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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 9 Equilibrium. Equilibrium Conditions. F = 0 Use the horizontal direction. Use the vertical direction.   = 0 Always start with a free body diagram. Hanging Sign. Mass of block is 60 kg. Beam is 3 m long. Beam has a mass of 10 kg.

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Physics 2053C – Fall 2001

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  1. Physics 2053C – Fall 2001 Chapter 9 Equilibrium Dr. Larry Dennis, FSU Department of Physics

  2. Equilibrium Conditions • F = 0 • Use the horizontal direction. • Use the vertical direction. • = 0 • Always start with a free body diagram.

  3. Hanging Sign • Mass of block is 60 kg. • Beam is 3 m long. • Beam has a mass of 10 kg. • Cable is attached 2 m from the hinge. • The angle between the cable and the bar is 37o. hinge

  4. hinge Hanging Sign Free Body Diagram T Fv Fh mg Mg

  5. T Fv Fh mg Mg Hanging Sign F = 0 Horizontal Direction:Fh – Tcos = 0 Vertical Direction :Fv + Tsin - (m+M)g = 0 = 0 Use the hinge as the pivot point: T*2L/3*sin - mgL/2 – MgL = 0

  6. T Fv Fh mg Mg Hanging Sign T*2L/3*sin - mgL/2 – MgL = 0 T = (m/2 + M)*gL/(2L/3*sin) T = 3/2*(m/2+M) *g/sin T = 3/2*(10/2+60)*9.8/sin37o T = 1590 N Fh – Tcos = 0 Fv + Tsin - (m+M)g = 0

  7. T Fv Fh mg Mg Hanging Sign Fh – Tcos = 0  Fh = Tcos Fh = 1590* cos37 = 1270 N Fv + Tsin - (m+M)g = 0  Fv = - Tsin + (m+M)g Fv = - 1590*sin(37) + (10+300)*g Fv = -270 N

  8. Painter M = 72 kg m = 16 kg D= 1.2 m L = 3.1 m

  9. Painter T1 T2 Mg mg F = 0 Vertical Direction:T1 + T2 - (m+M)g = 0 = 0 Use the left side as the pivot point:T2*L - mgL/2 – MgD = 0

  10. Painter T2L - mgL/2 – MgD = 0 T2L = mgL/2 + MgD  T2 = mg/2 + MgD/L T2 = 16*9.8/2 + 72*9.8*(1.2/3.1) T2 = 350 N

  11. Painter T1 + T2 - (m+M)g = 0 T1 = (m+M)g - T2 T1 = (16+72)9.8 – 350 = 510 N

  12. Painter: Check T1 + T2 - (m+M)g = 0 510 + 350 – (16+72)*9.8 = 2.4 N Is that close enough? Yes, it is.

  13. Painter: Check T2L - mgL/2 – MgD = 0 350*3.1 – (16*9.8*3.1)/2 – 72*9.8*1.2 = - 4.7 Nm Is that close enough? Error = 4.7/1085 = 0.4% within 3 significant figures. Yes, it is.

  14. F W N  f = N Ladder N – W = 0 F – f = 0 FL sin - Wx cos = 0

  15. F W N  f = N Ladder N = W F = f = N = W FL sin - Wx cos = 0 FL sin = Wx cos x = FL sin /W cos x = W L sin /W cos x =  L tan

  16. Quiz #6 • Sample Questions: Chap. 9: 5, 10 • Sample Problems: Chap. 9: 7,23, 25

  17. Next Time • Quiz on Chapter 9 • Begin Chapter 10. • Please see me with any questions or comments. See you on Wednesday.

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