Oregon Stte Universit PH 211 ll Term 2017 Prep 6-7 Recommended finish dte: Wednesd, November 8 The formts (tpe, length, scope of these Prep problems hve been purposel creted to closel prllel those of tpicl em (indeed, these problems were tken from pst ems. To get n ide of how best to pproch vrious problem tpes (there re three bsic tpes, refer to these Smple Problems.
1. Evlute the following sttements (T//N. As lws, eplin our resoning.. A 30-kg child stnding on the erth eerts grvittionl force on the erth tht is twice s much s tht of 15-kg child stnding net to her. True. The G cting on ech therefore b ech (this is Newton s Third Lw is simpl m child g. So twice the m child would give twice the force mgnitude. b. If one object is ccelerting, there must be nother object tht s ccelerting. True. If net force hs been eerted on mss 1 (since it s ccelerting it must hve been pplied b nother object (mss 2: pp.21 But b Newton s Third Lw, force ws pplied on mss 2 eull but oppositel: pp.12 = pp.21. So mss 2 must be ccelerting, lso. c. Units of ccelertion could be correctl written lso s N/kg. True. The units of force,, must be euivlent to m, so is euivlent to /m, which re N/kg in the SI sstem. d. A 6-kg block on erth would be 1-kg block on the moon. lse. The mss of n object is not ffected b which plnet it s ner. Mss in intrinsic to the mount of mtter contined in n object. If tht hsn t chnged, the mss hsn t chnged. e. A bod tht hs zero ccelertion m hve forces cting on it. True. Zero ccelertion indictes onl tht the (vector sum of ll forces cting on the object is zero. Tht sum m be (nd is often comprised of two or more force vectors. f. A cr is witing t rest t stop light on westbound rod. When the light turns green nd the cr ccelertes forwrd (to the west, its rer bumper eerts force on the rest of the vehicle, nd tht force is directed to the est. True. Anlze the bumper nd the rest of the cr s two connected msses ccelerting together. If the bumper is ccelerting to the west, the rest of the cr must be pulling on the bumper in tht direction. So, b Newton s 3rd Lw, the bumper is pulling oppositel (est on the rest of the cr. g. An object (m = 5.00 kg, initill t rest, eperiences net force of [1.5t 2, 2t] N (where t is in seconds throughout the intervl 0 t 4.00 s. Then ll force ceses. At t = 7.50 s, the object s speed is pproimtel 5.54 m/s. lse. P f = P i + DP, but P i = 0 (object strts t rest, so P f = DP. Thus: mv f = J So: v f = J /m J = (t dt = [ 1.5t2 dt, 2t dt] = [(1/2t3 0 4, t 2 0 4 ] = [32, 16] kg m/s Therefore: J = [32 2 + ( 16 2 ] (1280 So: v f = J /m = (1280/5.00 = 7.16 m/s <--- (cr (rer bumper 2
2. or ech item, be sure to show our work nd/or eplin our resoning.. A 150,000 kg jet irliner is moving with constnt velocit t n ngle of 12.7 bove the horizontl, t n ltitude of 10 km nd speed of 200 m/s. Wht is mgnitude of the totl force cting on the irplne? Zero. B Newton s irst Lw, if the object s ccelertion is zero (true here, so is the net force cting on it. b. A 2-kg object hs the ccelertion vector [ 1, 6] m/s 2. Wht dditionl force must be pplied to the object to put it into euilibrium? Epress this s mgnitude nd direction (using the +-is s 0. Right now the net force on the object is S = m = 2[ 1, 6] = [ 2, 12] N Euilibrium is when S = 0. To mke this true, we must dd the opposite of the current net force. Thus, we must dd dd = [2, 12] N = (2 2 + ( 12 2 [tn -1 ( 12/2] = 12.2 N 80.5 c. A 2-kg object is constrined to moving long stright line. The velocit s function of time is shown here. (i Wht is the net force cting on the object t t = 5s? Eplin our resoning. net = S = m. (5 is the slope of the v-t grph t t = 5: So (5 = [v(5 v(1]/(5 1 = 5 m/s 2. Thus net (5 = 2( 5 = 10 N (ii Wht is the verge net force cting on the object between t = 0s nd t = 7s? Eplin our resoning. ( net vg = m vg = m(v f v i /Dt = 2( 10 0/7 = 2.9 N d. Boes A nd B re sliding to the right cross frictionless tble. The hnd, H, is slowing them down. The mss of A is lrger thn the mss of B. Rnk, from lrgest to smllest, these horizontl force mgnitudes: AB, AH, BH, BA, HA, HB irst, note: The hnd isn t touching A, so AH = HA = 0. Then note: The hnd must eert enough force to ccelerte the combined mss, but B must eert on A onl enough force to ccelerte the mss of A: BA < HB = 0. inl rnking (using Third Lw euivlencies: BH = HB > BA = AB > AH = HA = 0. 3
3. or ech item, be sure to show our work nd/or eplin our resoning.. The figure shows two msses t rest. The string is mssless, nd the pulles re frictionless. The spring scle reds in kg. Wht is the reding of the scle? Be sure to full eplin our resoning. The reding is 5 kg. A scle reds onl the force it is eerting in one direction. If it were to dd or subtrct the forces it is eerting t both ends (es, BD will show this, it would red either double or zero, nd tht would mke it useless. b. In the sitution shown, the horizontl surfce is frictionless, nd the pulle is mssless nd frictionless. Is the tension in the cble less thn, greter thn, or eul to 88.2 N? Do full nlsis with BD to support our rgument. The hnging mss ( : The sliding mss (m 2 :.21.2 5 kg 9 kg.12 G.P1 G.P2 The hnging mss ( : The sliding mss (m 2 : But note: S 1. 1. S 2. 2..12 =.21 (These re mgnitudes nd so re eul. G.P1.21 1..12 2. 2. = 1. (Their mgnitudes nd signs re eul. g.21 1..12 2. So cll the bove simpl nd. Thus: g nd: Solve one eution for : = /m 2 Substitute nd solve for : g /m 2 g/(1 + /m 2 = (9(9.80/(1 + 9/5 = 31.5 N (less thn 88.2 N c. Rnk the tension in the string, from gretest to lest, in the cses shown here. Assume in ech cse tht the hnging mss is the sme, tht friction is negligible, nd tht the pulle nd rope re both mssless. Use BDs! The solution for here is the sme s in 2B, bove: g/(1 + /m 2 Here is the sme in ll three cses, but not m 2. And when m 2 is lrger the denomintor here is smller. Tht is, when m 2 is lrger, is lrger. Since m 2A > m 2B > m 2C, this mens tht A > B > C. d. An object of mss M is held in plce b n pplied force nd pulle sstem, s shown in the figure. The pulles re mssless nd frictionless, s re the cbles.. ind the tension in rope section T1..1 = = Mg/2 (See below for resoning. b. ind the tension in rope section T2..2 = = Mg/2 (See below for resoning. c. ind the tension in rope section T3..3 = = Mg/2 (See below for resoning. d. ind the tension in rope section T4..4 = 3Mg/2 (See below for resoning. e. ind the tension in rope section T5..5 = Mg (.5 holds M t rest ginst G.PM. f. ind the mgnitude of. = Mg/2 (.2 +.3 =.5. So: + = Mg or n mssless cble, the tension t n point in it is the sme (which ou cn prove b using BD to nlze n two prts of it. And if such cbles re connected to idel (mssless, frictionless pulles, their tension is redirected without ltertion. And looking t the whole pulle contrption s one object, we hve just one upwrd force (.4 nd two downwrd forces ( nd Mg cting on it. Thus:.4 = + Mg. T 1 T 4 T 2 T 3 M T 5 4
3. e. or the sitution shown, where four idel (mssless pulles support two msses hnging t rest, evlute (T//N the following sttement. As lws, ou must justif our nswer with vlid mi of words, drwings nd clcultions. B B /A = 8 A m 2 lse. Ech idel pulle trnsmits the tension force throughout n continuous wire. (This is how we know tht the tension in the wire connecting the second pulle to the ceiling is lso A. Thus, ech pulle eperiences twin upwrd forces; so the tension ech supports downwrd is twice tht. Hence the lbels shown here. Thus: B /A = 4 2A A B A 4A m 2 5
4.. Evlute the following sttements (T//N. As lws, eplin our resoning. (i If n elevtor is moving verticll upwrd, then the weight of person stnding in tht elevtor is lws greter thn her weight when she is t rest. lse. If she is slowing down or moving t constnt speed, her weight is either less thn or eul to her rest weight. (ii The weight of n stronut orbiting the erth is greter thn his weight when orbiting the moon. lse. Not greter thn eul to. The stronut is weightless in either cse: 0 = 0. (iii The sttic friction force eerted b one surfce on nother depends onl on the friction coefficient (m S nd the norml force tht eist between the two surfces. lse. The sttic friction force vries (from 0 to S m in response to the pplied force(s tht would otherwise be cusing the surfces to slide pst ech other. Onl S m is clculted vi m S. (iv A sttionr object cn eert kinetic friction force. (An nswer OK here. As some students hve lertl pointed out, there s no object in the universe tht is known to be sttionr. So b tht ssumption, this sttement hs no mening it s either flse or not enough informtion. However, if we designte something such s the erth s sttionr, then certinl it cn eert kinetic friction force: Slide book cross the floor nd wtch it slow down nd stop. (v A sttic friction force cn cuse ccelertion. True. or emple, look t the force S.12 in eercise 9 of Lb 5-I (the BD eercise set. (vi When block slides t 4 m/s on surfce with friction, there is twice the friction force cting on it s when it slides on tht surfce t 2 m/s. lse. No, for low speeds, m K (nd therefore K is bout eul for n non-zero speed (so long s hs not chnged. (vii A friction force between two surfces cts on ech surfce eull in mgnitude but opposite in direction. True. This is just re-sttement of Newton s Third Lw: An force cting between two objects cts on ech with the sme mgnitude but opposite direction. (viii The coefficient of kinetic friction is force. lse. m K is unitless rtio of two force mgnitudes: m K = K.S1 /.S1 (i A norml force is defined s force eerted verticll upwrd b surfce. lse. A norml force is defined s force eerted b surfce in direction perpendiculr to tht surfce. Tht direction is not necessril verticl or upwrd. ( A mn stnding on the level floor of n elevtor tht is moving verticll downwrd could hve weight greter thn the grvittionl force cting on him. True. The mn (nd elevtor could be ccelerting upwrd if the elevtor were slowing while moving downwrd. And n upwrd ccelertion would men tht the norml force upwrd b the elevtor floor (wht the mn feels to judge his weight must be must be greter thn the downwrd grvittionl force. (i The grvittionl force eerted b the erth on the orbiting spce shuttle is greter thn the grvittionl force eerted b the erth on the pilot s set inside tht shuttle. True. G.P1 g, where g is the locl vlue of grvittionl free-fll ccelertion. And the sme locl g vlue pplies to the entire spce shuttle nd to ll of its prts, so lrger mss eperiences lrger grvittionl force cting on it. 6
4. b. Helen is pushing bo so tht it slides cross the floor. The speed of the bo is incresing. Evlute the following sttements (T//N. As lws, eplin our resoning. (i The frictionl force is negligible (or effectivel zero. Not enough informtion. All we know from the given informtion is tht the bo is speeding up, so the horizontl component of her pushing force eceeds the kinetic friction force opposing the sliding. Tht is: App.HB cos > K, where is the ngle of her push, s mesured from the horizontl. In other words: App.HB cos > K. But we simpl don t know how much K is. (ii The frictionl force is blnced b Helen s pplied force. lse. The bo is ccelerting, so the friction force is not being blnced; it is being eceeded. (iii Helen is pushing with force greter thn the norml force on the bo. Not enough informtion. As rgued in (i, bove, App.HB cos > m K. And = mg App.HB sin. Thus, App.HB cos > m K (mg App.HB sin. But we don t know the vlue of either m K, m or. (iv Helen s pushing force on the bo is greter thn the frictionl force on the bo. True. As rgued in (i, bove, App.HB cos > K, nd so even if Helen s entire force is directed horizontll ( = 0, tht would still men App.HB > K. (v The bo s weight is less thn Helen s pplied pushing force. Not enough informtion. The bo s weight is the vector sum of ll contct forces ( App.HB, nd K being eerted on it. And we don t know the mgnitudes of n of those forces (nor the direction of App.HB. 7
5.. The block (m = 30 kg is on level surfce with friction coefficients m s = 0.750 nd m k = 0.500. If 1 = 100 N, 1 = 40.0, nd 2 = 25.0, find 2 so tht: the block is still t rest but just bout to strt sliding to the right. the block slides to the right t constnt velocit. BD of m sitution ( 1 2 1 2 1 2 m s m s becomes k in sitution (b. 1 2 m coord. es G -nlsis sitution ( -nlsis sitution ( S = m S = m 2 1 m s = m + 1 + 2 G = m 2 cos 2 1 cos 1 m s = 0 -nlsis sitution (b + 1 sin 1 + 2 sin 2 mg = 0 -nlsis sitution (b S = m S = m 2 1 k = m + 1 + 2 G = m 2 cos 2 1 cos 1 m k = 0 + 1 sin 1 + 2 sin 2 mg = 0 These re two eutions in two unknowns ( nd 2 solve simultneousl: Solve for : = mg 1 sin 1 2 sin 2 Substitute: 2 cos 2 1 cos 1 m s (mg 1 sin 1 2 sin 2 = 0 Collect terms: 2 cos 2 + m s 2 sin 2 = 1 cos 1 + m s mg m s 1 sin 1 Solve: 2 = (1 cos 1 + m s mg m s 1 sin 1 /(cos 2 + m s sin 2 = [(100(cos40 + (.75(30(9.80 (.75(100(sin40]/[cos25 + (.75(sin25] = 203 N or (b: 2 = (1 cos 1 + m k mg m k 1 sin 1 /(cos 2 + m k sin 2 = [(100(cos40 + (.50(30(9.80 (.50(100(sin40]/[cos25 + (.50(sin25] = 171 N 8
5. b. A 3-kg block is free to slide verticll up or down frictionless wll. A 20 N force is pplied s shown. ind the block s ccelertion (both mgnitude nd direction. S = m S = m 60 20 N = m sin60 = 0 G = m cos60 mg = m (cos60 /m g = = 6.47 m/s 2 (tht s 6.47 m/s 2 downwrd c. A 3-kg block is held t rest ginst wll b two identicl pplied forces, s shown. The wll hs coefficient of sttic friction of 0.65 with the block. ind the mgnitude of t which the block would be red to slip down the wll. S = m S = m = m + s m G = m = 0 + m s mg = 0 = + m s mg = 0 + m s mg = 0 (1+ m s = mg = mg/(1+ m s = 17.8 N d. A bo of mss m is initill sliding t speed of v on horizontl surfce. Wind is ppling constnt force mgnitude t n ngle down from the horizontl, in the opposite direction of the bo s motion. The bo comes to rest fter trveling distnce d. ind the coefficient of kinetic friction, m K, between the bo nd the surfce, epressed in terms of m, v,,, d nd g. irst, do kinemtics to find the -ccelertion vlue necessr to bring the bo to rest in distnce d: 0 2 = v 2 + 2 d Thus: = v 2 /(2d Now do BD nlsis (referring to digrms here using conventionl - nd - es: G S m G S = m S = m K = m G = m m K cos = mv 2 /(2d sin mg = 0 K Solve -eution for : Substitute into -eution: Solve for m K : = sin + mg m K (sin + mg cos = mv 2 /(2d m K = [cos mv 2 /(2d]/(sin + mg G 9
5. e. Two identicl msses re connected b mssless cord, s shown. The re being pulled to the right on level, frictionless surfce b horizontl force. The tension in the cord, the ngle of the cord, nd ll remin constnt. Evlute (T//N the sttement below. As lws, ou must justif our nswers with vlid mi of words, drwings nd clcultions. The difference in mgnitude between the two norml forces being eerted in this sitution is tn. True. m 2.1 G.1.21.12.2 G.2 S 1. 1. S 1. 1. S 2. 2. S 2. 2..21. 1..1 +.21. G.1 1..12. 2..2.12. G.2 2. cos = m.1 + sin mg = 0 cos = m.2 sin mg = 0.1 = mg sin.2 = mg + sin So: cos = cos Or: = /(2cos Thus:.2.1 = (mg + sin (mg sin = 2 sin = 2sin[/(2cos] = tn 10
6. In ech cse below, the ccelertion mgnitude on the 5-kg bo (M is 4 m/s 2, which is s lrge s possible without the bo sliding. or ech cse, find the coefficient of sttic friction between the bo nd the surfce beneth it.. The bo is on the bed of truck tht is ccelerting on level ground. S = m S = m s m = m G = m m s = m mg = 0 Therefore: m s mg = m Thus: m s = /g = 4/9.80 = 0.408 b. The bo is in ski gondol on its w to the top of the mountin. (The gondol does not swing during the motion; its ttchment to its cble remins verticl t ll times. G S m S = m S = m m s = m G = m m s = m(cos30 mg = m(sin30 Therefore: m s (msin30 + mg = m(cos30 Thus: m s = (cos30 /(sin30 + g S m = (4cos30 /(4sin30 + 9.80 = 0.294 G 11
7.. Block 2 is hnging freel (t rest b thred from block 1. Block 1 is not ttched to the ceiling, but it is being held t rest ginst the ceiling, due to the effects of the force,. ind the mimum force mgnitude,, which cn be pplied so tht block 1 won t slip. = 25.0 ; = 4.06 kg; m 2 = 1.37 kg m S.ceiling.block1 = 0.890 m S m 2 or m 2 : S or : S 0 = 0 m S T1.2 sin m S = 0 S m 2.1 G.2 S 1.2 G.2 S 2.1 G.1 G.1 1.2 m 2 g = 0 cos 2.1 g = 0 Thus: 1.2 g = 1.37(9.80 = 13.426 N Also: = sin/m S So: 2.1 = 1.2 = 13.426 N So: cos 2.1 g sin/m S = 0 Solve for : cos sin/m S = 2.1 + g (cos sin/m S = 2.1 + g = (2.1 + g/(cos sin/m S = [13.426 + (4.06(9.80]/[cos25 (sin25 /0.890] = 123 N b. Block 2 is ttched to the side of block 1 onl b thred. The blocks re ccelerting together s Block 1 slides long horizontl shelf. ind the coefficient of kinetic friction, m K. = 25.0 ; = 98.0 N; = 4.65 kg; m 2 = 1.73 kg or m 2 : S or : S m 2 m K 1.2 1.2. K 2.1. 1.2 sin N cos m K 2.1 sin G.2 S K S 1.2. G.2 2.1. G.1 T2.1 1.2 cos m 2 g = 0 2.1 cos g sin = 0 G.1 1.2 g/cos = 1.73(9.80/cos25 = 18.707 N 2.1 = 1.2 = 18.707 N = 1.2 sin/m 2 = 18.707sin25 /1.73 = 4.5699 m/s 2 Solve for m K : m K = (cos 2.1 sin / = [98cos25 18.707sin25 4.65(4.5699]/103.94 = 0.574 = 2.1 cos + g + sin = 18.707cos25 + 4.65(9.80 + 98sin25 = 103.94 N 12
8.. A block (m = 10.2 kg sits t rest on surfce tht hs µ S nd µ K vlues (with the block of 0.75 nd 0.36, respectivel. The surfce is initill level but then is grdull tilted until the block slides down the slope. At tht ngle of tilt, how long does it tke the block to slide for 1.84 m long tht slope mesured from the moment it begins to move? (Ignore n sudden jerks or irregulr motion to begin with tret the motion s smooth trnsition from rest. BD The first sitution to nlze is the block while it s still t rest but bout to slip: S M m coord. es G -nlsis -nlsis S = m S = m G S M = m G = m mgsin m S = 0 mgcos = 0 The unknowns here re nd. And it s rell the vlue we wnt to determine the slope ngle t which the block will slide down the incline so solve for first (using the -eution nd substitute this into the -eution: = mgcos mgsin m S (mgcos = 0 mgsin = m S (mgcos sin = m S cos tn = m S = tn -1 (m S = tn -1 (0.75 = 36.8699 (from the -eution (substituting into the -eution (rerrnging (simplifing (simplifing This is the slope t which the block will be sliding. 13
The net sitution to nlze is the block while it s sliding down the slope: BD m K coord. es G -nlsis -nlsis S = m S = m G K = m G = m mgsin m K = m mgcos = 0 The unknowns here re nd. And it s rell the vlue we wnt to determine the block s ccelertion so solve for first (using the -eution nd substitute this into the -eution: = mgcos mgsin m K (mgcos = m mg(sin m K cos = m (from the -eution (substituting into the -eution (simplifing g(sin m K cos = = (9.80[sin(36.8699.36 cos(36.8699] = 3.0576 m/s 2 Now it s just kinemtics problem. Knowing D = 1.84, v 0 = 0, nd = 3.0576, we cn solve for Dt: D = v 0 (Dt + (1/2(Dt 2 D = (1/2(Dt 2 2D/ = (Dt 2 (simplifing (rerrnging [2D/] 1/2 = Dt = [2(1.84/3.0576] 1/2 = 1.10 s 14
8. b. The block shown t right remins t rest while the force is present. ind the block s speed 2.50 seconds fter is suddenl removed. Be sure to show our work nd/or eplin our resoning. Does the block move when relesed? Is the slope sufficient 20 to cuse it to slip? Ask tht uestion first! If the grvit force down the slope eceeds the mimum sttic friction force the surfce cn eert, it will slip. S =? S = m m s G. =? G. = m m s mgsin =? mgcos = 0 = mgcos = 10(9.80cos20 = 92.09 So: s m = m s =.50(92.09 = 46.04 N mgsin = 10(9.80sin20 = 33.52 N Since the grvittionl pull down the slope is less thn s m, the block will not slip; its speed will remin zero. c. A 3-kg block is sliding t constnt speed down sloped surfce, s shown. The coefficient of kinetic friction (m k between the block nd the slope is 0.45. Wht is the ngle of the slope? Show our work nd/or eplin our resoning. 10 kg m s = 0.50 m k = 0.25 coord. es s k G S = m S = m G. k = m G. = m mgsin m k = 0 mgcos = 0 = mgcos G m k = mgsin/ = mgsin/(mgcos = tn Therefore: = tn -1 (m k = 24.2 d. A 3-kg block is sitting t rest on sloped surfce, s shown. The coefficient of sttic friction (m s between the block nd the sloped surfce is 0.55. Wht is the mgnitude of the totl force eerted b the sloped surfce on the block? Be sure to show our work nd/or eplin our resoning. 20 S = m S = m G. S = m G. = m mgsin S = 0 mgcos = 0 S = mgsin = 10.055 N = mgcos = 27.627 N The totl force b slope on block is vector sum: slope.block = + S But nd S re perpendiculr to ech other, so the form right tringle whose hpotenuse is the totl force; its mgnitude cn be clculted vi the Pthgoren Theorem: 29.4 N Also OK is clerl eplining: The onl object supporting the block upwrd is the sloped surfce, so the sum of ll forces eerted b the surfce on the block must ectl counter the downwrd grvittionl force on the block ( G = mg = 3 9.80 = 29.4 N. s G slope.block s 15
9. or ech item, be sure to show our work nd/or eplin our resoning.. A 200 g hocke puck is lunched up metl rmp tht is inclined t 30 ngle. The coefficients of sttic nd kinetic friction between the hocke puck nd the metl rmp re μ S = 0.40 nd μ K = 0.30, respectivel. The puck s initil speed is 3 m/s. (i ind the net force mgnitude nd direction on the puck while it is sliding up the rmp. So: Thus: ind : S 1 1 G.1. K.1 1 gsin m K.1.1 = S 1 1.1 G.1. 1.1 gcos = 0.1 gcos gsin m K.1 ( gcos = = g(sin + m K.1 cos = (9.80[sin30 + (0.30cos30 ] = 7.446 m/s 2 This is the totl ccelertion mgnitude of the puck (since = 0, nd it s in the negtive -direction, so = = 7.446 m/s 2 S 1 = (0.200( 7.446 = 1.49 N down the rmp (ii How fr up the rmp does the puck trvel? Just do kinemtics, since the -ccelertion vlue is now known: v f. 2 = v i. 2 + 2 D Thus: D = (v f. 2 v i. 2 /(2 = (0 2 3 2 /[2( 7.446] = 0.604 m b. A 20 kg block on tble is connected b string to 12 kg mss, which is hnging over the edge of the tble (modeled here s hnging over n idel pulle. The 20-kg block is 3.1 m from the tble edge. The coefficient of kinetic friction between the block nd the tble is 0.1. The coefficient of sttic friction between the block nd the tble is 0.5. (i Is the hnging block is hev enough to cuse the block on the tble to slide? Could S.1 m hold ginst the tension of the string, (=.21 =.12, if m 2 were hnging from rest? ind out: In tht specil, simple cse:.1 = G.P1 g And:.12 = G.P2 g Thus: m S.1 = m S (.1 = m S ( g = (0.5(20(9.80 = 98.0 N But:.21 =.12 g = (12(9.80 = 118 N The tension.21 would be greter thn S.1 m could hold. The block will slip. (ii Regrdless of our nswer to prt, ssume tht the block on the tble does slide nd determine how much time will pss before the block reches the end of the tble. The sliding mss ( : The hnging mss (m 2 : S 1. 1. S 1. 1. K.1.21 K.1 1. G.P2.12 2. K.1 m 2 g.21 = K.1 + g m 2 G.P1 So: K.1 + g m 2 Or: = (m 2 g K.1 /( + m 2 = (m 2 g m K g/( + m 2 = [(12(9.80 (0.1(20(9.80]/(20 + 12 = 3.063 m/s 2 Now kinemtics: D = v i. (Dt + (1/2 (Dt 2 Since v i. = 0, this simplifies to D = (1/2 (Dt 2 Thus: Dt = [2(D/ ] = [(2(3.1/3.063] = 1.42 s 30 S.1 m.1 K.1 G.P1 G.21 20 kg.12 G.P2.12 G.P2 16 12 kg
c. The figure shows block of mss m resting on 20 slope. The block nd this surfce hve coefficient of sttic friction of 0.86 nd coefficient of kinetic friction of 0.76. It is connected vi mssless string over mssless, frictionless pulle to hnging block of mss 2.0 kg. (i Wht is the minimum mss m tht will stick nd not slip?.1.21.12 S.1 m G.P1 G.P2 The sticking mss ( : The hnging mss (m 2 : S 1. 1. S 2. 2..21 G.P1. m S.1 1. 0 = 0 gsin m S.1 = 0 S 1. 1. S 2. 2..1 G.P1. 1..12 G.P2 2..1 gcos = 0 m 2 g = 0.1 gcos g So: m 2 g gsin m S gcos = 0 Or: /(sin + m S cos = 2/[sin20 + (0.86(cos20 ] = 1.7389 = 1.74 kg (ii If mss hlf of tht vlue were plced there insted, find the ccelertion of the blocks..1.12.21 K.1 G.P2 G.P1 Sliding mss ( = 1.7389/2 = 0.8694: Hnging mss (m 2 : S 1. 1. S 2. 2..21 G.P1. m S.1 1. 0 = 0 gsin m K.1 S 1. 1. S 2. 2..1 G.P1. 1..12 G.P2 2..1 gcos = 0 m 2 g = m 2.1 gcos g m 2 So: m 2 g m 2 gsin m K gcos Or: m 2 g gsin m K gcos = ( + m 2 Thus: = [m 2 g gsin m K gcos]/( + m 2 = [(2(9.80 (0.8694(9.80(sin20 (0.76(0.8694(9.80(cos20 ]/(0.8694 + 2 = 3.69 m/s 2 17
9. d. A bo (m = 10.0 kg on level surfce hs n pplied pushing force,, being eerted on it t n ngle,, s shown. The sttic nd kinetic friction coefficients between bo nd surfce re 0.700 nd 0.350, respectivel.. ind the friction force mgnitude cting on the bo bottom when = 75.0 N nd = 5.00. b. ind the friction force mgnitude cting on the bo bottom when = 75.0 N nd = 20.0. c. Wht minimum vlue of would be reuired to move the bo when = 40.0? d. Wht is the mimum ngle (0 90 t which ou could eert the force (of n mgnitude towrd the bo nd still get the bo to move? Here s sitution where there re two different vlues ( nd tht ou cn vr. or n given vlue of, if ou wnt it to be the force slip tht puts the block into red-to-slip mode, there s just one certin ngle (if n tht ou could ppl tht force ( certin vlue of cll this slip. So how do slip nd slip relte to ech other? ind out: Anlze the sitution s if the bo (cll it were bout to slip:.1 S 1. 1. S 1. 1. S.1 m m S.1 1..1 G.1 1. m S.1 slip cos slip = 0.1 slip sin slip g = 0 G.1 Thus: m S ( g + slip sin slip cos slip = 0 Thus:.1 g + slip sin slip Or: slip cos slip m S slip sin slip = m S g So: slip = m S g/(cos slip m S sin slip. ind slip for slip = 5.00 : slip = (0.7(10(9.80/[cos(5 (0.7sin(5 ] = 73.4 N Since the ctul pplied (75 N is greter thn slip for this, the block slips; the friction force is kinetic. But the bove -nlsis doesn t chnge for the sliding cse, so:.1 g + sin Thus: K.1 = m K (.1 = m K ( g + sin = (0.350[(10(9.80 + (75.0sin(5.00 ] = 36.6 N b. ind slip for slip = 20.0 : slip = (0.7(10(9.80/[cos(20 (0.7sin(20 ] = 98.0 N Since the ctul pplied (75.0 N is less thn slip for this ngle, the block does not slip, so the friction force is sttic but not mimum. So the nlsis becomes:.1 S 1. 1. G.1 S S 1. S cos = 0 Thus: S = cos = (75.0cos20 = 70.5 N c. ind slip for slip = 40.0 : slip = (0.7(10(9.80/[cos(40 (0.7sin(40 ] = 217 N d. Looking t slip = m S g/(cos slip m S sin slip, notice tht slip pproches infinit when the denomintor (cos slip m S sin slip pproches zero. In other words when (cos slip m S sin slip goes to zero, it doesn t mtter how gret the force ou ppl t tht slip, the block will still be onl bout to slip still t rest. So the mimum ngle, slip.m, is given b: cos slip.m m S sin slip.m = 0 Tht is: tn slip.m = 1/m S Or: slip.m = tn -1 (1/m S = tn -1 (1/0.7 = 55.0 18
10.. The figure shows 100 kg block being relesed from rest from height of 1.0 m. The block tkes 0.64 s to rech the floor. Wht is the mss of the block on the left? irst, do kinemtics to find the -ccelertion vlue so tht the bo lnds in 0.64 s: D = v i. (Dt + (1/2 (Dt 2 Or: = 2[D v i. (Dt]/(Dt 2 = 2[ 1 0(0.64]/(0.64 2 = 4.883 m/s 2 Now do BD nlsis (referring to digrms here using conventionl - nd - es: Let the unknown mss be nd the 100-kg mss be m 2. And note tht the mgnitudes of their ccelertions will be the sme ( = 4.883 m/s 2 ; nd tht the tension forces eerted will lso be eul (becuse the pulle is idel mssless. Thus: : m 2 : 2.1 S 1 1 2.1 G.1 1 g 1.2 S 2 2 G.1 Then: m 2 g m 2 g Or: m 2 (g ( + g 1.2 G.2 2 m 2 g = m 2 So: g m 2 So: (g /( + g = 100(9.80 4.883/(4.883 + 9.80 = 33.5 kg b. The two blocks shown here re sliding down the incline. ind the tension in the (mssless string. Do BD nlsis (referring to digrms here using es s shown: Let the 1-kg mss be nd the 2-kg mss be m 2. Note tht the mgnitudes of their ccelertions will be the sme (. Thus: G.2 : m 2 :.1.2 K.1.21 K.2.12 G.1 G.2 S 1 1 2.1 + G.1. K.1 1 + gsin m K.1.1 S 1 1 S 2 2 G.2. 1.2 K.2 2 m 2 gsin m K.2.2 S 2 2.1 G.1. 1.2 G.2. 2.1 gcos = 0.2 m 2 gcos = 0 So:.1 gcos And:.2 gcos Thus: + gsin m K.1 ( gcos And: m 2 gsin m K.2 (m 2 gcos ind : = / + gsin m K.1 (gcos And: = gsin /m 2 m K.2 (gcos So: ind : / + gsin m K.1 (gcos = gsin /m 2 m K.2 (gcos (1/ + 1/m 2 = (m K.1 m K.2 (gcos = [(m K.1 m K.2 (gcos]/(1/ + 1/m 2 = [(0.20 0.10(9.80(cos20 ]/(1/1.0 + 1/2.0 = 0.614 N 19
10. c. Two blocks (msses = 26.5 kg; m 2 = 38.9 kg re moving together horizontll to the right. Block 2 is touching, but not ttched to, block 1. The coefficient of sttic friction, m s, between the two blocks is 1.47 (es, the re uite stick. ind the minimum coefficient of kinetic friction, m k, between block 1 nd the floor so tht block 2 does not slip. m 2 irst, do complete BD nd Newton s Lws nlsis of ech mss, s follows: BD of -nlsis S.1.21 k.1.21.21 m k.1 k.1 -nlsis coord. es s.21 m G.1 S.1 s.21 m G.1.1 m s.21 g = 0 -nlsis BD of m 2 S s.12 m.12.12 -nlsis.12 S coord. es G.2 s.12 m G.2 m s.12 m 2 g = 0 20
Now solve for m k, s follows: I. Solve m s.12 m 2 g = 0 for.12 :.12 g/m s II. Solve.12 for (the -ccelertion of both msses: =.12 /m 2 = (m 2 g/m s /m 2 = (g/m s III. Solve for.21 : B Newton s Third Lw, the mgnitudes of.21 nd.12 must be eul..21 =.12 g/m s IV. Solve.1 m s.21 g = 0 for.1 :.1 = m s.21 + g = m s (m 2 g/m s + g = ( + m 2 g V. Solve.21 m k.1 for m k : m k = (.21 /.1 = [m 2 g/m s + (g/m s ]/[( + m 2 g] = 1/m s 21
10. d. As shown in this side (cutw or -r view, sphere (m sphere = 2.00 kg rests in one comprtment of two-comprtment bo (m bo = 5.00 kg. The bo rests on level, frictionless surfce. All surfces of the bo nd sphere re lso frictionless. When the bo nd sphere re t rest (s shown, there re two norml forces cting on the sphere, nd one of those forces hs mgnitude 20% greter thn the other. But when tension force, (not shown here is then eerted horizontll (to the right on the right side of the bo, the two norml forces cting on the sphere become eul in mgnitude. ind the mgnitude of. Let the two surfces of the bo tht re eerting norml forces on the sphere be clled Wll (W nd Divider (D. Then here is the nlsis of the sphere ( when everthing is t rest, s shown bove (nd note the definition of now on tht digrm:.d1 S 1. 1. S 1. 1..W1.W1.D1. 1..D1. G.1 1..W1.D1 cos = 0.D1 sin g = 0 Thus:.D1 =.W1 /cos But we lso know:.d1 = 1.20(.W1 Therefore: G.1 1.20(.W1 =.W1 /cos Solve for : = cos -1 (1/1.20 = 33.557 Now nlze the combined mss (bo nd sphere together cll this &2, when the force is pplied:.1&2 S 1&2. &2 1&2. S 1&2. &2 1&2. &2 1&2..1&2 G.1&2 &2 1&2. &2.1&2 &2 g = 0 Thus: = /&2 G.1&2 Now nlze just the sphere ( s it is ccelerting to the right inside the bo:.d1 S 1. 1. S 1. 1..W1.W1.D1. 1..D1. G.1 1..W1.D1 cos.d1 sin g = 0 G.1 Thus:.D1 g/sin Thus:.D1.D1 cos ( /&2 Or: ( g/sin(1 cos ( /&2 Solve for : = (&2 / ( g/sin(1 cos = (7.00/2.00[(2.00(9.80/sin(33.557 ][1 cos(33.557 ] = 20.7 N
11.. Two crtes ( is known; m 2 is unknown re being drgged together (one linked to the other vi horizontl cble, s shown cross the floor of lrge freight elevtor. The crtes horizontl velocit is constnt. The elevtor is ccelerting upwrd t known rte,. The pulling force is known, s is the ngle bove the horizontl t which pulls. The coefficient of kinetic friction, m K, between ech crte nd the elevtor floor is lso known. ind m 2, epressed in terms of,,, m K nd g elevtor m 2 (cble irst, do complete BD nd Newton s Lws nlsis of ech mss, s follows: BD of -nlsis S.1 k.1.21.21 k.1 cos m k.1.21 = 0 -nlsis coord. es G.1 S +.1 G.1 sin +.1 g -nlsis BD of m 2 S.2.12 k.2.12 m k.2 = 0.12 -nlsis k.2 S coord. es G.2.2 G.2.2 m 2 g 23
Now solve for m 2, s follows: I. Solve the -force eution (sin +.1 g for.1 :.1 + g sin II. Solve the -force eution (cos m k.1.21 = 0 for.21 :.21 = cos m k.1 = cos m k ( + g sin III. Solve for.12 : B Newton s Third Lw, the mgnitudes of.21 nd.12 must be eul..12 =.21 = cos m k ( + g sin IV. Solve the -force eution (.12 m k.2 = 0 for.2 :.2 =.12 /m k = cos/m k ( + g sin = (cos/m k + sin ( + g V. Solve the -force eution (.2 m 2 g for m 2 : m 2 =.2 /(g + = (cos/m k + sin/( + g 24
11. b. Wlking through our neighborhood, ou see Thor set down his hmmer while working on the roof. The hmmer slides, strting from rest, distnce d down the roof, which is ngled from the horizontl. The hmmer leves the roof from height h, nd lnds horizontl distnce from the point where it becomes projectile. Wht is the coefficient of kinetic friction between hmmer nd the roof? You m consider these vlues s known: d,, h,, g. This is n ODAVEST problem, but keep in mind tht ou re not being sked to ctull solve for the finl epression. In fct, ou re not being sked to do n mth t ll not even n lgebr. Rther, for the Solve step, ou re to write series of succinct instructions on how to solve this problem. Pretend tht our instructions will be given to someone who knows mth but not phsics. And for the Test step, ou should consider the sitution nd predict how the solution would chnge if the dt were different chnging one dt vlue t time. Objective: An object strts from rest nd slides distnce d down roof surfce tht is inclined t n ngle below the horizontl. The object leves the roof t height h bove the ground nd first impcts the ground t horizontl distnce from the point where it left the roof. We need to determine m K (the coefficient of kinetic friction between the object nd the roof surfce. Dt: v i = 0 The object begn its slide t rest. d h g The distnce over which the object slid on the roof surfce. The ngle of incline of the roof surfce, with respect to the horizontl. The verticl distnce from the impct point to where the object left the roof. The horizontl distnce from the impct point to where the object left the roof. The locl free-fll ccelertion mgnitude. Assumptions: Objects We will tret the hmmer (object s point mss hving no etent or rottion. Surfces Slide Projectile We will ssume tht the surfces of both the roof nd the hmmer re uniform. We will lso ssume tht the roof is plnr. We ssume tht the slide ws in stright line nd directl down the slope (long the steepest possible pth. We ssume the g is constnt for ll relevnt heights here. We will disregrd n effects of wind or ir drg. Visul Reps.:.S1 ree-bod digrm of object sliding ginst kinetic friction down the roof slope: K.S1 Kinemtics digrm (nd visul dt inventor of the slide: G.P1 v i = 0 D = d v f ( ( D = d v i. = 0 v f. = =?? Dt = 25
Kinemtics digrm (nd visul dt inventor of the projectile motion: D = h v i i D = v f f ( ( D = D = h v i. = v i cos v i. = v i sin v f. = v f. = = 0 = g Dt = Eutions: I. gsin m k.s1 II..S1 gcos = 0 III. v f. 2 = 0 2 + 2 d IV. = v i cos(dt V. h = v i sin(dt (1/2g(Dt 2 Solving: Solve IV for Dt. Substitute tht result into V. Solve V for Dt. Substitute tht result into IV. Solve IV for v i. Substitute tht result s v f. into III. Solve III for. Substitute tht result into I. Solve II for.s1. Substitute tht result into I. Solve I for m k. Testing: Dimensions: The solution would need to be unitless (just numericl vlue. Dependencies: In generl, this number should be somewhere between 0 nd 1 (or possibl little bove 1 if the roof is ver stick or rough. A greter v i tht produces the sme projectile motion would impl tht the roof slows down the slide little more greter m k. A greter slide distnce d tht produces the sme projectile motion would impl tht the ccelertion is less (chieves sme lunch speed over longer slide: greter m k. A higher fll, h, with the sme s before, would impl slower lunch speed: greter m k. Without eplicitl solving the bove eution set s prescribed, the effects of greter roof ngle re not cler. A greter trvel for the sme fll height would impl fster lunch speed: smller m k. Without eplicitl solving the bove eution set s prescribed, the effects of greter vlue of g re not cler. It would ffect the friction, the slide nd the fll. (It s possible, therefore, tht it might even hve no implictions t ll for m k. 26
12. A 90-kg mn bords n elevtor nd stnds on n ccurte bthroom scle (which just hppens to be there. During his elevtor ride, the scle reds sted 700 N. Also during this ride, he hppens to drop coin from height of 1 m bove the elevtor floor. rom the moment he releses the coin, how long does it tke to hit the floor of the elevtor? elevtor BD of m m (person (scle coord. es G -nlsis -nlsis S = m S = m 0 = 0 G = m mg = m The -direction hs no forces nd thus zero ccelertion. In the -direction, is esil solved for: = ( mg/m = [700 (90(9.80]/90 = 2.022 m/s 2 This is the ccelertion of the elevtor floor, the mn, nd the coin in his hnd the ll move s one until the moment when he releses the coin. After tht moment, the coin s ccelertion is g. But the elevtor floor s ccelertion continues to be ( 2.022. Essentill then, the uestion becomes this: How long does it tke the coin to ctch up to the elevtor floor s the both ccelerte downwrd t different rtes? Object A (coin: D A = v A.i (Dt + (1/2( g(dt 2 Object B (elevtor floor: D B = v B.i (Dt + (1/2( (Dt 2 But D A = D B 1 (It s 1 becuse D downwrd is negtive; A trvels downwrd 1 meter more thn B. And v A.i = v B.i (When the coin is relesed, its velocit is the sme s the elevtor floor s. Substituting: v A.i (Dt + (1/2( g(dt 2 = v A.i (Dt + (1/2( (Dt 2 1 Solve for Dt: (1/2( (Dt 2 (1/2(g(Dt 2 = 1 (Dt 2 [(1/2( + (1/2(g] = 1 (Dt = {1/[(1/2( + (1/2(g]} 1/2 = {1/[(1/2( 2.022 + (1/2(9.80]} 1/2 = 0.507 s 27