We have completed, for now, the study of kinematics. This will give us the basis for studying the movement of matter. Now we turn to forces, which cause the movement of matter we have studied.

A force is a difficult thing for a physicist to define because it is so fundamental that there are no terms more fundamental with which to explain it. Although a definition in terms of motion will be provided in the next chapter, for now it is probably best to just think of a force as you always have a push or a pull on something. The units of force are familiar, pounds in the English system, and Newtons in the international system.

In this chapter we will study forces in equilibrium. That is, forces which balance each other out, or are equal in all directions. Here's a more technical way to say it: for any line passing through an object, the sum of all forces acting one way parallel to the line will be exactly balanced by the sum of the forces acting oppositely.

Think about it. For a body at rest this has to be true. If it weren't, there would be a greater force in one direction and the object would begin to accelerate.

In the following problems, sketch the diagrams and draw the forces acting on each object. Check frequently with the solutions so you will become familiar with the notation I use for various forces.

1. An object of weight W sits on a table.

The upward force of table balances the downward force of gravity. F = W

2. An object of weight W hangs on a string.

As before, T = W (They must be equal, or the weight will start to accelerate)

3. Two objects weighing W₁ and W₂ are stacked on a table.

F1 = W1 F2 = W1 + W2

4. Two weights hang by a string.

W2 = T2 W1 + W2 = T1 NOTE: T2 pulls up on w2 and down on w1.

5. A skydiver falls at terminal velocity. (That is, air drag is enough to keep him from falling any faster.)

6. A block is pushed along a frictional table at constant speed.

F2 = Ff F1 = W

7. Find F.

up forces = down forces so 20 nt + F = 30 nt F = 10 nt

8. Find T in newtons.

38 nt = 3 lb * (4.45 nt / 1 lb) + T T = 24.7 nt

9. Find F_x and F_y.

H) Fx = (30 nt * (1 lb / 4.45 nt) + 8 lb Fx = 14.7 lb V) 6 lb + Fy = 8 lb + 3 lb Fy = 5.0 lb

10. Find F_x and F_y.

H) Fx = 25 cos32 Fx = 21.2 lb V) Fy = 25 sin32 Fy = 13.3 lb

11. Again, F_x and F_y in newtons.

H) Fx = 36 cos40 Fx = 27.6 nt V) Fy + (2 lb * (4.45 nt / 1 lb)) = 36 sin40 Fy = 14.2 nt

12. One last time, F_x and F_y.

H) Fx = 27 cos50 Fx = 17.4 nt V) Fy = 27 sin50 + 6 Fy = 26.7 nt

As in kinematics, a two-dimensional problem is really two separate problems coexisting under the guise of a single problem. If you approach it as two problems a horizontal one and a vertical one, you will find the 2-D problems no harder than 1-D. (More complex, maybe, but no harder.)

13. First find F₁, then find F_y.

H) F1 cos33 = 20 nt F1 = 20 / cos33 F1 = 23.9 nt V) Fy + 8 = F1sin33 Fy = (23.9) sin33 - 8 Fy = 5.0 nt

14. Again, F₁ and then F_x.

V) 32 nt = (3.2 lb * (4.45 nt / 1 lb)) + F1 sin52 F1 = 22.5 nt H) Fx = F1 cos52 Fx = 22.5 cos52 Fx = 13.9 nt

15*. Find F₁ and F_y.

H) F1 cos48 = 40 cos30 F1 = 40 cos30 / cos48 F1 = 51.8 nt V) F1 sin48 + 40 sin30 = 16 + Fy Fy = 42.5 nt

16*. Find F_x and q.

V) (5 lb * (4.45 nt / 1 lb)) sin63 = 8 + 18 sinØ sinØ = .657 Ø = 41.1 H) Fx = (5 * 4.45) cos63 + 18 cosØ Fx = 23.7 nt

17. Find T and q.

Vertical: T sinØ = 18 nt T = 18 / sinØ Horizontal: T cosØ = 6 nt T = 6 / cosØ Equating the T's: (18 / sinØ) = (6 / cosØ) or 18 / 6 = sinØ / cosØ = tanØ Ø = 71.6° T = 18 / sin(71.6) T = 19.0 nt

18. Find T and q.

Vertical: T sinØ +53 sin24 = 20 + 15 cos22 so T = (20 + 15 cos22 - 53 sin24) / sinØ Horizontal: T cosØ + 15 sin22 = 53 sin24 so T = (53 cos24 - 15 sin22 / cosØ Equating, (20 + 15 cos22 - 53 sin24) / sinØ = (53 cos24 - 15 sin22) / cosØ tanØ = .29 Ø = 16.1° T = 44.6 nt

In problems 22-27, write down the horizontal and vertical equations that are true. DON'T attempt to solve anything at this stage, just write out the set-ups. W is the weight of the block.

19.

H) F1 = F2 cosØ V) F2 sinØ = W

20.

H) F2 cosØ = F3 cosØ3 V) F2 sinØ2 + F1 = F3 sinØ + W

21*.

W2: H) F2 cosØ2 = F1 cosØ1 V) F2 sinØ2 + F1 sinØ1 = W2 + T W1: H) F3 = F4 sin"Ø4 V) T = F4 cos"Ø4 + W1 Note that here Ø4 is measured from the vertical, so sin and cos are reversed.

22*.

20 nt: H) T1 cos46 = T2 cos15 V) T1 sin46 = 20 + T2 sin15 W: H) T2 cos15 = T3 cos50 V) T2 sin15 + T3 sin50 = W

23**.

W1: H) T1 cos40 = T3 + T4 cos40 V) T1 sin40 = W1 + T4 sin40 W2: H) T3 + T5 sin25 = T2 cos20 V) T2 sin20 = W2 + T5 sin25 W3: H) T4 sin40 = T5 sin25 V) T4 cos40 = T5 cos25 = W3

Now find numerical solutions or algebraic expressions for the following:

24*.

H) 15 cosØ = Fx V) 15 sinØ = 8 so sinØ = 8/15 Ø = 32.2° Fx = 15 cos(32.2) Fx = 12.7 nt

25.

H) F cosØ = 76 cos31 V) F sinØ + 76 sin31 = 85 tanØ = (85 - 76 sin31) / 76 cos31 Ø = 35.1° F2 = (76 cos31)2 + (85 - 76 sin31)2 F = 79.7 nt

26.

H) F1 cos52 = F2 cos40 V) F1 sin52 + F2 sin40 = 46 Solve (H) for F1 and substitute into (V) (F2 (cos40) / (cos52)) sin52 + F2 sin 40 = 46 F2 = 46 / ((cos40 / cos52) sin52 + sin40) F2 = 28.3 nt F1 = 28.3 (cos40 / cos52) F1 = 35.3 nt

27.

H) T1 cos62 = T2 cos21 V) T1 sin62 = 33 + T2 sin21 T1 = T2 (cos21 / cos62) [T2 (cos21 / cos62]sin62 = 33 + T2sin21 T2 = 23.6 lb T1 = 47.0 lb

28.

H) 30 cosØ1 = 20 cosØ2 V) 30 sinØ1 + 20 sinØ2 = 16 This is hard to analyze, but squaring both equations, adding, and using sin2Ø + cos2Ø = 1 will get you an answer.

The nature of balanced forces gives us insight into the way pulleys operate. Here are the rules of pulleys, or at least the rules for the massless, frictionless pulleys commonly used by physicists.

1) When a pulley is at rest, the string going over it will have equal tension on both sides. (This must be so, or the pulley would be jerked in one direction or the other.)
2) The upward forces acting on each pulley exactly balance the downward forces.

29. One end of a rope to a pulley is attached to the ceiling. What tension, T, must be supplied to the other end to lift the weight?

Because the bottom string has a weight of 80 lb, the whole pulley has a tension of 80 lb. Thus, if the tension is divided between two strings, each string has a tension of 40 lb. T = 40 lb

30. Find the tensions in each of the following:

a)	b)	c)	d)	e)
a) T = 11/2 T = 5.5 lb	b) T = 7 nt	c) T = 12 + 12 T = 24 nt	d) T = 15/2 T = 7.5 nt	e) T + 6 = 18 T = 12 lb

Now that everything is going smoothly with forces, I want to introduce a small wrinkle. If you pick up a rock on Earth, it will have a certain weight. This weight will not be the same if you carry the rock to the moon, or to Jupiter, or just float with it out in space.

The reason is obvious. The pull of gravity differs from place to place, so the same amount of mass will weigh more in a strong gravitational field, less in a weak one, and zero if there is no gravity present.

Without going more deeply at this time, let me just say that the two are related by the equation

W = mg weight = mass x gravity

The unit of mass in the international system is the kilogram (we'll come back to the gram later), the unit of mass in the English system is the slug. That's right, the slug.

31. What is the weight of a 12.6 kg mass on Earth?

W = mg W = (12.6 kg) (9.8 m/sec2) W = 123 nt note: a newton is just a kg-m / sec2

32. What is the tension is a string supporting a 4.6 slug mass? (Unless otherwise stated, assume you're on Earth)

T = W = mg T = (4.6 slug)(32 ft/sec2) T = 146 lb note: a pound is just a slug-ft / sec2

33. A block of wood of density 730 kg/m³ had dimensions 1.2 m by 0.4 m by 0.7 m. What is the tension in the string if it is lifted on the moon? (Our moon's gravity is 1.6 m/sec²)

W = mg = pVg W = (730)(.7 * .4 * 1.2)(1.6) W = 392 nt Remember, this is the moon were talking about!

34. Find the tensions in the pulley systems shown:

a)	b)	c)
a) The tension at the marked spot equals weight = .8 * 32 lb. T = 6.4 lb.	b) T = 5.0 nt	c) By symmetry, each string carries 15 nt, thus: T = 15 nt

On these problems, where the weight is held by multiple strings, you must work backward. Because there is asymmetry in each of the systems, you cannot assume that the two strings supporting the load have equal tensions. Begin at T, and express all other tensions in terms of T. Finally equate the sum of tensions to the weight and solve.

35. Find the tension in the string.

a)	b)	c)
a) T + T/2 = 30 lb (3/2)T = 30 T = 20 lb	b) 2T + T/2 = (18.7 * 9.8) 2.5T = 183 nt T = 73.2 nt	c) 2(T + 32) + 2T = 6 * 32 4T + 64 = 192 T = 32 lb

In reality the support cords would have to be placed assymetrically on the blocks to balance the forces acting on them.

Now for a little tougher one.

36*.

The two lines of logic converge at the marked position. FUP = FDOWN 8T + 16 = 44 + T - 2 T = 3.71 nt