hw9_sol_s11

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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online