Physics 12 Assignment KEY Equilibrium & 2-D Dynamics

Transcription

1 Physics Assignment KEY Equilibrium & -D Dynamics. Deine the llwing terms: Newtn s laws mtin - three undamental laws mtin which are the basis Newtnian mechanics are: ) an bject will remain at rest r in straight-line mtin unless acted n by an utside rce; ) the acceleratin an bject is prprtinal t the rce acting n it and inersely prprtinal t its mass; ) r eery actin rce n an bject, the bject exerts and equal and ppsite reactin rce rictin - a rce that acts t slw the mtin an bject r eep it at rest ensin - the magnitude the rce exerted n and by a cable, rpe, r string Static equilibrium - the state an bject when the ectr sum all the rces acting n it is zer and the sum all the trques acting n it is zer ranslatinal equilibrium a state equilibrium where the ectr sum all rces is zer Rtatinal equilibrium - a state equilibrium where the ectr sum all trques is zer rque the tendency a rce t rtate an bject abut an axis, deined by the prduct the rce n a bject and the leer (mment) arm (NOE: trque is nt rce) Leer arm - the perpendicular distance between the line alng which the rce is acting and the pit pint a rtatin Center mass - the pint at which an bject can be balanced ulcrum (r pit pint) - the pint arund which rtatinal mtin ccurs. a) A bx sits at rest n a rugh 0 inclined plane. Draw ree bdy diagram, shwing all the rces acting n the bx. b) Hw des the diagram change i the bx were sliding dwn the plane? c) Hw des the diagram change i the bx were sliding up the plane? (a) (b) (c) In (a) the rictin is static and ppses the impending mtin dwn the plane. In (b) the rictin is inetic and ppses the mtin dwn the plane. In (c) the rictin is inetic and ppses the mtin up the plane.. A bungee jumper mmentarily cmes t rest at the bttm the die bere she springs bac upward. At that mment is the bungee jumper in equilibrium? Explain. he bungee jumper is nt in equilibrium, because the net rce n the jumper is nt zer. I the jumper were at rest and the net rce were zer, then the jumper wuld stay at rest by Newtn s st law. he jumper has a net upward rce when at the bttm the die, and that is why the jumper is then pulled bac upwards.

2 4. A ladder, leaning against a wall, maes a 60 angle with the grund. hen it is mre liely t slip: when a persn stands n the ladder near the tp r near the bttm? Explain. hen the persn stands near the tp, the ladder is mre liely t slip. In the accmpanying diagram t the right, the rce the persn pushing dwn n the ladder,, causes a clcwise trque abut the cntact pint with the grund, with leer arm d x. he nly rce causing a cunterclcwise trque abut that same pint is the reactin rce the wall n the ladder,. hile the ladder is in equilibrium, will be the same magnitude as the rictinal rce at the grund, Gx. Since Gx has a maximum alue, will hae the same maximum alue, and s will hae a maximum cunterclcwise trque that it can exert. As the persn climbs the ladder, their leer arm gets lnger and s the trque due t their weight gets larger. Eentually, i the trque caused by the persn is larger than the maximum trque caused by, the ladder will start t slip it will nt stay in equilibrium. 5. An earthen retaining wall is shwn in the igure (a) belw. he earth, particularly when wet, can exert a signiicant rce n the wall. (a) hat rce prduces the trque t eep the wall upright? (b) Explain why the retaining wall in the igure (b) belw wuld be much less liely t erturn. (a) I we assume that the pit pint rtatin is the lwer let crner the wall in the picture, then the graity rce acting thrugh the CM prides the trque t eep the wall upright. Nte that the graity rce wuld hae a relatiely small leer arm (abut hal the width the wall) and s the sideways rce wuld nt hae t be particularly large t start t me the wall. (b) ith the hrizntal extensin, there are actrs that mae the wall less liely t erturn: he mass the secnd wall is larger, and s the trque caused by graity (helping t eep the wall upright) will be larger r the secnd wall. he center graity the secnd wall is urther t the right the pit pint and s graity exerts a larger trque t cunteract the trque due t. he weight the grund abe the new part the wall prides a large clcwise trque that helps t cunteract the trque due t.

3 6. A child slides dwn a slide with a 8 incline, and at the bttm her speed is precisely hal what it wuld hae been i the slide had been rictinless. Calculate the ceicient inetic rictin between the slide and the child. Examine bth situatins, ne withut rictin and ne with rictin: ithut rictin: m a sinθ net a g sinθ ithut rictin: ith rictin: - m a sinθ net sin θ - µ cs θ m a a g(sinθ µ cs θ ) d is the displacement the child (i.e. Super Raccn Mari) dwn the slide, regardless the presence rictin + ad i 0 d a a a Since "d" is the same in bth situatins, set rati a sle r µ. rictin nne rictin nne a a i rictin rictin rictin ( ) a a in each situatin equal t each ther and nne a rictin g(sinθ µ cs θ ) 4 a g sinθ nne µ.99 rictin nne a a rictin nne ith rictin: Nte: the child in this BD is actually Racn Mari! 7. In the design a supermaret, there are t be seeral ramps cnnecting dierent parts the stre. Custmers will hae t push grcery carts up the ramps, and it is biusly desired that this nt be t diicult. An engineer has dne a surey and und that almst n cmplains i the rce required is n mre than 50 N. ill a slpe 5 be t steep, assuming a 0-g grcery carts (ull grceries)? Assume rictin (wheels against grund, wheel n the axles, and s n), can be accunted r by a c-eicient µ.0.

4 It must be determined i the sum the rces dwn the ramp are less than 50 N while ming at cnstant speed. I s, n ne shuld cmplain abut ming their shpping cart up the ramp. 50 N - - net 50 N + sinθ + µ csθ 50 N (0 g)(9.8 m/s )(sin cs5 ) N 50 N N Since the ppsing rces are greater than 50 N, the ramp is deemed t steep. 8. A 8.0 g blc is cnnected t an empty.00 g bucet by a crd running er a rictinless pulley (See igure t the right). he ceicient static rictin between the table and the blc is 0.450, and the ceicient inetic rictin between the table and the blc is 0.0. Sand is gradually added t the bucet until the system just begins t me. (a) Calculate the mass sand added t the bucet. (b) Calculate the acceleratin the system. (a) he mass the sand can be determined by examining the rces at the mment the system is abut t accelerate (a ). r the sand bucet... - net G ( m + m ) g G bucet sand r the blc... - net r the Atwd's Machine, is the same thrughut the cnnectr, thus... ( m + m ) g µ m s bucet sand blc g ( m + m ) g bucet sand m µ m m sand s blc bucet.6 g (b) he acceleratin the system is und by r the sand bucet... - m a net G system ( m + m ) g µ m g ( m + m + m ) a bucet sand blc bucet sand bucet ( mbucet + msand ) g µ mblc g a.88 m/s ( m + m + m ) bucet sand bucet

5 9. ind the center mass the three-mass system shwn belw. Assume the spheres are pint masses. Chse the reerence pint, x m, at the letmst particle. hus, 0. he CM an empty 050-g car is.50 m behind the rnt the car. Hw ar rm the rnt the car will the CM be when tw peple sit in the rnt seat.80 m rm the rnt the car, and three peple sit in the bac seat.90 m rm the rnt? Assume that each persn has a mass 70.0 g.. A square unirm rat, 8 m by 8 m, mass 6800 g, is used as a errybat. I three cars, each mass 00 g, ccupy its NE, SE, and S crners, determine the CM the laded errybat.

6 . hree rces are applied t a tree sapling, as shwn in the igure belw, t stabilize it. I 8 N and 55 N, ind in magnitude and directin. rm the rce diagram r the sapling, we can write: Σ sin 0 8 N - (55 N)(sin 0 csα 6 N x Σ sinα 4 N y cs 0 csα sinα ) - csα hus, we hae: 70 N α tan - θ 80 (6 N) 4 N 64 6 N (4 N) 6 \. Calculate the trque abut the rnt supprt a diing bard, in the igure t the right, exerted by a 60-g persn.0 m rm that supprt. e chse the ulcrum at the rnt supprt the diing bard, with psitie trques clcwise. he persn s weight prduces a psitie trque abut the ulcrum: τ r r (60 g)(9.8 N/g)(.0 m) 800 N m g 4. w crds supprt a chandelier in the manner shwn in the igure t the right except that the upper wire maes an angle 45 with the ceiling. I the crds can sustain a rce 00N withut breaing, what is the maximum chandelier weight that can be supprted? here are actually three rces t cnsider:,,and the weight, g, the chandelier (as it deelps tensin in t cnnecting ertical crd). I we l at the pint where the crds cme tgether, we can express the equilibrium that pint as llws: (he ectr sum,, and g - where is the weight the chandelier)

7 In the x directin: x + x + gx - cs(45 )+ x + 0 S we nw that x (It has n y cmpnent) cs(45 ) In the y directin y + y + gy sin(45 )+ 0 - g S we nw that ( g )/sin(45 ) And cs(45 ) (( g )/sin(45 ))cs(45 ) g [since sin(45 ) cs(45 )] S is always ging t be larger than and which are bth equal t the weight, since sin(45 ) is less than ne, and ( g )/sin(45 ). Since the maximum tensin is 00 N, set equal t 00 N: In the x directin: x + x + x -(00 N)cs(45 )+ x N + x + 0 x 99.4 N In the y directin y + y + y (00 N)sin(45 )+ 0 - g g g 99.4 N 90 N 5. A shp sign weighing 5 N is supprted by a unirm 5-N beam as shwn in the igure belw. ind the tensin in the guy wire and the hrizntal and ertical rces exerted by the hinge n the beam. e chse the ulcrum at the pint where the hrizntal beam maes cntact with the wall, called the hinge (see BD belw): Σ x hingeh hingeh x x csθ

8 Σ y hingev hingev hingev + y + + y sin θ 0 Στ τ hingeh 0 Since the leer arm at the hinge is zer, the trque due the hinge rce is zer. hus, Στ τ ( )(.5 m)(sin 4 ) (5 N)(0.85 m) + (5 N)(.70 m) (5 N)(0.85 m) + (5 N)(.70 m) (.5 m)(sin 4 ) 54 N τ τ herere, csθ (54 N)(cs 4 hingeh hingev ) 409 N 5 N + 5 N (54 N)(sin4 ) -5.7 N 6. Cnsider a ladder in the igure belw, with a painter climbing up. I the mass the ladder is.0 g, the mass the painter is 60.0 g, and the ladder begins t slip at its base when she is 70% the way up the length the ladder, what is the ceicient static rictin between the ladder and the lr? Assume the wall is rictinless. Chse the pit pint t be the pint n the grund where the ladder maes cntact. he ladder is in equilibrium, s the net trque and net rce must be zer. By stating that the ladder is n the erge slipping, the static rictinal rce at the grund, Gx, is at its maximum alue and s: Gx µ Gy. Gien the height the ladder is 4.0 m and the distance rm the wall is.0 m, the angle that the ladder maes with the grund is θ tan - (4.0 m/.0 m) 5.. he length the ladder is L. Σ Gx x Gx

9 Σ Gy y Gy 706. N Στ τ G Since the leer arm at the pit pint is zer, the trque due the pit pint rce, s, is zer. hus, Στ τ τ ( τ )( L)(sinθ ) ( )(csθ )( L) + ( )(csθ )(0.7L) ( )(csθ )( L) + ( )(csθ )(0.7L) ( L)(sinθ ) ( )(csθ )( ) + ( )(csθ )(0.7) sinθ ( )(ctθ )( ) + ( )(ctθ )(0.7) 5.6 N µ Gx Gy.5