1. Copyright Sautter 2003

2. The next slide is a quick promo for my books after which the presentation will begin Thanks for your patience! Walt S. Wsautter@optonline.net More stuff at: www.wsautter.comwww.wsautter.com

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4. Statics When objects are subjected to forces and the net force does not equal zero translational (place to place) motion occurs. When a rigid body is subjected to a net force which is not zero, translation occurs. In addition, rotational motion may occur and even when the net force does equal zero! Rotational motion is the result of applied torques often times called moments. Torques means the “twist” experienced by a body. This twisting is the result of an applied force multiplied by the perpendicular distance to pivot point or center of rotation. This pivot point is often referred to as a fulcrum in some applications. The following slide shows torques as applied to a seesaw device. When the sum of the torques equals zero (clockwise torques equals counterclockwise torques) the angular acceleration equals zero. If the system is not initially rotating then no rotation occurs when equal torques act.

5. Σ Counterclockwise Torque F cc x r cc Clockwise Torque F c x r c r cc rcrc F cc FcFc fulcrum Small mass Large perpendicular distance Large mass Small perpendicular distance Σ

6. Analyzing Force Systems Concurrent refers to the fact that all the forces are acting at a point. In order to analyze a system of concurrent forces we must first draw a “Free Body Diagram” for the system. A Free Body Diagram selects a point of force interaction and then shows each of the individual forces acting at that point using vector arrows. Each force is then resolved into its horizontal (x) and vertical (y) components using cosine and sine functions. In order for the point to be in translational equilibrium the sum of forces acting on the point must equal zero, that is all x forces must add to zero and all y forces must add to zero. When a rigid body is contained in the system, torques must also be considered. The sum of torques around any point on the body must be zero to ensure rotational equilibrium.

7. Σ 100 lbs W =100 lbs Rope tension pull Θ1Θ1 Θ2Θ2 FREE BODY DIAGRAM SYSTEM Point of Force Interaction

8. 100 lbs W =100 lbs Rope tension pull Θ1Θ1 Θ2Θ2 PXPX TXTX PYPY TYTY ΣF X = 0 T X = P X ΣF Y = 0 T Y + P Y = W Σ

9. Σ W =100 lbs pull Θ1Θ1 Θ2Θ2 PXPX TXTX PYPY TYTY ΣF X = 0 T X = P X T Cos Θ 1 = P Cos Θ 2 T = P Cos Θ 2 / Cos Θ 1 ΣF Y = 0 T Y + P Y = W T Sin Θ 1 + P Sin Θ 2 = W T X = T Cos Θ 1 T Y = T Sin Θ 2 P x = P Cos Θ 2 P Y = P Sin Θ 2 By substitution P( Cos Θ 2 / Cos Θ 1 ) Sin Θ 1 + P Sin Θ 2 = W With given angles and weight pull P can now be found. With the value of P found, tension T can also be found rope tension` VERTICAL FORCES HORIZONTAL FORCES

10. Problems in Statics A 100 lb weight is suspended from the center of a wire which make an angle of 20 0 with the ceiling. Find the tension in the wire. ceiling 100 lbs 20 0 Free Body Diagram Σ F X = 0 T X1 = T X2 T 1 Cos Θ 1 = T 2 Cos Θ 2 Θ 1 = Θ 2 T 1 = T 2 20 0 100 lbs T X1 T X2 T1T1 T2T2 Σ F Y = 0 T Y + P Y = W T Sin Θ 1 + P Sin Θ 2 = W By substitution T Sin Θ + T Sin Θ = W 2 T Sin Θ 2 = W T =100 / 2 Sin 20 0 T = 146 nt

11. Problems in Statics A 100 lb weight hangs from the ceiling by two wires making angles of 30 and 45 degrees respectively with the ceiling. Find the tension in each of the wires. ceiling 100 lbs 30 0 45 0 30 0 45 0 100 lbs T1T1 T2T2 Free Body Diagram Σ F X = 0 T X1 = T X2 T 1 Cos 30 0 = T 2 Cos 45 0 T 1 = T 2 Cos 45 0 / Cos 30 0 Σ F Y = 0 T Y1 + T Y2 = W T 1 Sin Θ 1 + T 2 Sin Θ 2 = W By substitution T 2 (Cos 45 0 / Cos 30 0 ) Sin 30 0 + T 2 Sin 45 0 = W T 2 (0.707 / 0.866) (0.5)+ T 2 0.707 = 100 T 2 = 89.7 lbs T 1 = T 2 Cos 45 0 / Cos 30 0 T 1 = 89.7(.0.707 / 0.866) T 1 = 73.2 lbs

12. Statics Problems with Rigid Supports A stick or rigid support is often called a strut or sometimes a boom. These struts used in statics problems are of two types, weighted and weightless. When weightless struts are used, any torques created by the weight of the strut itself are neglected. When weighted struts are used, the torque created by the strut’s weight must be considered. Since the strut can be considered as a rectangular solid, its center of mass is located at the geometric center (midpoint or at one half its length) The following problem is solved on the next slide: A weightless strut of length l, supports a 100 pound weight at its end and is held perpendicular to the wall by a cable which makes an angle of 45 degrees with the strut. The cable is attached to the wall above the strut. Find the tension in the cable.

13. Problems in Statics - (weightless struts) 100 lbs tension Push of strut weight Θ =45 0 r Clockwise torque Counter Clockwise torque pivot τ c = l x w Length of strut ( l ) τ cc = r x T Σ τ = 0 τ cc = τ c r x T = l x w Sin Θ = r / l r = l Sin Θ l Sin Θ T = l x w canceling l gives T = (l x w) / l Sin Θ T = w / Sin Θ T = 100 lb / sin 45 0 = 141 lbs T P W Σ F X = 0 Σ T cos Θ = P 141 cos 45 0 = P P = 99.7 lbs Free Body Diagram of Forces acting on body Analyzing Torques in Rigid Bodies r = torque arm of cable tension

14. Statics Problems with Rigid Supports The following problem involving a weighted strut is solved on the next slide: A strut of uniform density and of length l weighing 50 pounds, supports a 100 pound weight at its end and is held perpendicular to the wall by a cable which makes an angle of 45 degrees with the strut. The cable is attached to the wall above the strut. Find the tension in the cable, the compression (push) of the strut and the forces acting on the hinge (pivot point).

15. Problems in Statics - (struts of uniform mass) 100 lbs tension Θ =45 0 r Clockwise torque of strut τ s = w strut x l / 2 Length of strut = l Weight of strut at Center of Mass Clockwise torque of weight τ w = w hanging x l Counter Clockwise torque of cable Σ τ = 0 τ cc = τ c τ cc = s τ + τ w r x T = (w s x l / 2) + (w h x l) r = l sin Θ l sin Θ x T = (w s x l / 2) + (w h x l) canceling l gives T = ((w s / 2) + (w h )) / sin Θ τ cc = r x T pivot T= ((50/2) + 100) / sin 45 0 T = 125 /.707 = 177 lbs weight of strut = 50 lbs Center of Mass at mid point r = torque arm of cable tension Continued on next slide

16. FREE BODY DIAGRAM FROM THE PROBLEM ON THE PREVIOUS SLIDE Tension in cable = 177 lbs Θ =45 0 Weight = 100 lbs Push of strut opposite pull of cable lift of cable Σ F Y = 0 push of strut = pull of cable pull of cable = T cos 45 0 Push of strut = 177 x 0.707 P = 125 lbs The up pull is that of the cable and the hinge (pivot) They must balance the weight of 100 lbs. Pull of cable = T sin 45 0 = 177 (0.707) Pull of cable = 125 lbs Upward force of hinge = 25 lbs Forces on hinge 125 lbs 25 lbs Vector sum Force net = (25 2 + 125 2 ) 1/2 = 127 lbs Θ = tan -1 (25 / 125) = 11.3 0

17. Forces & Torques ( Translational & Rotational Equilibrium ) When the sum of forces applied to a body equals zero the body experiences no acceleration. If it is at rest initially, it will remain at rest. If it is in motion, its motion will continue at a constant rate. When the sum of forces on a body initially at rest equals zero, although the body will not translate (move linearly) it still may rotate (spin) ! To insure rotational equilibrium, the sum of torques acting on the body must also be zero. All torques tending to rotate the body clockwise must be balanced by all torques tending to rotate the body in the counterclockwise direction. Torque is the product of applied force times the perpendicular distance between the line of action of the force (the straight direction of the force) and the pivot point (point around which rotation occurs). When applying torques to a body, not only the force but the location of the the force must be considered.

18. AB 10 cm90 cm center of mass 50 cm weight of plank upward push of A upward push of B Σ F = 0 F A + F B = weight of plank τAτA τBτB τ wt Σ τ = 0, c τ c = τ c τ A + τ B = τ wt Force Equilibrium Torque Equilibrium Arbitrary Reference point

19. TORQUES & FORCES ON A RIGID BODY A 1.0 meter plank of mass 10 kg is supported at each end. The supports are placed 10 cm from one end and 10 cm from the other. A 60 kg man stands 30 cm from one end, Find force on each support. Arbitrary Reference point AB 10 cm90 cm50 cm weight of plank upward push of A τAτA τ plank Wt of man 30 cm τ man upward push of B τBτB Σ F = 0 F A + F B = wt of plank + wt man Force Equilibrium F A + F B = (10 x 9.8) + (60 x 9.8) F A + F B = 686 nt Continued on next slide

20. TORQUES & FORCES ON A RIGID BODY (cont’d) Arbitrary Reference point AB 10 cm90 cm50 cm weight of plank upward push of A τAτA τ plank Wt of man 30 cm τ man upward push of B τBτB Torque Equilibrium Wt man = 588 nt, Wt plank = 98 nt (10 x F A ) + (90 x F B ) = (50 x 98) + (30 x 588) 10 F A + 90 F B = 22540 From previous slide F A + F B = 686 F A = 686 - F B 10 (686 - F B )+ 90 F B = 22540 F B = 15680 / 80 = 196 nt F B = 686 – 196 = 490 nt Στ = 0, c τ c = τ c τ A + τ B = τ p +τ man

21. Center of Mass for Complex Bodies The center of mass of a body is the point where all the mass could be concentrated and give the same effect as the actual distribution of mass within the body. The center of mass of a regularly shaped, symmetrical body (such as a sphere, a cylinder, a rectangular solid, etc.) of uniform density, is located at the geometric center of the object. For bodies consisting of several connected points of mass the center of mass can be found by adding the torques caused by each of the masses using a selected reference point to measure the torque arms and then dividing that sum the the sum of the masses. The reference used may be any point on or even outside the body but all measurements of the torque arms (sometimes called moment arms) must be made from this point.

22. The Center of Mass of a body is a point where all the mass of the object could be concentrated and give the same results as the actual distribution of mass within the body. Center of Mass is a simplifying technique used in problem solving. The Center of Mass of a symmetrically shaped object of uniform density is located at its geometric center. Finding the Center of Mass of a Body consisting of several Individual connected points of mass. m1m1 m2m2 m3m3 Arbitrary reference point x1x1 x2x2 x3x3

23. 4 ft Find the center of mass of the object shown. It is of uniform density. 6 ft 2 ft Center of mass segment #1 ( 1,2 ) Center of mass segment #2 ( 2,5 ) Reference point x cm = ((m x 1 ) + (m x 2)) / (m + m) X cm = 3m / 2m = 1.5 ft y cm = ((m x 2 ) + (m x 5)) / (m + m) y cm = 7m / 2m = 3.5 ft Center of mass Of object (1.5, 3.5)

24. Torques & Non Perpendicular Forces Since torque is defined in terms of a perpendicular torque arm (distance from the line of action of the force to the center of rotation), when forces are applied at angles other than 90 0, only the perpendicular component of the force must be used to calculate the torque! To determine the components of a force we use sine and cosine functions as described in previous programs on forces and vectors. Forces applied to a body whose line of action pass through the center of rotation have no perpendicular distance between the pivot point and the direction of the force and therefore cause no torque since any force value times zero gives zero.

25. c 1 ft 45 0 30 0 F 1 = 2 lbs F 2 = 5 lbs F 3 = 1 lb F 4 = 3 lbs F 5 = 2 lbs τ C 3 = 1 x 1 τ C 4 = 1 x 3 τ C 5 = 2 cos 30 0 x 1 τ 2 = 0 x 5 τ 1 = 0 x 2 2 cos 30 0 c = center of rotation ( pivot point ) Line of action passes through pivot for F 1 & F 2 Torque arm = 0 Perpendicular component of F 5 is used to calculate torque Torque = force x perpendicular distance from pivot

26. Push of wall Push of floor Force of friction Wt of ladder Wt of man ½ Length of ladder Θ Angle = Selected pivot point Moment arm r wall Moment arm r ladder Moment arm r man τ C wall τ CC ladder τ CC man

27. LADDER PROBLEMS A uniform ladder rests against a wall. It weighs 50 lbs and makes a 30 0 angle with the floor. A 200 lb climbs ¼ of the way up the ladder. What force of friction with the floor is required ? Push of wall floor friction Wt of ladder Wt of man ½ Length of ladder Θ pivot point r wall r ladder r man τ C wall = P wall x r wall τ CC ladder = wt ladder x r ladder τ CC man = wt man x r man L = length of ladder r wall = L sin 30 0 r ladder = ½ L cos 30 0 r man ¼ L cos 30 0 Σ τ = 0, τ cc = τ c τ CC ladder + τ CC man = τ C wall Continued on next slide

28. LADDER PROBLEMS (cont’d) wt ladder x ½ L cos 30 0 + wt man x ¼ L cos 30 0 = P wall x L sin 30 0 The only horizontal forces acting on the system are the push of the wall and the force of friction. They must be equal and opposite ! F friction = P wall P wall = (wt ladder x ½ L cos 30 0 + wt man x ¼ L cos 30 0 ) / L sin 30 0 P wall = ((50 x ½ x 0.866) + (200 x ¼ x 0.866)) / 0.500 P wall = F friction = 130 nt Up ward force on the floor = wt of man + wt of ladder F floor = 200 + 50 = 250 lbs F ladder = (( 130 2 ) + (250 2 )) 1/2 = 282 nt Θ = tan -1 (130 / 250 ) = 27.5 0 130 nt 250 lbs Force on Foot of ladder Θ

30. A picture hangs from a wire which makes a 30 0 angle with the Vertical. The picture weighs 8.0 nt. Find the tension in the wire ? (A) 4.0 nt (B) 4.6 nt (C) 8.0 nt (D) 9.2 nt A 100 kg box is hung from two ropes which make angles of 20 and 40 degrees respectively. Find the tension in the 20 0 rope. (A) 57 nt (B) 727 nt (C) 74.2 nt (D) 39.5 nt A 30 lb weight is hung from the end of a 6 foot beam which weighs 20 lbs. Where is the balance point as measured from the 30 lb weight ? (A) 2.4 ft (B) 2.6 ft (C) 0.8 ft (D) 3.5 ft A ladder weighs 240 nt and rests against a wall at a 60 0 angle. A 600 nt man stands ¾ of the distance from the top of the ladder. What is the force of the ladder against the wall ? (A) 240 nt (B) 840 nt (C) 440 nt (D) 606 nt A pipe 2.0 meters long has another similar pipe 1.0 meter long connected to its end to form a T. Find the center of gravity from the T intersection. (A) 10 cm (B) 67 cm (C) 16 cm (D) 84 cm Click here for answers