PHYS-211-WK-7-CH-6-STATIC-PPT.pptx for psychology

PHYSICS

1. : PHYSICS 211 CH 6 - S TATIC FIRST CONDITION EQUILIBRIUM

2. At the end of the chapter the students are expected: 1. Distinguish confidently the state of equilibrium of bodies at rest and those forces acting. 2. Decide confidently whether a body will be in rotational rotational motion.

3. Static - involves analysis of the forces acting on objects that are at rest or moving . It is in equilibrium. ( acceleration = 0 )   Static equilibrium – object at rest . ( remains at rest ) Dynamic equilibrium - object in motion ( motion continues unchanged ) Concurrent forces – forces that that at a common point . ex. Component method forces ( vector topic) Non-concurrent – forces that do not intersect at a common point.

4. : Two Conditions for a body to be in equilibrium :   1. First condition of equilibrium in statement :   “ It is the vector sum of all forces acting on a body must be zero” in formula : for horizontal forces for vertical forces ΣFx = 0 ΣFy = 0

5. : Two Conditions for a body to be in equilibrium :   1. First condition of equilibrium in formula : ( acceleration = 0 ) for horizontal forces for vertical forces Forces are to the LEFT OR RIGHT Forces are going UP or DOWN Note : assume sign of DIRECTION – one direction Positive, other negative ΣFx = 0 ΣFy = 0

6. : Two Conditions for a body to be in equilibrium : 1. First condition of equilibrium ( Ch 6 ) in FORMULA :   ANALYSIS OF FORCES ON OBJECT for horizontal forces ( x-axis) for vertical forces ( y – axis ) Forces to the LEFT OR RIGHT forces going UP or DOWN F F F F ΣFx = 0 ΣFy = 0 object

7. : ACTUAL FIGURE HOW TO ANALYZE FORCES IN A GIVEN OBJECT OR BODIES

8. : ROPES HAS TENSION, T ACTUAL FIGURE HOW TO ANALYZE FORCES IN A GIVEN OBJECT OR BODIES

9. : CABLES HAS TENSION. T ACTUAL FIGURE HOW TO ANALYZE FORCES IN A GIVEN OBJECT OR BODIES object has Weight, W

10. : HOW TO ANALYZE FORCES IN A GIVEN OBJECT OR BODIES arrow representation of the forces acting on the body. RED ARROW ARE THE FORCES

11. : FORMULA : 1 ST CONDITION OF EQUILIBRIUM Formula : Formula :    

12. : HOW TO USE THE FORMULA Formula :   ΣFx = 0 ͵ Fx 1 + Fx 2 + Fx 3 ……. = 0 ( for horizontal forces )   Formula : ͵   ΣFy = 0 ͵ Fy 1 + Fy 2 + Fy 3 ……. = 0 ( for vertical forces )  

13. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ?   * HOW TO USE THE FORMULA IN A PROBLEM       OBJECT

14. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? FOR HORIZONTAL FORCES : USING Formula :   ΣFx = 0 ͵ Fx 1 + Fx 2 + Fx 3 ……. = ( for horizontal forces )   OBJECT

15. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? ..   FOR VERTICAL FORCES ; USING Formula :   ΣFy = 0 ͵ Fy 1 + Fy 2 + Fy 3 ……. = 0 ( for vertical forces )   OBJECT

16. : PROBLEM APPLICATION ..  

17. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? Problem : Calculate : Force a) F5 and b) F3 using the figure above by 1 st condition of equilibrium OBJECT

18. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? Problem : Calculate : a) F5 using 1 st condition of equilibrium Solution : Formula : ΣFx = 0 ͵ ( IDENTIFY all horizontal forces in the problem _)   .. OBJECT

19. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? Solution : ΣFx = 0 ͵ write horizontal forces given in the problem   ΣFx = 0 ͵ F5 F4 = Note : apply assume direction of forces ( if one direction positive, the other is the opposite ) OBJECT

20. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? Solution : Formula : ΣFx = 0 ͵ write horizontal forces given in the problem   + ΣFx = 0 ͵ + F5 – F4 = (assume direction – red arrow ) ( assume   FORCES TO THE LEFT IS POSITIVE ) .. OBJECT

21. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? Solution : ΣFx = 0 ͵ using all horizontal forces in the problem ( left /right forces)   + ΣFx = 0 ͵ + F5 – F4 = ( substitute and transpose) + F5 – 10kg = 0 F5 = 10 kg answer OBJECT

22. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? Problem : b) Calculate F3 using 1 st condition of equilibrium formula : ΣFy = ͵ IDENTIFY all vertical forces in the problem & assume direction . . OBJECT

23. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? Solution : Formula : ΣFy = ͵ write all vertical forces given in the problem   + Σfy = ͵ – F1 + F2 – F3 = ( ASSUME DIRECTION ) ( assume   : FORCES UPWARD to be positive ) .. OBJECT

24. : F2 = 7kg F5 =? F4 = 10 kg F1 = 4kg F3 = ? Solution : Formula : ΣFy = ͵ using all vertical forces in the problem ( up /down forces)   + Σfy = ͵ – F1 + F2 – F3 = ( substitute and transpose F3 ) – 4kg + 7kg – F3 = 0 F3 = 3 kg answer OBJECT

25. : SUMMARY OF THE FORCES : 7 kg – 7kg = 0 10 kg – 10 kg = 0 , acc = 0 F1 = 7 kg F5 = 10 KG F4 = 10 KG F = 7 KG since forces acting horizontally and vertically are equal. therefore body will stay in equilibrium. OBJECT

26. : F2 F = 5kg F1 30 ᵒ Note : if given forces is at a certain angle . ( FORECE, F in the problem ) Resolve the slanted force vector ( use in component method ) RECALL : FORMULA TO RESOLVE THE VECTOR at AN ANGLE. A ) If angle starts at x axis B) if angle starts at y axis Fx = Fcos θ Fx = Fsin θ Fy = Fsin θ Fy = Fcos θ OBJECT

27. : F2 F = 5kg F1 30 ᵒ x -axis Problem : Find a) F1 and b) F2 by 1 st condition of equilibrium * Resolve force vector ,F and formula A since θ starts at x-axis. FORMULA TO RESOLVE THE SLANTED VECTOR A ) angle , θ = 30 ᵒ starts at x axis . Fx = Fcos θ = 5kg ( cos 30 ᵒ ) = 4.33 kg ( USE THE COMPONENTS ) Fy = Fsin θ = 5Kg ( sin 30 ᵒ ) = 2.50 kg. OBJECT

28. : problem : F2 Fy = 2.50 kg F1 Fx = 4.33 kg Problem : Find F1 and F2 by 1 st condition of equilibrium * Resolve force vector ,F & formula A since θ starts at x-axis. FORMULA TO RESOLVE THE SLANTED VECTOR A ) angle , θ = 30 ᵒ starts at x axis . Fx = Fcos θ = 5kg ( cos 30 ᵒ ) = 4.33 kg Fy = Fsin θ = 5Kg (sin 30 ᵒ ) = 2.50 kg. apply the 1 st condition of equilibrium formula. OBJECT

29. : problem : F2= ? Fy = 2.50 kg F1 = ? Fx = 4.33 kg . applying the 1 st condition of equilibrium formula. ΣFx = 0 ͵ F1 – Fx = ( for horizontal forces , substitute (-)   F1 – 4.33 kg = 0 ( transpose ) F1 = 4.33kg answer ΣFy = ͵ – F2 + Fy = 0 ( for vertical forces, substitute ) (+) – F2 + 2.5 kg = 0 ( transpose ) F2 = 2.50 kg answer OBJECT

30. :   Worded problem : An object weighing 100 lbs and suspended by a rope A is pulled by horizontal rope B and held so that rope makes an angle of 30 ̊ with the vertical. Find Tensions a) in rope A (Ta ) and b) roрe B ( Tb ).   Draw figure ( FBD or Force diagram ) and analyze the forces , solution  

31. :   Example : an object weighing 100 lbs and suspended by a rope A is pulled by horizontal rope B and held so that rope makes an angle of 30 ̊ with the vertical. Find Tensions a) in rope A (Ta ) and b) roрe B ( Tb ).   Draw figure (FBD) and analysis , solution actual figure, apply analysis of forces  

32. :   Example : an object weighing 100 lbs and suspended by a rope A is pulled by horizontal rope B and held so that rope makes an angle of 30 ̊ with the vertical. Find Tensions a) in rope A (Ta ) and b) roрe B ( Tb ).   Draw figure (FBD) and analysis , solution  

33. :   Example : an object weighing 100 lbs and suspended by a rope A is pulled by horizontal rope B and held so that rope makes an angle of 30 ̊ with the vertical. Find Tensions a) in rope A (Ta ) and b) roрe B ( Tb ).   Draw figure (FBD) and analysis , solution ( simple drawing of forces )   rope A Ta = ? wall 30 ᵒ Tb = ? object rope B W = 100 lbs.

34. :   Example : an object weighing 100 lbs and suspended by a rope A is pulled by horizontal rope B and held so that rope makes an angle of 30 ̊ with the vertical. Find Tensions a) in rope A (Ta ) and b) roрe B ( Tb ).   Draw figure (FBD) and analysis , solution ( simple drawing of forces )   rope A Ta = ? Observed that Ta at an angle wall 30 ᵒ Tb = ? object rope B W = 100 lbs.

35. :   Example : an object weighing 100 lbs and suspended by a rope A is pulled by horizontal rope B and held so that rope makes an angle of 30 ̊ with the vertical. Find Tensions a) in rope A (Ta ) and b) roрe B ( Tb ).   Draw figure (FBD) and analysis , solution ( simple drawing of forces )   rope A Ta = ? Observed that Ta at an angle wall 30 ᵒ Tb = ? Resolve by component object rope B Fx = Fsin θ , Fy = FCos θ ( θ start at y axis ) W = 100 lbs.

36. :   Example : an object weighing 100 lbs and suspended by a rope A is pulled by horizontal rope B and held so that rope makes an angle of 30 ̊ with the vertical. Find Tensions a) in rope A (Ta ) and b) roрe B ( Tb ).   Draw figure (FBD) and analysis , solution ( simple drawing of forces )   rope A Ta = ? observed that Ta is at an angle wall 30 ᵒ Tb = ? Resolve by component object rope B Fx = Fsin θ , Fy = FCos θ ( θ start at y axis ) W = 100 lbs. F = Ta

37. : Resolve force vector Tension A ( slanted ) Fy = Tacos 30 ᵒ Tb = ? Fx = Tasin30 ᵒ W = 100 lbs By 1 st condition of equilibrium ΣFx = 0 ͵ Tb – Fx = 0 ( for horizontal forces , substitute ) (- )   Tb – Ta sin 30 ᵒ = equation 1 there are two unknown variables, called this as equation 1

38. : Resolve force vector Tension A ( slanted , θ start with y-axis ) Fy = Ta cos 30 ᵒ Tb = ? Fx = Tasin30 ᵒ W = 100 lbs By 1 st condition of equilibrium ( use the other formula ) Σfy = ͵ Fy – W = 0 ( for horizontal forces , substitute ) (+)   Ta cos 30 ᵒ – 100 = Ta = 100/ cos 30 Ta = 115.47 lbs answer

39. : Resolve force vector Tension A ( slanted , θ start with y-axis ) Fy = Tacos 30 ᵒ Tb = ? Fx = Tasin30 ᵒ W = 100 lbs ( To solve for Tb, go back to equation 1 in the previous page.) Tb – Ta sin 30 ᵒ = 0 ( equation 1 in previous number ) Tb = 115.47 ( 0.5) = 57.7 lbs answer

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