First, we will find the components of the force of gravity: Perpendicular Forces (using away from the ramp as positive) ma F

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1 1.. In Clss or Homework Eercise 1. An 18.0 kg bo is relesed on 33.0 o incline nd ccelertes t m/s. Wht is the coeicient o riction? m 18.0kg 33.0? m / s irst, we will ind the components o the orce o grvit: g mg sin (18.0)(9.80) sin g mg cos (18.0)(9.80) cos Perpendiculr orces (using w rom the rmp s positive) m m g m g 148 Prllel orces (using down the rmp s positive) RRHS Phsics Pge 75

2 m (18.0)(0.300) 96.1 g g m m 90.7 (148) A bo (mss is 455 g) is ling on hill which hs n inclintion o 15.0 o with the horizontl.. Ignoring riction, wht is the ccelertion o the bo down the hill? m 455g 0.455kg ? irst, we will ind the components o the orce o grvit: g mg sin g mg cos (0.455)(9.80) sin15.0 (0.455)(9.80) cos Since there is no riction this time, we do not need to look t the perpendiculr orces: Prllel orces (using down the rmp s positive) RRHS Phsics Pge 76

3 m m m (0.455) 1.15 g g.53 m / s b. I there is coeicient o riction o 0.0, will the bo slide down the hill? I so, t wht ccelertion? m 455g 0.455kg g g ? Since there is riction this time, we will need the norml orce. We must thereore look t the perpendiculr orces: Perpendiculr orces m m g m g 4.31 Prllel orces (using down the rmp s positive) RRHS Phsics Pge 77

4 (0.0)(4.31) 0.86 Since g, the bo will ccelerte down the hill. m (0.455) g g m m 0.64 m / s c. How much orce is required to push the bo up the rmp t constnt speed? Since the bo is now going up the hill, riction must be down the hill: m 0.455kg g Using up the rmp s positive nd looking t the prllel orces, RRHS Phsics Pge 78

5 m m g p m g p p g p g A 165 kg pino is on 5.0 rmp. The coeicient o riction is Jck is responsible or seeing tht nobod is killed b runw pino.. How much orce (nd in wht direction) must Jck eert so tht the pino descends t constnt speed? Initill, we do not know wht direction Jck hs to ppl his orce so tht the pino descends t constnt speed. I we look t the ree Bod Digrm without Jck s orce m 165kg p 0 0? We see tht we must compre g nd g mg sin (165)(9.80) sin g mg cos (165)(9.80) cos Perpendiculr orces: RRHS Phsics Pge 79

6 m g g m m 1470 (0.30)(1470) 440 Since g, the pino will ccelerte down the rmp i Jck does nothing; thereore Jck must push up the rmp: Using up the rmp s positive, Prllel orces: m m 0 p g p g p g The orce must be directed up the rmp. b. How much orce (nd in wht direction) must Jck eert so tht the pino scends t constnt speed? RRHS Phsics Pge 80

7 g p ? Using up the rmp s positive, m m 0 p g p g p g The orce must be directed up the rmp. 4. A cr cn decelerte t -5.5 m/s when coming to rest on level rod. Wht would the decelertion be i the rod inclines 15 o uphill? The coeicient o riction will be the sme in ech sitution (sme surces). We must ind this rom the level surce prt o the problem. Level Rod 5.5 m / s? Using the orwrd direction s positive, RRHS Phsics Pge 81

8 m m m m mg 5.5 (9.80) 0.56 Incline ? Since there is no perpendiculr ccelertion, g. Looking t the prllel orces nd using up the rmp s positive, m g g m mg sin mg cos m m (9.80) sin15 (0.56)(9.80) cos m/ s 5. A sled is sliding down hill nd is 5 m rom the bottom. It tkes 13.5 s or the sled to rech the bottom. I the sled s speed t this loction is 6.0 m/s nd the slope o the hill is 30.0, wht is the coeicient o riction between the hill nd the sled? Using down the rmp s positive, RRHS Phsics Pge 8

9 d 5m v 6.0 m / s i t 13.5s 30.0? d v t t 1 i 5 (6.0)(13.5) (13.5) 1.6 m / s 1 m m mg sin mg cos 1.6 (9.80) sin 30.0 (9.80) cos g g g m m m 6. A 5.0 kg mss is on rmp tht is inclined t 30 o with the horizontl. A rope ttched to the 5.0 kg block goes up the rmp nd over pulle, where it is ttched to 4. kg block tht is hnging in mid ir. The coeicient o riction between the 5.0 kg block nd the rmp is Wht is the ccelertion o this sstem? igure 19: Digrm or Question 6 RRHS Phsics Pge 83

10 m 5.0kg m 1 4.kg 0.10? We must irst determine the direction o ccelertion o the sstem b compring g nd g1. g1 m1g sin (5.0)(9.80) sin g m g (4.)(9.80) 41. Since g g 1, the sstem will ccelerte clockwise nd riction on the 5.0 kg object will be down the rmp. I we tret the whole sstem s one object (s in Unit 3), our linerized ree bod digrm will look like this: The tension in the rope cn be ignored since we re treting the sstem s one big object. g mg cos (0.10)(5.0)(9.80) cos Using clockwise s the positive direction (towrd the 4. kg mss), RRHS Phsics Pge 84

11 m m t g g1 m t g g1 (9.) m / s 7. A orce o pplied to rope held t 30.0 o bove the surce o rmp is required to pull bo weighing t constnt velocit up the plne. The rmp hs bse o 14.0 m nd length o 15.0 m. Wht is the coeicient o riction? igure 0: Digrm or Question 7 p g ? g mg m(9.80) m 10kg RRHS Phsics Pge 85

12 We must brek the pplied orce up into components tht re prllel nd perpendiculr to the rmp. cos sin p p (500.0)(cos 30.0 ) (500.0)(sin 30.0 ) p p et we must ind the ngle tht the rmp mkes with the horizontl so tht we cn ind the components o the orce o grvit: 14 cos 15 1 g mg sin (10)(9.80) sin g mg cos (10)(9.80) cos Perpendiculr orces m p g 0 p g m 680 Prllel orces (using up the rmp s positive) m p g m 73 p g 73 (680) I bicclist (75 kg) cn cost down 5.6 o hill t sted speed o 7.0 km/h, how much orce must be pplied to climb the hill t the sme speed? We cn ssume the sme mgnitude or the orce o riction on the w up nd the w down the hill: Down the Hill RRHS Phsics Pge 86

13 m 75kg v 7.0 km / h? Prllel orces (Using down the rmp s positive) m 0 g g m mg sin (75)(9.80) sin(5.6 ) 7 Up the Hill m 75kg v 7.0 km / h p 7? RRHS Phsics Pge 87

14 Prllel orces (Using down the rmp s positive) m m 0 g p g p p g RRHS Phsics Pge 88

A wire. 100 kg. Fig. 1.1

A wire. 100 kg. Fig. 1.1 1 Fig. 1.1 shows circulr cylinder of mss 100 kg being rised by light, inextensible verticl wire. There is negligible ir resistnce. wire 100 kg Fig. 1.1 (i) lculte the ccelertion of the cylinder when the