Phic 110 Spring 006 orce in 1- nd -Dienion heir Solution 1. wo orce 1 nd ct on 5kg. I the gnitude o 1 nd re 0 nd 15 repectivel wht re the ccelertion o ech o the e elow?.. 0; ( 0 ) + ( 15 ) 1 5kg 15 @ θ tn @ θ tn inθ 16in 60 13 @ θ tn @ θ tn 15 0 5 + 1, @ θ tn + coθ 0 + 15 co 60 7.5 + + + @ θ 31 0 ( 7.5 ) + ( 13 ) 13 0 5kg @ θ tn 7.5 6.08 @ θ 5.3
. You tnd on the et o chir nd then hop o.. During the tie ou re in light down to the loor, the Erth i oving up towrd ou with n ccelertion o wht order o gnitude?. he Erth ove up through ditnce o wht order o gnitude?. hu,. Erth Erth Both ou nd the Erth eert equl nd oppoite orce on ech other. 70kg 9.8 4 6.0 10 kg Erth Erth 1 Erth ou t ou t Erth ou Erth Erth Erth Erth 1. 10 ou ou ou 1. 10 0.50 9.8 ou Erth 5.8 10 ou Erth or chir tht i,50c high, ou oth ove or equl tie. 3. he ditnce etween two telephone pole i 50. When 1kg ird lnd on the telephone wire hlw etween the pole, the wire g 0... Drw ree od digr o the ird.. 5 θ 0. 4 ou 1 g. How uch tenion doe the ird produce in the wire? 0. o. tnθ 0.008 θ 0.458 5 : coθ coθ 0 1 : 1 1 inθ g g 1kg 9.8 inθ in ( 0.458) 61.5 0 1 1 inθ g
4. A g o ceent o weight 35 hng ro three wire hown elow. wo o the wire ke ngle θ 1 60 o nd θ 5 o with repect to the horizontl. I the te i in equiliriu, wht re the three tenion orce in the wire? θ 1 60 θ 60 1 g ro the ree od digr, 3 g 35 (1) Σ : 1 co θ 1 - co θ 0 () Σ : 1 in θ 1 + in θ g (3) coθ1 ro () we hve 1 co. Sutituting thi into (3) we coθ1 hve inθ in 35 1 + θ 1 1 1 coθ 95. 7. hereore, 163.1 5. A ire/recue helicopter crie 60kg ucket o wter t the end o 0 long cle. A the ircrt lie ck ro ire t contnt peed o 40/, the cle ke n ngle o 40 o with repect to the verticl.. Wht i the orce due to ir reitnce on the ucket?. Ater illing the ucket with wter, the pilot return to the ire t the e peed with the ucket now king 7 o ngle with the verticl. Wht i the o the wter in the ucket? http://www..ed.u/r4/c/ire/005/photo/llcrk/helicopter.jpg. ro the ree od digr we hve
40 o : co50 g : in 50 g 0 ir ro () we hve tht ro (1) we hve tht ir 0 ( 1) ( ) g 60kg 9.8 7931.7 in 50 in 50 7931.7 co50 5098.4 7 o g. ro the ree od digr we hve : co83 ir 0 ( 1 ) : in 83 ( + ) g 0 ( ) w ir 5098.4 ro (1) we hve tht, ince ir i contnt, co83 co83 in 83 g ro () we hve olving or w : w 3617.1kg g 41835 6. A lock lide down rictionle plne hving n inclintion o θ 15 o.. I the lock trt ro ret, wht i the ccelertion o the lock down the incline?. Wht i the peed o the lock t the otto o the incline i it trvel ditnce o down the incline? c. Suppoe now tht riction doe eit etween the lock nd incline. Wht i the new ccelertion i the coeicient o kiic riction, µ k 0.5 (OE µ K HAS BEE CHAGED ROM HE ORIGIAL ASSIGEME.)? d. Wht i the new peed o the lock t the otto o the incline?
. ro cl we ve een, tht or coordinte te with the -i perpendiculr to the plne nd the -i down the plne we hve the µ ccelertion ginθ 9.8 in15.54.. I the lock trvel then the inl peed i given v.54 10. 15. c. When we included riction, in cl, the ccelertion i lower nd i µ µ given ginθ µ k g coθ 9.8 in15 0.5 9.8 co15 0. 17. d. he inl peed i given v 0.174 0. 68 7. In the te hown elow horizontl orce ct on the 8kg. Conider two ce, one where the horizontl urce i rictionle nd one where there i riction nd the coeicient o kiic riction i µ k 0.75.. or wht vlue o doe the kg ccelerte upwrd?. or wht vlue o i the tenion in the cord zero? c. Plot the ccelertion o the 8kg veru. Include vlue o ro -100 to 100. (Quetion #7 will not e on e #1! Solution will e poted ter the e.) 8. A won t the irport i towing her 0kg uitce t contnt peed pulling on trp t n ngle o θ ove the horizontl. She pull the trp with orce o 35 nd the rictionl orce on the uitce i 0.. Wht i the ngle θ tht the trp ke with the horizontl?. Wht norl orce doe the ground eert on the uitce?
θ 35 r 0 : : g coθ 0 + inθ g ( 1) 0 ( ) -1 0 o ro () 1 we get the ngle : θ co 55. 35 ro we the norl orce : g inθ 167.3 9. A o drg hi 60 led t contnt peed up 15 o hill. He doe o pulling with 5 orce on rope ttched to the led. I the rope i inclined t 35 o to the horizontl,. Wht i the coeicient o kiic riction etween the led nd the now?. At the top o the hill he jup on the led nd lide don the hill. Wht i the gnitude o the ccelertion o the o nd led?. ro the ree od digr we ee tht the nd orce re : 5 co 0 60 in15 0 1 ro : ( ) () + 5 in 0 60 co15 0 49. 43. ( ) ( ) hu ro 1 we get the coeicient o kiic riction ince K µ 0.16 r we get the norl orce : r µ K.. ro quetion 6c, the ccelertion i given gin g co 9.8 µ k in15 0.16 9.8 µ θ µ θ co15 1. 0 10. A lock o kg i releed ro ret h 0.5 ro the urce o tle, t the top o θ 30 o incline. he rictionle incline i ied on tle o height H.. Wht i the ccelertion o the lock it lide down the incline?. Wht i the velocit o the lock it leve the incline? c. How r ro the tle will the lock lnd? d. How uch tie h elped etween the lock w releed nd when it hit the loor? e. Doe the o the lock ect n o the ove clcultion?
. Redo the ove clcultion including rictionl orce etween the lock nd the incline with coeicient o kiic riction o µ k 0.57.. ro cl, the ccelertion down the incline i given µ ginθ 9.8 in 30 4.. 9 v 4.9 1 3.13. he velocit i given, c. where, in θ h/d o d h/inθ 1. v v i i R.71 o clculte R, we need to know the tie. hi we get ro the verticl otion. H -1.57 t 3.13 3.13 co30.71 t in 30.57 t 4.9 { 0.499,-0.819}, t which give the qudrtic orul, nd we chooe t 0.499. hu, R 1.35 d. he totl tie i the u o the liding otion nd the lling otion. he tie to ll i ro prt c nd i 0.499. he tie to lide i ound 3.13 ro v vi + t t 0. 639. hereore the totl tie 4.9 ro relee to lnding i 0.499 + 0.639 1.14. e. o, the doe not ect n reult ince it doe not pper in n o the ove clcultion.. Including riction, the ccelertion i given gin g co 9.8 µ in 30 k 0.57 9.8 µ θ µ θ co 30 0. 06. he velocit i given v 0.06 1 0.35.
v v i i R.71 o clculte R, we need to know the tie. hi we get ro the verticl otion. H -.176 t 0.35 0.35 co 30 0.303 t t 4.9 { 0.434,-0.470}, in 30 0.176 t which give the qudrtic orul, nd we chooe t 0.434. hu, R 0.13. he totl tie i give ro the u o the ll tie nd the lide tie. 0.35 he lide tie i given v vi + t t 5. 65, thu the 0.06 totl tie i 0.434 + 5.65 6.08. Agin, h no eect on the reult. 11. A oile hng in child roo ored upporting our etl utterlie o equl ro tring o length L. he point o upport re evenl pced ditnce l prt. he tring or n ngle θ 1 with the ceiling t ech point. he center ection o the tring i horizontl.. Wht i the tenion in ech ection o the tring in ter o θ 1, nd g?. Wht i the ngle θ, in ter o θ 1, tht the ection o the tring etween the outide utterlie nd the inide utterlie or with the horizontl. c. or cold pring night: Show tht the ditnce D etween the end point L 1 o the tring i D { coθ1 + co[ tn ( tn 1) ] θ + 1}. 5 (Quetion #11 will not e on e #1! Solution will e poted ter the e.) 1. Beore 1960 it w elieved tht the iu ttinle coeicient o ttic riction or n utooile tire w le thn 1. hen out 196, three
copnie independentl developed rcing tire with coeicient o 1.6. Since then, tire hve iproved. According to the 1990 Guinne ook o Record, the tet tie in which piton-engine cr initill t ret h covered ditnce o ¼ ile i 4.96. hi record w et Shirle Muldowne in Septeer 1989.. Auing tht the rer wheel lited the ront o the pveent, wht iniu vlue o µ i necer to chieve thi record tie?. Suppoe Muldowne were le to doule her engine power, keeping ll other ctor equl, how would thi chnge the elped tie? (Quetion #1 will not e on e #1! Solution will e poted ter the e.)