f = µ mg = kg 9.8m/s = 15.7N. Since this is more than the applied

Transcription

1 Phsics 141H lutins r Hmewrk et #5 Chapter 5: Multiple chice: 8) (a) he maimum rce eerted b static rictin is µ N. ince the blck is resting n a level surace, N = mg. the maimum rictinal rce is ( ) ( ) ( ) = µ mg = kg 9.8m/s = 15.7N. ince this is mre than the applied hrizntal rce, rictin will eert as much rce as it needs t neutralize the eternal rce, and the blck will nt mve. (B) (b) With the blck in mtin, the rce eerted b kinetic rictin will be: ( ) ( ) ( ) = µ N = µ mg = kg 9.8m/s = 11.8N k k his will be in the ppsite directin ppsite the mtin, but is less than the applied eternal rce. Hence the net rce will be 1.9N in the directin mtin. (A) 1) a) ince the speed the mtrccle is cnstant as it ges arund the lp, the magnitude v its acceleratin,, is als cnstant. herere the net rce acting n the mtrccle r als has cnstant magnitude. ince the net rce is N + W +, the crrect answer is (D). b) Cnsider the rces acting n the mtrccle at a randm pint arund the lp: N W With these crdinates, the center the lp is in the directin. he velcit is in the directin, and since it is cnstant we knw there can be n acceleratin in, and therere

2 n net rce in that directin. N has n cmpnent, and therere the sum + W must als have n cmpnent. In ther wrds, + W is alwas directed tward the center the circle. (C) Questins: 7) B having the air eert a dwnward rce, the nrmal rce between the tires and the rad is increased. his is turn means that rictin can eert a greater rce n the car, allwing it t turn at a tighter radius than it culd therwise. 15) Accrding t the equatins r the velcit and perid the cnical pendulum (5-14 and 5-15), the speed wuld be ininite and the perid 0 r a 90 angle. his can never happen because i the string s tensin acts nl in the hrizntal directin, there will be nthing t balance the vertical rce gravit pulling the ball dwn. r θ = 0, the velcit is zer. his is the case the ball hanging straight dwn and rtating abut its ais. here is n deinite perid assciated with this tpe mtin. 16) As the turntable spins aster, the rce between the cin and the turntable needed t keep the cin mving in a circle increases. his rce is supplied b rictin, s nce it eceeds the maimum pssible rictinal rce, the cin will mve tward the uter edge the turntable and eventuall all. Eercises: 13) We start as alwas b drawing a rce diagram r the bar: 1 N 1 N 1 and r the mass: 1 mg ince we want the mass t be static, we must have: + = = mg 1

3 r the bar, the rces N 1 and N are each equal t (rm Newtn s hird Law), s the cancel. hat leaves the net rce equatins as: = N N = 0 1 = + mg 1 B smmetr, we knw that 1 =. Each these rictinal rces has a value and since we re interested in appling the least pssible, we ll take the case = = µ. We want the net rce in t be zer, s: 1 µ, = + mg = 0 1 µ = mg mg 75kg 9.8m/s = = = 900N µ ) We pull the crate at sme angle, as shwn belw: N W θ ind the nrmal rce, we set the cmpnent the net rce t 0: = N + sinθ W = 0 N = W sinθ the largest rce rictin can appl is µ N µ ( W sinθ ) = =. r the b t just start mving, the hrizntal cmpnent the tensin must be equal t this rce: ( ) = csθ = csθ µ W sinθ = 0 csθ W = + sinθ µ Nw we need t knw the angle t pull t maimize the weight we can mve. his means taking the derivative with respect t θ, and setting it equal t 0:

4 dw sinθ = + csθ= 0 µ sinθ = µ csθ tanθ = µ θ = tan µ = tan 0.35 = Putting this int ur equatin r W, we ind that the maimum weight that can be mved is: W csθ cs19 = + = µ + = 0.35 sinθ 1.kN sin kN 30) irst we need t determine the rictinal rce acting n the blck. In this case rictin acts n tw the blck s suraces, and we need t ind the nrmal rce r each. d s, let s lk at the side view irst: N z mg ince the blck is nt accelerating in the z directin, N (the net nrmal rce) must be equal t mg cs θ. Nw cnsider the rnt view: N 1 N mgcsθ B smmetr, we knw that N1 = N, and that the cmpnents are equal and ppsite. the net nrmal rce is:

5 N = N cs 45 + N cs 45 = mg csθ N1 = mg csθ mg csθ N = = N 1 1 µ Kmg csθ the rictin rce acting n each surace is 1, = µ K N 1, =, which means the ttal rictinal rce is = 1 + = µ Kmg csθ. rm the side-view diagram, this makes the net rce in equal t ( K ) mg sinθ µ mg csθ = mg sinθ µ cs θ. herere the acceleratin is g ( θ µ K θ ) K sin cs. 33) he rces n the car are (lking rm the rear the car): N mg he rictinal rce is the nl rce acting in, s it alne must prvide enugh rce t accelerate the car in circular mtin.

6 a) he rce needed is: ( ) ( ) mv 10.7kN/9.8m/s 13.4m/s = = = 3.kN r 61.0m b) ince the nrmal rce is equal t the car s weight 10.7kN, the minimum ceicient static rictin needed is: = µ N 3.kN µ = = = 0.30 N 10.7kN Prblems: 3) rm Newtn s ecnd Law, we knw that the acceleratin the sstem will be a = m j where I ve taken the directin t be the directin the rce. B smmetr, we als knw that each particle will have the same acceleratin and equal but ppsite acceleratins. Nw cnsider the rces acting n ne particle: θ sinθ a = = m = sinθ m a csθ csθ = = = tanθ m m sinθ m he act that the particle is held b the string means that: + = L L tanθ = =

7 ubstituting this int the equatin r a gives: a = tanθ = m m L he case = L is the situatin shwn in ig In this case, an nnzer will result in an ininite acceleratin in, meaning that the tensin in the string will be ininite (r that instant). An real string wuld stretch under these circumstances, making + > L, since appling an ininite tensin is impssible. 11) he ke t this prblem is t realize that the tensin in the string is dierent n the tw sides the dwel. see wh, cnsider the rces acting n a small segment the string n the dwel (this segment subtends an angle ): N 1 Each tensin vectr is at an angle d θ with respect t the ais. Assume the sstem is at rest. hen the rces must balance: = N 1 sin sin = 0 = 1 cs + cs = 0 ince we knw that is a ver small angle, we have the apprimate relatins: N = = 1 + µ ( + ) = 1 1 ( ) = d = µ + 1 1

8 ince d is small, we can apprimate 1 + as, where is the average the tw tensins. we have: d = µ ind the ttal change in tensin between the tw sides the string, we integrate ver the entire length cntact (rm θ = 0 t π): π 0 d 1 1 = π = = = ln = ln i µ µ µ i i i In ur case, we want the initial tensin t balance W and the inal tensin t balance, s: 17) 1 π = ln µ W = e W = We µπ µπ a) he rces n the ball are: 1 θ 1 mg θ here is n acceleratin in, s: csθ = csθ + mg 1 1 csθ mg 35N cs kg 9.8m/s = = = 8.74N 1 1 csθ cs 60 b) he net rce is in the directin:

9 = = = 1 sinθ1 sinθ 35Nsin Nsin N c) he net rce accelerates the ball in a circle, s its magnitude must be equal t he radius the circle is given b: mv r. r d d / 1.70m tan 60 r = = 1.47m tan 60 =, where is the distance between where the tw rpes are attached t the rd herere, the velcit is given b: mv r = 37.9N 37.9N 1.47m v = = 6.44m/s 1.34kg Chapter 6: Multiple chice: 3) he rm Newtn s ecnd Law as written b Newtn tells us that the rate change the mmentum an bject is equal t the rce acting n it. r circular mtin, the rce (and therere the rate change mmentum) is prprtinal t v. (C) 7) I the impulse is 0, then the change in mmentum must als be 0. Hwever, it is pssible r the psitin the bject t change. r eample, cnsider an bject at rest, acted n b a rce 1N in the + directin r 1s, and then b a rce 1N in the directin. he irst rce will cause the bject t accelerate in the + directin, and the secnd will bring it t rest again, but wn t mve it back t = 0. p is alwas 0, but r ma nt be. (C) 10) As ar as we knw, the law cnservatin mmentum is alwas valid. (A) Questins: 3) he initial velcit imparted t the baseball depends n the batter. his means that sme batters are able t cause a greater change in the mmentum the pitched ball than thers can, b appling a greater impulse during the ball-bat cllisin.

10 8) here are tw cmpeting eects at wrk here. r ne thing, the scale is nt supprting the grains sand that are alling, s that tends t reduce the reading. On the ther hand, the sand striking the pile n the bttm eerts an impulse n the bttm pile, and that impulse must be balanced b the scale. his eect tends t increase the scale s reading. We knw that each grain sand begins at rest in the tp hal the glass, and ends at rest in the bttm hal, s the net impulse acting n it must have been 0. his means that the tw eects discussed abve eactl cancel, s the hurglass weighs the same whether the sand is alling r nt. Eercises: ) I we take Nrth t be the + directin and East t be the + directin, we have: km 4 km Initial mmentum = mv = 000kg40 j= 8 10 kg j hr hr km 5 km inal mmentum = 000kg50 = kg hr i hr i the change in mmentum is: 1) 5 km 4 km p pi = kg i 8 10 kg j hr hr a) ince the hand travels.8cm bere cming t rest, and we are assuming cnstant rce n it, we can determine the time rm: 1 = vt at v = 0 v = at 1 v 1 = vt t = vt t t = = = v 9.5m/s 0.08m s b) We knw that the impulse equals the change in mmentum the hand.,

11 avg t p m v m v 0 avg -3 ( 0) = = = mv 0.540kg 9.5m/s = = = 870N t s 18) he initial mmentum the car + man sstem as measured b an bserver net t the tracks is v pi = M car + mcar v0 = W + w g 0 ( ) ( ) Ater the man starts running the same bserver measures the car s velcit as v and the man s velcit as v v. rel herere, the inal mmentum is measured as: p ( v ) Wv w v = + g g rel ince n eternal rces act n the sstem, pi = p, r: v0 Wv w v ( W + w) = + g g g ( ) ( ) v ( W + w) + wv 0 rel ( v ) v W + w = v W + w + wv v = W + w wvrel v = v v0 = W + w rel 0 rel