We have just studied one kind of equilibrium, called translational equilibrium. The condition for translational equilibrium was that forces acting in one direction on an object be exactly balanced by forces acting oppositely. We again meet Harry Hamburger, demonstrating this principle by lifting a box.

Harry (as well as anyone else who has lifted a box) knows, however, that not only must he lift with the right force (F = W), but it must be exerted in the right place.

We come, then, to the problem of rotational equilibrium. You are probably familiar with the famous teeter-totter problem, where A. Orta who weighs 240 pounds must balance B. C. Enyu who weighs 80 pounds.

You probably know from experience that B must be three times as far from the pivot point as A is.

The general statement of this was discovered by Archimedes, about 2000 years ago. The principle reads: F₁L₁ = F₂L₂. That is, the twist on the teeter-totter in one direction must be balanced by an equal twist in the opposite direction, and the amount of twist is given by the product of the force and its distance from the pivot. Try some.

1.

a) F1L1 = F2L2 (30 nt)(1.8 m) = (45 nt)(x) x = 1.20 m

b) F1L1 = F2L2 (120 nt)(1.6 m) = (F)(3.0 m) F = 64.0 nt

2.

a) F1L1 = F2L2 (18 nt)(2.6 m) = (F)(5.7 m) F = 8.21 nt

b) F1L1 = F2L2 (F)(6.5 ft) = (33 lb)(6.5 ft + 2.1 ft) F = 43.7 lb note: remember to always calculate forces from pivot

c) F1L1 = F2L2 T(.4 m) = (150) nt)(2.1 m) T = 787 nt

Our argument needs some improvement. Although force times lever arm balances in the diagram below (80 nt x 2 m = 40 nt x 4 m), they obviously will not be in rotational equilibrium.

Think about it. It's not the force that's important, it's the force component perpendicular to the lever arm that makes the real difference.

Author's note: Anyone adept at opening doors knows these principles. First, the doorknob is placed as far from the hinges (pivot point) as possible. Second, the knob itself allows you to rotate it with minimum force. Finally, when it comes time to swing the door open, the force on the knob should be exerted perpendicularly to the door. Any force exerted parallel to the door will help rip the hinges off the wall, but will not get the door open.

So keep truckin'.

3.

a) F1L1 = F2L2 (35 nt)(1.6 m) = (Fsin50)(2.2) F = 33.2 nt

b) F1L1 = F2L2 (135sin60)(4.6 m) = (5 kg * 9.8 m/sec2)(x) x = 10.9 m

c) F1L1 = F2L2 (62sin70)(1.7 ft) = (Fsin38)(2.8 ft) F = 57.5 lb

4.

a) The angle at the top is the compliment of 33°, or 57°. Hence: (Fsin33)(4.7) = (16sin57)(5.3) F = 27.8 nt

b) The 20 lb force must make a 70° angle to the beam, so: (Tsin20)(4) = (20sin70)(6) T = 82.4 lb

5.

a) [1.0 * 32(sin31)](2 + x) = (41sin58)(2) x = 2.22 ft

b) (Tsin62)(1.4) = (30sin50)(3.7) T = 68.8 lb

6*.

a) Let t = tension in vertical string (t)(3.0) = (2 * 32)(3.0 + 2.4) t = 115.2 lb (Tsin65)(1.2) = (115.2)(3.7 + 1.2) T = 519 lb

b) T1 = 2 * 18 T1 = 36 lb (36sin40)(4) = (T2sin50)(9) T2 = 13.4 lb (13.4)(5) = (T3sin33)(9) T3 = 13.7 lb

When we get more than two forces acting, it becomes necessary for us to generalize the principle. We will define a new term, torque, as the perpendicular force exerted times the distance from the pivot: t = F L , where t is the Greek letter tau for torque and L is used to denote the distance from the pivot to the applied force. It is understood that F is the perpendicular component of the force.

Rotational equilibrium is established when the clockwise torques balance the sum of the counter-clockwise torques1.

7.

counter-clockwise torques = clockwise torques (Fsin68)(5.6 + 3.2) = (20sin60)(3.2) + (16sin70)(4.2) F = 14.5 nt

8.

Tension in string is w/2 (w/2)(x + 4.1 + 3.2) = (w)(4.1 + 3.2) + [(w/2)sin40](3.2) w divides out: 1/2(x + 7.3) = 7.3 + (1.6sin40) x = 9.36 ft

We have ignored, until now, the weight of the beam itself. Fortunately, this is easily included. We imagine the entire weight of the beam to be concentrated at the precise center of the beam. All calculations then proceed normally.

9. A 25 pound beam is supported by a string as shown. What is the tension?

counter-clockwise torques = clockwise torques (20)(6) + (25)(3) = (Tsin33)(4) T = 89.5 lb note: the full 25 lb can be considered at the center of the bar

10. In each case the bar weighs 320 nt and is 6 meters long. Find the tension in the supporting strings.

a)	b)
c)	d)
e)	f)

a) F1L1 = F2L2 (320)(3) = T(6) T = 160 nt	b) F1L1 = F2L2 (320)(3 - 1.3) = T(6 - 1.3) T = 116 nt note: (3 - 1.3) = distance from the pivot to the center of the mass
c) F1L1 = F2L2 (320)(3) = (Tsin33)(6) T = 294 nt	d) F1L1 = F2L2 (Tsin72)(2.8) = (320)(2) T = 240 nt
e) F1L1 = F2L2 (Tsin59)(2.8 + 1.6) = (320)(3) + (8 * 9.8)(1.6) T = 288 nt	f) F1L1 = F2L2 (320)(1) + (Tsin65)(3) = (20 * 9.8)(2) T = 26.5 nt

11. Now here is one with an angled beam of 6 m length and 320 nt weight. Remember, the weight still pulls straight down, so its perpendicular component will have to be calculated.

By alternate interior angles, the force is applied at 68° F1L1 = F2L2 (320sin68)(3) = (Tsin39)(5.2) T = 272 nt

12. In each of the following we will assume the beam weighs 45 nt and is 2.4 m long. Find the unknowns.

a)	b)
c)	d)
e)	f)

a) The center of the mass is in the center of the bar, so it's 0.4 m from the pivot F1L1 = F2L2 (45)(.4) = W(.8) W = 22.5 nt	b) F1L1 = F2L2 (90)(1) = (1.2 - 1.0)(45) + W(1.4) W = 57.9 nt
c) F1L1 = F2L2 (20)(2.4 - x) + (45)(1.2 - x) = 40x 48 - 20x + 54 -45x = 40x x = 0.971 m	d) F1L1 = F2L2 (Tsin37)(2.4) = (20)(1.6) + (45)(1.2) T = 59.5 nt
e) F1L1 = F2L2 (190sinØ)(.7) = (20)(2.4) + (45)(1.2) sinØ = .767 Ø = 50.1°	f) F1L1 = F2L2 (Tsin41)(2.0) = (20sin62)(2.4) + (45sin62)(1.2) T = 68.6 nt

13. The spring is stretched 6 cm, and the bars are weightless. Find T₁ and T₂.

F1L1 = F2L2 T1(1.1) = (150)(2.7) T1 = 368 nt F3L3 = F4L4 (368)(.8 + .5) = (T2sin38)(.8) T2 = 972 nt

14*. What is the tension, T? Each bar is 3.0 m long and weighs 90 nt.

First get the tension, T1, in the string connecting the levers: F1L1 = F2L2 (2 * 3 * 9.8)(2.2) + (90)(.7) = (T1)(.8) T1 = 240.5 nt (240.4)(2.3) + (90)(1.5) = (Tsin36)(3) T = 390 nt

15. A 12 foot board is used as a bridge. It weighs 37 lb and supports a 92 lb girl who stands 3 feet from one end. What force is exerted on each end of the board?

Imagine that the left end of the board is hinged. Then: (F2)(12) = (37)(6) + (92)(9) F2 = 87.5 lb Now imagine the right end being hinged. Then: (F1)(12) = (37)(6) + (92)(3) F1 = 41.5 lb note: F1 + F2 = 129 lb, the weight of the bridge and the girl

16. A 9.2 kg watermelon is supported by a 212 nt scaffolding. What is the tension in each rope?

Choose the right end of the scaffold as a pivot: F1L1 + F2L2 = F3L3 (9.2 * 9.8)(1.6 + 3.2) + (212)(4.8/2) = (T1)(3.2) T1 = 294 nt Now choose the point where T1 attaches as the pivot: (9.2 * 9.8)(1.6) + (T2)(3.2) = (212)(4.8/2 - 1.6) T2 = 7.9 nt Check: T1 + T2 = 302 nt = weight of mellon and board

17. Mandy Lifeboats stands on the edge of a 35 lb diving board. If Mandy tips the scales at 112 lb, what is the tension in the rope?

F1L1 = F2L2 + F3L3 (Tsin65)(1.2) = (112)(3.2) + (35)(2.2 - 1.2) T = 362 lb

18. Cole LaDrinque snags a big one, which exerts a 30 pound tension in his line. What force must he apply with the upper hand to support his 25 lb pole as well as the fish? (Cole holds the pole at 60° to the horizontal)

Pivot at lower end of the pole: (F)(.8) = (30sin80)(3) + (25sin30)(1.5) F = 134 lb

19. In an attempt to build a better mouse trap, an enterprising inventor lays a 5 nt board over the edge of a table as shown. He places a 0.7 nt piece of Mozzarella at the end. How far will a 2.2 nt mouse be able to get from the table fore he falls to an untimely end?

Use the edge of the table as a pivot: F1L1 = F2L2 + F3L3 (5)(.1) = (.7)(.2) + (2.2)(x) x = .163 m

20. A painter sits on a 20 kg scaffolding, 3 m long. If the tension in the right hand rope is 340 nt, what is the weight of the painter? What is the tension in the left rope?

First choose the left end of the scaffold as the pivot. We do this so that our ignorance of the tension in that rope won't enter into the equation. (After all, any force applied at the pivot exerts a zero torque--have you tried to open a door by pushing the hinge?) F1L1 = F2L2 + F3L3 (340)(3) = (20 * 9.8)(1.5) + (W)(3 - 1.2) W = 403 nt Now take the pivot at the right end: T(3) = (20 * 9.8)(1.5) + (403)(1.2) T = 259 nt