Oberlin College Physics 110, Fll 2011 Model Solutions to Assignment 4 Additionl problem 56: A girl, sled, nd n ice-covered lke geometry digrm: girl shore rope sled ice free body digrms: force on girl by ground (friction) force on girl by sled force on sled by girl. he only force on the sled is the leftwrd force due to the girl. (o be bsolutely precise, the leftwrd force on the sled is the force due to the rope, not due to the girl. However we hve seen tht when the mss of the rope is negligible, then the force on the sled due to the rope is nerly equl to the force on the rope due to the girl. We sy tht the rope trnsmits the force without chnge. Similrly for the rightwrd force on the girl due to the sled.) he ccelertion of the sled is S = F S,net m S = 5.2 N 8.4 kg = 0.62 m/s2. b. he two forces on the girl re the rightwrd force due to the sled (ctully due to the rope plus sled), nd the leftwrd force of friction due to the ground. hese two forces blnce so tht the net force vnishes whence the ccelertion is zero. 1
geometry digrm: girl shore rope sled ice free body digrms: force on girl by sled force on sled by girl c. Sme s prt (). d. By Newton s third lw, the force by the girl on the sled is equl nd opposite to the force on the sled by the girl. hus G = F G,net m G = 5.2 N 40 kg = 0.13 m/s2. e. At ny instnt of time, the distnces trveled by the girl nd the sled from their strting points re so x G = 1 2 Gt 2 nd x S = 1 2 St 2, x S = S = m G. x G G m S his mkes sense: the high-mss (more sluggish, more inert) girl moves less thn the low-mss sled does but if the two were eqully mssive, then by symmetry they would move equl mounts. In prticulr, we wnt to know x G when the girl nd sled touch, tht is, when x G + x S = 15 m. his is when: x G + x S = 15 m x G + m G x G = 15 m [ m S x G 1 + m ] G = 15 m m [ S ] ms + m G x G = 15 m m S [ m S x G = m S + m [ G 8.4 kg x G = 48 kg 2 ] (15 m) ] (15 m) = 2.6 m.
As expected, the low-mss sled moves more thn the high-mss girl. f. In prt (b), the two forces cting on the girl re blnced: you know they re equl nd opposite becuse the ccelertion is zero, so the forces must dd up to zero. But in the forces mentioned in prts (c) nd (d) re equl nd opposite by Newton s third lw. hey re equl nd opposite even when the girl ccelertes. he two members of third-lw pir lwys ct upon different objects. Morl of the story: I drop bll. he grvittionl force on the bll due to the erth is equl nd opposite to the grvittionl force on the erth due to the bll. But the bll ccelertes whole lot more thn the erth does becuse the bll hs much smller mss. he word force, like most words, hs multiple menings. (he Oxford English Dictionry lists 53 menings for the noun force.) In terms of the everydy mening of the word force, it seems bsurd tht the huge, enormous erth cn only muster up s much force s the tiny bll cn. But in the physics mening of the word force this is exctly wht hs to hppen, in order for the ccelertions to be so different. I cution you tht intuition developed from the militry or legl senses of the word force probbly doesn t pply to the physics sense of the word force. Additionl problem 58: Monkey business he monkey hs mss m M, the bnns hve mss m B. he drwings re: monkey bnns m M m g M b g. [8 points] he bnns don t move, so for the bnns F = m becomes = mb g. For the monkey, F = m becomes m M g = m M whence m B g m M g = m M whence = m B m M m M g. Given the numbers of the problem, = 0.50g = 4.9 m/s 2. b. [2 points] It is cler from our eqution tht if both msses re doubled, the ccelertion won t chnge. HRW problem 5-54: Pulled penguins free body digrms: 111 N 111 N 222 N rer two penguins mss: 12 kg + m front two penguins mss: 35 kg 3
he front two penguins experience net force of 111 N to the right, nd this force results in certin ccelertion. he rer two penguins experience net force of 111 N to the right, nd they lso undergo the sme ccelertion. hus they must hve the sme mss... the mss of the remining penguin is 23 kg. HRW problem 5-55: Blocks Only horizontl forces re relevnt, so only those re shown: geometry digrm: hnd m 1 free body digrms: m 1 FH F B F B Note tht both blocks undergo the sme ccelertion. For left block: Fx = m 1 = F H F B = m 1 For right block: Fx = = F B = We know m 1,, nd F H. Above re two equtions in two unknowns, nmely nd F B. Solving for F B : ( ) FB F B = F H m 1 = F H m 1 nd finlly Using numbers: ( F B 1 + m ) 1 F B ( m1 + ) F B = = F H = F H ( m2 m 1 + ) F H. (1). F B = ( ) 1.2 kg (3.2 N) = 1.1 N. 3.5 kg 4
b. his sitution just swps m 1 nd, so now F B = 2.1 N. c. Result (1) sys tht F B diminishes s shrinks: From F B = F H if m 1 = 0 down to F B = 0 if = 0. HRW problem 5-57: Wedge As usul, we solve this problem first with symbols (m r, m h, nd θ), nd only then plug in numbers. his is becuse (1) it is esier to do lgebr with symbols thn numbers nd (2) we will be ble to red the eqution to check it for resonbleness. geometry digrm: m r m h θ free body digrms: N m r m h W h W r (Note: we could hve chosen the two positive ccelertion directions pointing the other wy... this would hve just resulted in finl ccelertion of the opposite sign.) Apply F = m to both blocks: F = m Forces on hnging block m h g = m h (1) Forces on rmp block, prllel to surfce m r g sin θ = m r (2) Forces on rmp block, perpendiculr to surfce N m r g cos θ = 0 (3) Equtions (1) nd (2) re two equtions for the two unknowns nd. Solve them simultneously to find Now, do these equtions mke sense? = m h m r sin θ g, m h + m r m h m r = (1 + sin θ)g. m h + m r 5
he dimensions re correct. Check. When m r = 0, these equtions give = g nd = 0. Check. When θ = 90 this is exctly the sitution of the dditionl problem on sliding slmi, nd these equtions give exctly the sme nswers. Check. If m h m r, the ccelertion is positive; if m r m h, the ccelertion is negtive. Check. I cn t think of ny more circumstnces in which I hve ny intuition. Cn you? Plugging in the numbers given, nd using significnt figures properly, we find tht for the sitution of this problem: () = 0.736 m/s 2 ; (b) the hnging block ccelertes downwrd; (c) = 20.9 N. HRW problem 5-58: Window wsher mn W he eqution F = m becomes 2 mg = m, so = 1 2m( + g). () When = 0, = 1 2 (95.0 kg)(9.81 m/s2 ) = 466 N. (b) When = 1.30 m/s 2, = 1 2 (95.0 kg)(1.30 m/s2 + 9.81 m/s 2 ) = 528 N. mn W 6
In this cse the eqution F = m becomes mg = m, so = m( + g). (c) When = 0, = (95.0 kg)(9.81 m/s 2 ) = 932 N. (d) When = 1.30 m/s 2, = (95.0 kg)(1.30 m/s 2 + 9.81 m/s 2 ) = 1060 N. Discussion: Suppose we wnt the cb to rise t constnt speed. he mn in the cb ccomplishes this by pplying force of 466 N. His coworker on the ground needs to pply twice s much force! It seems tht the mn in the cb is getting something for nothing. However, consider this spect: When the coworker on the ground pulls two meters of rope through his hnds, the mn in the cb rises by two meters. But when the mn in the cb pulls two meters of rope, the cb rises by only one meter. (One meter is eliminted between the cb nd the pulley, the other meter is eliminted between the pulley nd the hnd.) So the mn in the cb pulls with hlf the force, but hs to pull twice s much rope in order to rise the sme distnce. We will return to this issue when we discuss work. Problem 5-88: Lnding on Cllisto. he weight is 3260 N. b. When the thrust is 2200 N, the net force is 1060 N (downwrd), so the mss of the spcecrft is m = F net = 1060 N 2 = 2700 kg. 0.39 m/s thrust (2200 N) spcecrft weight (3260 N) c. he ccelertion of grvity is weight mss = 3260 N 2700 kg = 1.2 m/s2. 7
ride tody! HRW problem 6-52: Amusement prk geometry digrm: free body digrm: W F on cr by boom he eqution F = m becomes (tking the positive direction to be downwrd) So For the two cses given: W F on cr by boom = m = m v2 r. F on cr by boom = W m v2 r = W ) (1 v2. rg. If v = 5.0 m/s, then F = +3.7 kn (tht is, the force points upwrd, s shown in the digrm). b. If v = 12 m/s, then F = 2.3 kn. (he negtive sign mens tht the force points downwrd.) It mkes sense tht t low speeds, the force should point upwrd, becuse t zero speed the force is of course upwrd. As the speed increses, the boom needs to hold on to the cr lest it fly wy. 8
HRW problem 6-57: Orbiting puck free body digrm: he cylinder will remin t rest when the string tension supplies ll the needed centripetl ccelertion v 2 /r. his occurs when = Mg = m v2 M or v = r m rg. (With lower puck speed, the cylinder drops. With higher puck speed, it rises.) HRW problem 6-54: Amusement prk design he net force on the pssenger must be F = mv2 r.. If there is smll chnge dr in the rdius, with no chnge in v, then the force will chnge by bout df = df dr dr = mv2 r 2 dr. b. If there is smll chnge dv in the speed, with no chnge in r, then the force will chnge by bout c. If there is smll chnge d in the period df = df dv 2mv dv = r dv. = 2πr v, with no chnge in r, then the ssocited chnge in velocity is dv = dv d = 2πr d 2 d. Using the result of prt (b), this chnges the force by df = 2mv r dv = 2mv 2πr r 2 d = mr 8π2 3 d. 9