Transcript Document
Chapter 8: Equilibrium and Mechanical Advantage Equilibrium: stability, steadiness, balance etc. Mechanical Equilibrium: absence of change in motion => Net Force = 0 ! (usually, no motion) F 0 sum of x force components = sum of y force components = F F x 0 y 0 p150c8:1 How to approach an equilibrium problem: • Draw the forces that act on the object (i.e. draw a free-body diagram) • Choose a convenient set of coordinate axis and resolve all forces into components. Watch carefully for appropriate use of +/- signs. • Set the sum of the force components along each axis equal to 0. • Solve the resulting equations for the unknown quantity or quantities. • Substitute numerical values of the known quantities to find the answer. p150c8:2 Hypothetical Example: mass hanging from two ropes as shown. TB q TBy f B A TA f q TAy TBx TAx M W = Mg TAx TA cosq TAy TA sin q F F TBx TB cos f TBy TB sin f W Horizontal Component s Vertical Components x TAx TBx TA cosq TB cos f 0 Horizontal direction y TAy TBy TA sinq TB sin f W Vertical direction p150c8:3 Example: mass hanging from two ropes as shown. Determine the tension in each rope. 37o Draw the forces that act on the object (i.e. draw a free-body diagram) 60o B A 10kg Choose a convenient set of coordinate axis and resolve all forces into components. Watch carefully for appropriate use of +/- signs. Set the sum of the force components along each axis equal to 0. Solve the resulting equations for the unknown quantity or quantities. Substitute numerical values of the known quantities to find the answer. p150c8:4 Example: A 100-N box is suspended from the end of a (pivoted) horizontal strut, as shown. Find the tension in the cable and the force exerted on the strut. 30o w p150c8:5 Rotational Equilibrium Rotational Equilibrium: absence of change in rotation => net torque is zero (usually: no rotation) 0 for all torques lying in the same plane Watch signs for torque F L Positive torque for counterclockwise rotation: = F L L F Negative torque for clockwise rotation: = F L p150c8:6 Example: The weight of an unknown mass located 15 cm from the pivot on a beam is exactly balanced by the weight of a 10 kg mass located at 45 cm. Determine the unknown mass. 10 kg ? kg p150c8:7 Kinds of equilibrium: (demos with cone) Stable Equilibrium: object tends to return to equilibrium after a small disturbance. Unstable Equilibrium: object tends to change rapidly away from equilibrium after a small disturbance. Neutral Equilibrium: object has no tendency towards or away from equilibrium after a small disturbance. p150c8:8 Center of Gravity (CG): the point of an object from which it could be suspended without tending to rotate. The point where all the mass of an object can be considered to be located. CG does not need to be located within the physical object! Horseshoe, for example usually easily identified from symmetry. p150c8:9 Example 8.7: A 60 kg woman stands at the end of a uniform 4m, 50 kg diving board supported as shown. Determine the forces exerted by the two supports. 4.00 m .800 m p150c8:10 Example: A horizontal beam 4 m long is supported at one end by a vertical post and at the other end by a cable which makes an angle of 40o with the beam. A load of 1500 Kg is suspended from the outer end of the beam, which itself has a mass of 250 kg. Find the tension in the cable. 40o w p150c8:11 Example: A ladder 4.00 m long is leaning against a frictionless wall with its lower end 1.6 m away from the wall. If the ladder weighs 150N, what forces do the wall and the ground exert on it? 4.0 m q 1.6 m p150c8:12 Finding the center of gravity: weight of “pieces” provides torque = torque by total mass at CM m1 gx1 m2 gx 2 M tot gX CM location m1 x1 m2 x 2 X m1 m2 (this can be extended to 2 and 3 dimensions with y,z) example: balance in lab Forces through CM do not cause rotation p150c8:13