Lecture10 - Chapter 12 Sta,c Equilibrium and...

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Unformatted text preview: Chapter 12 Sta,c Equilibrium and Elas,city Equilibrium and sta,c equilibrium Systems with constant linear and angular momentum are said to be in equilibrium. Examples: •  A book res,ng on a table (P = L = 0) •  A hockey puck sliding on a fric,onless surface (P =const., L=0) •  The rota,ng blades of a ceiling fan (P=0, L=const.) •  Wheel of a bicycle traveling along a straight path at constant speed (P=L=const.) If both P and L are zero (the first example above), i.e. the object is neither moving by transla,on nor rota,on, the object is in sta,c equilibrium. Sta,c equilibrium Remember that there are three kinds of sta,c equilibria: Stable: Body returns to equilibrium aOer having been displaced by a force. Neutral: Body can be easily moved to a new equilibrium posi,on. Unstable: Even a small force will move body out of equilibrium. What makes some objects more stable than others ? At which point do they become unstable ? What’s needed for equilibrium Equilibrium: L = p = const. If P is constant, then Torque and angular form (“Newton’s second law”): If L is constant then Condi,ons for equilibrium: That is: 1.  2.  The vector sum of all external forces ac,ng on the body must be zero. The vector sum of all external torques that act on the body, measured about any possible point, must also be zero. and Newton’s second law: τ net = 0 For sta,c (sta,onary) equilibrium: The linear and angular momentum must also be zero. If all forces act in the xy ­plane (2D problem), condi,ons 1 & 2 simplify to: ∑F ∑F ∑τ x y z =0 =0 =0 The center of gravity The gravita,onal force Fg (=mg) on a body effec,vely acts on a single point, called the center of gravity (CG) of the body. Proof: Consider all the torques produced by the gravita,onal force on all parts of an object: Now assume that this is the same as the gravita,onal force ac,ng on just on one point (the CG): O τ CG = xCG Mg Se]ng the two equal yields: x τ net = τ CG ⇒ xCG = 1 M ∑m x ii = xCM The CG is the same as the center of mass (CM), unless the accelera,on due to gravity is different for different parts of the body, which may happen for very large objects, such as the moon. Example A uniform beam of length L and mass m = 1.8kg is at rest on two scales. A uniform block, with mass M=2.7 kg, is at rest on the beam, with its center a distance L/4 from the beam’s leC end. What do the scales read ? Step 1: Draw a force diagram Step 2: Write down the force balance: Fnet = 0 ⇒ Fl + Fr − Mg − mg = 0 ⇒ Fl + Fr = Mg + mg.......(1) Step 3: Write down the torque balance about a convenient rotaMon axis, e.g. the leC end of the beam: τ net = 0 ⇒ 0 ⋅ Fl − ( L / 4 ) ⋅ Mg − ( L / 2 ) ⋅ mg + L ⋅ Fr = 0 1 1 ⇒ Fr = Mg + mg.....(2 ) 4 2 Note how the choice of the rotaMon axis eliminated one of the unknown forces. Step 4: Solve the equaMons (eqns(1) and (2) here): Fr = 15.4 N Fl = 28.7 N Example A firefighter (M = 72 kg) is walking up a ladder of length L = 12 m and mass m =45 kg that leans against a fricMonless wall. The ladder’s upper part is h=9.3 m above the pavement on which the ladder rests (the pavement is not fricMonless). The ladder’s CM is L/3 from the lower end. What is the force (magnitude) on the ladder from the wall and from the pavement when the person is half ­way up the ladder ? Force balance: Fnet , x = 0 ⇒ FW − Fpx = 0 Fnet , y = 0 ⇒ Fpy − Mg − mg = 0 Solve for Fpy: Torque balance (about O): use τ net ,O = 0 ⇒ − FW h + Mg(a / 2 ) + mg(a / 3) = 0 Get a from geometry: Solve for FW: 2 2 Fp = Fpx + Fpy = 1217 N Example: Problem 19 (Ch 12, textbook) Chalkboard stuff ...
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This note was uploaded on 11/16/2010 for the course PHY 2170 taught by Professor Blank during the Spring '08 term at Wayne State University.

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