25lecture2a08

25lecture2a08 - Physics 2A Lecture 25: Mar 7 Vivek Sharma...

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Physics 2A Lecture 25: Mar 7 Vivek Sharma UCSD Physics
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Learning From Equilibrium Conditions C Box, mass M is hanging by a rope from a boom consisting of a hinged beam and a horizontal cable. Beam is uniform has mass = m, cable & rope are massless. (a) tension T in c ab (b) magnitude Find F le of force on beam due to hinge
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static equilibriu is in Beam F= 0 m 0 , τ = G G O 2 x y 2 Force balancing: & F0 hc vr h v FT F mg T F F F = = = =+ = + Calculate torque about the hinge at point O: 0 ( ) ()( ) / 2 ) ( ) 0 cr aT bT b m g =⇒ = G (/ 2 sin e ) c r c gb M m T M a Tg = + =
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Romeo trying to reach Juliet by climbing 5m long uniform ladder weighing 180N. Romeo weighs 800N. Clim hungry alliga bs up 1/3 of ladder then ponders the do Bottom wn of in the ladder moat. rests tors on a o s horizontal surface & leans across from moat ( =53.1 ) in equilibrium against frictionless vertical w Find normal & friction forces on ladder at it (b) find minimum friction needed to pre s base all. θ μ vent slipping at base Freebody diagram of Romeo & Ladder o cos(53.1 )=0.6 h l θ Romeo & Juliet
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uniform ladder CG = geometric center x ladder =1.5m Romeo’s weight acts thru x romeo =1m frictionless wall exerts normal force n 1 2 2 orces at base of ladder are upward normal & static friction ; coeff. of static friction (equilibrium n condition) s s s f n μ μ= G 1 2 2 1 equilibrium condit 0 ; ion () 0 0 0 xs s rome y romeo ladde ol a d e r r d F nf nw w Ff n Fn w w τ =+ = == = =− = G G
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Calculate Equilibri torque a um no net rotati bout ladder's 0 bas on ! of Rom ( why a smart choic eo e? ) e B +ladder τ = G gives fewest terms with fewest unkno (? wns! ) s f = 1 s2 2 1 1 1 1 ( )/ 268N (0) 0 268N & / 0.27 ladder ladder romeo romeo ladder ladder romeo romeo Bs s s nn f d n f wx x d n fn w μ =+ + = −− + = ⇒= = == = G B 2 ˆˆ ˆ ˆ Contact force F (268 980 ) s i j i j = + G
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Romeo’s Fate ? Romeo take 2A ! did not (such is foolish love !) 1s lever arm and torque due to his weight about B increas s he climbs higher on ladder, increasing values of n ,f es & s μ ladder s at top of ladder, x 3 requiring 0.7 for equilibrium!
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25lecture2a08 - Physics 2A Lecture 25: Mar 7 Vivek Sharma...

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