Trigonometry. Pythagorean Theorem. Force Components. Components of Force. hypotenuse. hypotenuse

Transcription

1 Pthagorean Theorem Trigonometr B C α A c α b a + b c a opposite side sin sinα hpotenuse adjacent side cos cosα hpotenuse opposite side tan tanα adjacent side AB CB CA CB AB AC orce Components Components of orce cosθ sinθ + tanθ 10 lbs * cos (30 ) 10 * lbs * sin (30 ) 10 * lbs 10 lbs lbs 5 lbs soh cah toa sine (angle) opposite/hpotenuse cosine (angle) adjacent/hpotenuse tangent (angle) opposite/adjacent 1

2 Components of orce Similar Triangles How can we tell if forces are balanced? ind ind Hpotenuse Use pthagorean theorem (3 +4 ) * 0 5* 4* 0 5* 3* 0 5* 4* 0 5* lbs 16lbs 3 0 lbs 4 The and components cancel What is the balancing force? 10 lb force 0 lbs 10 * cos (30 ) -8.7 lbs 10 lbs 10 * sin (30 ) 5 lbs 7 0 lb force 3*0/7.6 8 lbs 7*0/ lbs Combine components X: lbs Y: lbs lbs 3 Negative of these components represent balancing force X -0.7 lbs, -3.4 lbs esultant ( ) 3.4 lbs Angle: tan-1(-3.4/-0.7) 88 in third quadrant Translational Equilibrium In order to be in translational equilibrium, the total forces of the object must be zero. Set the horizontal and vertical components of the forces to zero. 0 0

3 otational Equilibrium Torque or Moment In order to be in rotational equilibrium, the total torques acting on an object must be zero. 0 or M 0 d or M d Pole Vaulter ind the forces and. pivot c.g. ft. 5 lbs 6 ft. The pole vaulter is holding pole so that it does not rotate and does not move. It is in equilibrium. Pole vaulter con t Pole vaulter con t otation r + r Translation ft + 6 ft 5lb + 15lbs 15 lbs pivot c.g lbs + 5lbs 10lbs ft. 10 lbs 6 ft. 5 lbs 3

4 Think Pair - Share What eerts more pressure (in pounds per square inch) when walking: a 100 lb woman in high heels or a 6,000 lb elephant in bare feet? [At the moment when onl the heel rests on the ground.] Stiletto heels have an area of about 1/16 of a square inch. Elephants, unlike humans, walk with two feet on the ground at a time. Each foot is about 40 square inches. Answer Pressure P A Woman s heels 100lbs lbs P psi 1 in in 16 Elephant s feet 6000lbs P 75 psi 40in The heels eert more pressure! Equations E Strain E Δ Strain E Strain E is the modulus of elasticit Units of pressure (N/m or psi) Strain E Δ and Strain eample 1 How far can ou stretch a 0 ft steel cable before it snaps? E for steel is *10 11 N/m Ultimate tensile strength for steel is 5.*10 8 N/m * E Strain Δ Δ * Δ E 8 0 ft *5. 10 N m Δ 0.05 ft 0.6inches N m 4

5 Middle Third ule Must sta within the middle third of the block to prevent the opposite side from going into tension this is the important MIDDE THID UE. Wh is it important? Because bricks/stone/etc are much stronger in compression than in tension; need to keep them in compression or the can break! To get around this problem reinforce concrete with a steel bar the steel can take the tension when the concrete can t. We ll go over this in more detail in Chapter 8. Cables Principal Elements for practical suspension sstems Vertical supports or towers Main cables Anchorages Stabilizers Vertical Supports or Towers Provide essential reactions that keep the cable sstem above the ground Ma be simple vertical or sloping piers or masts, diagonal struts, or a wall. Ideall, the aes of the supports should bisect the angle between the cables that pass over them 5

6 Main Cables Primar tensile elements Carr roof with a minimum of material Steel used in cable structures has breaking stresses that eceed 00,000 psi Anchorages Because the cables are not vertical onl, horizontal force resistance is required. In suspension bridges, the massive concrete abutments provide the horizontal reaction force Stabilizers ightweight roof or bridge sstems are susceptible to pronounced undulation or fluttering when acted upon b wind forces. Cables resist load through tension. The destructive force is vibration or flutter Cables with a Single Concentrated oad T T ( 1' ) + ( 3' ) 1.4' A T 1 [ 0] T W 0 W T 3 C l W B l 4 h 3 T T T T W T T W 3 3 T h 6