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Physics 2053C – Fall 2001

Physics 2053C – Fall 2001. Chapter 9 Equilibrium. Equilibrium Conditions. An object at rest in in equilibrium. Newton’s Law still holds!!! F = ma = 0 for an object at rest. Newton’s Law for rotational motion still holds!!!  = I = 0 for an object at rest. . F 2. F 1. M = 300 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 • An object at rest in in equilibrium. • Newton’s Law still holds!!! • F = ma= 0 for an object at rest. • Newton’s Law for rotational motion still holds!!! • =I = 0 for an object at rest.

  3. F2 F1 M = 300 kg = 53° W = Mg Example Problem • F = ma= 0 • =I = 0 First Draw a Free Body Diagram

  4. F2 F1 W = Mg Example Problem  Use Newton’s Law in both the Horizontal direction: F1 – F2 cos = 0 Vertical direction: F2 sin – W= 0

  5. F2 F1 W = Mg Example Problem  Solve for the unknowns, F1 & F2 F1 – F2 cos = 0 F2 sin – W= 0  F2 = W/sin F2 = (200 kg * 9.8 m/s2 )/sin53° F2 = 2450 N F1 – F2 cos = 0  F1 = F2 cos F1 = 2450 * cos53° = 1480 N

  6. The mass of beam acts at the center of mass (L/2) x L Using Torques Two boys, one with mass 47 kg (ML), the other with a mass of 63 kg (MH), are trying to balance a 3 m long teeter-totter (mass M = 40 kg ). Where should the pivot be to balance the beam? N MHg Mg MLg The normal force acts upward at the pivot.

  7. Using Torques =I = 0 Two boys, one with mass 47 kg, the other with a mass of 63 kg are trying to balance a 3 meter long teeter-totter (mass = 40 kg). Where should the pivot be to balance the beam? N MHg Mg MLg Note the pivot is closer To the heavier boy. Select the pivot as our axis of rotation.

  8. Using Torques -MHg * x + MLg * (L-x) + Mg(L/2-x) = 0 -gx*(MH + ML + M) + gL(ML + M/2) = 0 Two boys, one with mass 47 kg, the other with a mass of 63 kg are trying to balance a 3 meter long teeter-totter (mass = 40 kg). Where should the pivot be to balance the beam? N x Torque due to the normal force is zero. MHg Mg MLg =I = 0 -H + L + Beam= 0

  9. Using Torques Two boys, one with mass 47 kg, the other with a mass of 63 kg are trying to balance a 3 meter long teeter-totter (mass = 40 kg). Where should the center be to balance the beam? N MHg MLg Mg • -MHg * x + MLg * (L-x) + Mg(L/2-x) = 0 • -gx*(MH + ML + M) + gL(ML + M/2) = 0 • gx*(MH + ML + M) = gL(ML + M/2) • x = L(ML + M/2) /(MH + ML + M)

  10. Using Torques Two boys, one with mass 47 kg, the other with a mass of 63 kg are trying to balance a 3 meter long teeter-totter (mass = 40 kg). Where should the center be to balance the beam? N MHg MLg Mg x = L(ML + M/2) /(MH + ML + M) x = 3 m *(47 + 40/2)/( 63 + 47 + 40) x = 1.34 m from the heavier boy (MH) and 1.66 m from the lighter boy (ML).

  11. Next Time • Chapter 9 • Begin Discussion of CAPA • Please see me with any questions or comments. See you on Wednesday.

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