create equations for each mass: F = ma 44 - T2 = 7a T2 - T1 = 4a T1 = 6a Add the equations togeather (T1 and T2 cancel): 44 = (7 + 4 + 6)a a ~= 2.59 m/sec2

b) Plug in the acceleration to find the tensions: 44 - T2 = 7(2.59) T2 ~= 25.9 nt T1 = 6(2.59) T1 ~= 15.5 nt

35. Two masses are separated by a spring of 244 nt/m stiffness, and a 90 nt force is applied as shown. Again, draw the forces on each mass and determine how far will the spring compress.

Create the two equations: F = ma 90 - 244x = 12a 244x = 7a Add the equations togeather: 90 = 12a + 7a a ~= 4.74 m/sec2 so, 244x = 7(4.74) x ~= .136 m

36. 160 lb Malcolm Tent is shoved forward by a force of 75 lb. He is holding a 13 lb birthday present. What horizontal force acts between Malcolm and the present? (One might also ask what force acts between Malcolm and the past.)

Here's one approach: F = ma 75 = [(160 + 13)/32]a a ~= 13.9 ft/sec2 So, it could be solved like: F = ma 75 - F = (160/32)(13.9) F ~= 5.64 lb

37. Given the masses and accelerations, find the tensions.

a)	b)	c)	d)
a) F = ma T - 11 * 9.8 = 11 * 3 T = 141 nt	b) F = ma but mg = (4/3)(3.14159)(6/12)3(6)(32) mg = 100.5 lb 100.5 - T = (3.14 slug)(7) T ~= 78.5 lb	c) F = ma T - (85 * 980) = 85 * 110 T ~= 9.27 * 104 dyne	d) F = ma 9.8 * 8 - T = 8 * 9.8 T = 0 nt

38. A 73 kg man wishes to lower himself from a 36 m high apartment that is burning out of control. He has a rope which will support only 700 nt.
a) At what acceleration can he lower himself without the rope breaking?
b) With what speed will he hit the ground?

a) F = ma a = F/m [(73 * 9.8) - 700]/73 a ~= .211 m/sec2

b) v2 = 2as 2(.211)(36) = 15.2 v ~= 3.90 m/sec

39. A 45 kg crate of pickled herrings is to be rapidly lifted to 13.0 m from the ground. What tension should be applied to the rope to lift the crate in 4.0 sec?

First find the acceleration: S = (1/2)at2 a = 2s/t2 a = (2 * 13)/42 a ~= 1.625 m/sec2 But F=ma T - (45 * 9.8) = (45)(1.625) T ~= 514 nt

40. A macho-crazed mountaineer slides down a rope. If his weight is 815 nt, and he applies a constant force of 55 nt to the rope, what is his downward acceleration?

F = ma a = F/m a = (815 - 55)nt/(815/9.8)kg a ~= 9.14 m/sec2 Note that the resistance to acceleration comes from an objects mass, not its weight. Remember that F = ma

41. Two small rocket engines of 8 nt and 14 nt thrust are attached to a 2 kg organically grown grapefruit as shown. With what acceleration, and in what direction will the grapefruit move?

Break the problem into vertical and horizontal components: F = ma a = F/m V) ay = Fy/m ay = 8 nt/2 kg ay = 4 m/sec2 H) ax = Fx/m ax = 14 nt/2 kg ax = 7 m/sec2 Using trig, we can find the the vector sum of the two components (hypotenuse) and the angle. a2 = b2 + c2 a ~= 8.06 m/sec2 tan(Ø) = 4/7 Ø = 29.7°

42. Now find the accelerations and directions of these if they are in weightless space.

a) F = ma a = (2002 + 3002)1/2/80 a = 4.51 cm/sec2 tan(Ø) = 200/800 Ø = 33.7°

b) F = ma a = (302 + 372)1/2/1.4 a = 34.0 ft/sec2 tan(Ø) = 37/30 Ø = 51.0°

c) F = ma a = (122 + 212)1/2/[(4/3)(3.14159)(.14)3(348)] a = 6.05 m/sec2 tan(Ø) = 21/12 Ø = 60.3°

43*. Think back to how we solved statics problems when you do these. Ignore gravity. a) b)

a) H) Fx = 6cos(50) Fx = -4.14 nt V) Fy = 6sin(50) Fy = 4.60 nt F = ma a = (-4.142 + 4.602)1/2/7 a = .884 m/sec2 tan(Ø) = 4.6/4.14 Ø = 48.0°

b) H) Fx = 20cos(23) - 8cos(50) Fx = 13.27 lb V) Fy = 20sin(23) + 8sin(50) Fy = 13.9 lb F = ma a = (13.272 + 13.92)1/2/2 a = .9.61 ft/sec2 tan(Ø) = 13.9/13.27 Ø = 46.3°

Keep in mind, when applying Newton's second law, that the F is the net, or unbalanced force acting on the mass.

44. A 65 kg man on a skateboard slows from 2.0 m/sec to 0.8 m/sec in 4.5 sec. What is his acceleration? What frictional force is slowing him?

a) a = change(v)/change(t) a = (.8 - 2.0)/4.5 a ~= -.267 m/sec2

b) F = ma F = (65 kg)(-.267 m/sec2) F ~= -17.3 nt

45*. A car starts from rest, 50 ft from a cliff. It accelerates at 19 ft/sec² until it goes over the cliff to the valley 85 ft below. How far from the base does it land?

a) First find the speed at the edge: v2 = 2as v2 = 2(19 ft/sec2)(50 ft) v ~= 43.6 ft/sec Next, find the time in the air: s = (1/2)at2 t = (2s/g)1/2 t = [(2 * 85)/32]1/2 t ~= 2.30 sec Finally, find the distance from the base: s = vt s = (43.6)(2.3) s ~= 100 ft

46*. A stunt man and his moped mass a combined 408 kg. If he plans to jump the 17 m gap to the 4 m lower cliff on the opposite side,
a) How fast must he be going when he takes off?
b) How fast must his motorcycle accelerate?
c) What force must the wheels supply?

a) V) s = (1/2)at2 t = (2s/g)1/2 t = [(2 * 4)/9.8]1/2 t = .904 sec H) s = vt v = s/t v = 17 m/.904 sec v ~= 18.8 m/sec

b) v2 = 2as a = v2/2s a = 18.82/(2 * 12) a ~= 14.7 m/sec2

c) F = ma F = (408 kg)(14.7 m/sec2) F = 6.00 * 103 nt