Statics

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Explains interaction of forces in non accelerating systems
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Statics

  1. 1. Copyright Sautter 2015
  2. 2. 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. 2
  3. 3. Σ Counterclockwise Torque Fcc x rcc Clockwise Torque Fc x rc rcc rc Fcc Fcfulcrum Small mass Large perpendicular distance Large mass Small perpendicular distance Σ 3
  4. 4. 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. 4
  5. 5. Σ 100 lbs W =100 lbs Rope tension pull Θ1 Θ2 FREE BODY DIAGRAMSYSTEM Point of Force Interaction 5
  6. 6. 100 lbs W =100 lbs Rope tension pull Θ1 Θ2 PXTX PY TY ΣFX = 0 TX = PX ΣFY = 0 TY + PY = W Σ 6
  7. 7. Σ W =100 lbs pull Θ1 Θ2 PXTX PY TY ΣFX = 0 TX = PX T Cos Θ1 = P Cos Θ2 T = P Cos Θ2 / Cos Θ1 ΣFY = 0 TY + PY = W T Sin Θ1 + P Sin Θ2 = W TX = T Cos Θ1 TY = T Sin Θ2 Px = P Cos Θ2 PY = P Sin Θ2 By substitution P( Cos Θ2 / Cos Θ1) Sin Θ1 + P Sin Θ2 = 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 7
  8. 8. Problems in Statics A 100 lb weight is suspended from the center of a wire which make an angle of 200 with the ceiling. Find the tension in the wire. ceiling 100 lbs 200 200 Free Body Diagram Σ FX = 0 TX1 = TX2 T1 Cos Θ1 = T2 Cos Θ2 Θ1 = Θ2 T1 = T2 200 200 100 lbs TX1 TX2 T1 T2 Σ FY = 0 TY + PY = W T Sin Θ1 + P Sin Θ2 = W By substitution T Sin Θ+ T Sin Θ = W 2 T Sin Θ 2 = W T =100 / 2 Sin 200 T = 146 nt 8
  9. 9. 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 300 450 300 450 100 lbs T1 T2 Free Body Diagram Σ FX = 0 TX1 = TX2 T1 Cos 300 = T2 Cos 450 T1 = T2 Cos 450 / Cos 300 Σ FY = 0 TY1 + TY2 = W T1 Sin Θ1 + T2 Sin Θ2 = W By substitution T2 (Cos 450 / Cos 300) Sin 300 + T2 Sin 450 = W T2 (0.707 / 0.866) (0.5)+ T2 0.707 = 100 T2 = 89.7 lbs T1 = T2 Cos 450 / Cos 300 T1 = 89.7(.0.707 / 0.866) T1 = 73.2 lbs 9
  10. 10. 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. 10
  11. 11. Problems in Statics - (weightless struts) 100 lbs tension Push of strut weight Θ=450 r Clockwise torque Counter Clockwise torque pivot τc = l x w Length of strut ( l ) τcc = r x T Σ τ= 0 τcc = τc 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 450 = 141 lbs T P W Σ FX = 0 Σ T cos Θ = P 141 cos 450 = P P = 99.7 lbs Free Body Diagram of Forces acting on body Analyzing Torques in Rigid Bodies r = torque arm of cable tension 11
  12. 12. 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). 12
  13. 13. Problems in Statics - (struts of uniform mass) 100 lbs tension Θ=450 r Clockwise torque of strut τs = wstrut x l / 2 Length of strut = l Weight of strut at Center of Mass Clockwise torque of weight τw = whanging x l Counter Clockwise torque of cable Σ τ = 0 τcc = τc τcc = sτ+ τw r x T = (ws x l / 2) + (wh x l) r = l sin Θ l sin Θ x T = (ws x l / 2) + (wh x l) canceling l gives T = ((ws / 2) + (wh)) / sin Θ τcc = r x T pivot T= ((50/2) + 100) / sin 450 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 13
  14. 14. FREE BODY DIAGRAM FROM THE PROBLEM ON THE PREVIOUS SLIDE Tension in cable = 177 lbs Θ=450 Weight = 100 lbs Push of strut opposite pull of cable lift of cable Σ FY = 0 push of strut = pull of cable pull of cable = T cos 450 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 450 = 177 (0.707) Pull of cable = 125 lbs Upward force of hinge = 25 lbsForces 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 14
  15. 15. 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. 15
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