hw9_sol_s11

hw9_sol_s11 - Phys 2101 Homework 9 Solution Spring `11...

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These so appear in however. 1. Three of the wa downwar rope and The horiz (a) We so obtain (b) We so obtain N F 2. The ob causing a problem accelerat string and Therefor lutions use t n your person . forces act on all N F r (actin rd). Since th the vertical. zontal comp olve the first mg T = olve the seco 22 Tr Lr = = + bject exerts a a “kink” sim in the text). tion is zero) d its “relaxe e, T = F /(2si Phys 210 the paramete nal homewo n the sphere ng horizontal he sphere is i . Then, the v onent is F N t equation fo ( L + = ond equation 2 mg L r L + a downward ilar to that s By analyzin 2 T sin θ = F , d” position ( in ) = 7.92 01 Home er values fro ork assignme : the tension lly away from n equilibrium vertical comp T sin = 0 or the tension 0.85 kg)(9.8 n for the nor 2 r = + force of ma shown for pr ng the forces , where is (when the tw tan = × 10 3 N. work 9 Sol m the proble ent. The logic n force r T of m the wall), m they sum ponent of Ne 0. n: T = mg /co 2 8 m/s ) (0.0 0.080 m rmal force: F (0.8 mgr L = agnitude F = roblem 10 (se s at the “kink the angle (ta wo segments 1 0.35m 1.72m ⎛⎞ = ⎜⎟ ⎝⎠ lution Spr ems printed i c used to get the rope (ac and the forc to zero. Let ewton’s seco os . We sub 2 080 m) (0. m + . sin N T = . 5 kg)(9.8 m/ (0.080 3160 N at th ee the figure k” where r F aken positive s are collinea 11.5 . ° ring `11 in the book, t to the answ cting along th ce of gravity be the ang ond law is T bstitute cos 2 042 m) 9 = . Usingsin 2 /s )(0.042 m 0 m) he midpoint that accom is exerted, w e) between e ar). In this pr not those th wer is the sam he rope), the y mg r (acting gle between cos mg = + LL / 2 .4 N =+ rL r / 2 ) 4.4 N. = of the rope, panies that we find (sinc each segmen roblem, we h hat me, e force g the = 0. r 2 to 2 , we , ce the nt of the have
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3. P. 12-0 diving bo the weigh forces eq (a) The s which sh (b) Since (c) The f which sh (d) The r (e) The f pedestal (f) The fo compress 4. (a) An (b) Equil (c) Comp 07. We take oard. We tak ht of the div qual to zero ( econd equat hould be roun e F 1 is negati first equation hould be roun result is posi force of the d on the divin orce of the d sed. nalyzing the librium of ve puting torque v F d the force of ke the force o er, located a (with upward tion gives 1 F = − nded off to F ive, indicatin n gives 2 F = nded off to F tive, indicat diving board ng board), so diving board horizontal fo ertical forces es about poi 23 d Fb Fa =+ f the left ped of the right p at x = L . The ds positive), 1 F Fd Ld W d = − 1 1.2 10 =− × ng that this f 1 58 WF −= 3 2 1.7 10 ing that this d on the left p this pedesta on the right orces (which s leads to F v nt O , we obt ( 10 d ⇒= estal to be F pedestal to b following tw and the sum 12 () FW WL d + = + = 3.0m (5 1.5m ⎛⎞ ⎜⎟ ⎝⎠ 3 0 N . Thus, force is down 0 N+1160 N N . Thus, | force is upw pedestal is u al is being st t pedestal is d h add to zero = F 1 + F 2 = tain )( ) 0 N3 .
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hw9_sol_s11 - Phys 2101 Homework 9 Solution Spring `11...

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