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Transcript
11. To solve the problem, we note that acceleration is the second time derivative of the position function; it is a vector and can be determined from its components. The net force is related to the acceleration via Newton’s second law. Thus, differentiating twice with respect to t, we get Similarly, differentiating twice with respect to t yields (a) The acceleration is At , we have with a magnitude of Thus, the magnitude of the force is (b) The angle or We choose the latter ( makes with ) since is is in the third quadrant. (c) The direction of travel is the direction of a tangent to the path, which is the direction of the velocity vector: At makes with , we have is We choose the latter ( Therefore, the angle ) since is in the third quadrant. 13. (a) – (c) In all three cases the scale is not accelerating, which means that the two cords exert forces of equal magnitude on it. The scale reads the magnitude of either of these forces. In each case the tension force of the cord attached to the salami must be the same in magnitude as the weight of the salami because the salami is not accelerating. Thus the scale reading is mg, where m is the mass of the salami. Its value is (11.0 kg) (9.8 m/s2) = 108 N. 15. (a) From the fact that T3 = 9.8 N, we conclude the mass of disk D is 1.0 kg. Both this and that of disk C cause the tension T2 = 49 N, which allows us to conclude that disk C has a mass of 4.0 kg. The weights of these two disks plus that of disk B determine the tension T1 = 58.8 N, which leads to the conclusion that mB = 1.0 kg. The weights of all the disks must add to the 98 N force described in the problem; therefore, disk A has mass 4.0 kg. (b) mB = 1.0 kg, as found in part (a). (c) mC = 4.0 kg, as found in part (a). (d) mD = 1.0 kg, as found in part (a). 57. We take +y to be up for both the monkey and the package. !"#$%&$'()&$*!$'+$,&$-.$/+0$,+'1$'1&$2+3)&4$(35$'1&$.(6)(7&#$ (a) The force the monkey pulls downward on the rope has magnitude F. According to Newton’s 8(9$ :1&$ /+06&$ '1&$ 2+3)&4$ .-;;<$ 5+=3=(05$ +3$ '1&$ 0+.&$ 1(<$ 2(73>'-5&$"#$ ?66+05>37$ '+$ third law, the rope pulls upward on the monkey with a force of the same magnitude, so Newton’s @&='+3A<$ '1>05$ ;(=B$ '1&$ 0+.&$ .-;;<$ -.=(05$ +3$ '1&$ 2+3)&4$ =>'1$ ($ /+06&$ +/$ '1&$ <(2&$ second law for forces acting on the monkey leads to 2(73>'-5&B$<+$@&='+3A<$<&6+35$;(=$/+0$/+06&<$(6'>37$+3$'1&$2+3)&4$;&(5<$'+ F – mmg = mmam, "$C$##$$D$# #%#B where # mm$><$'1&$2(<<$+/$'1&$2+3)&4$(35$% is the mass of the monkey and am $><$>'<$(66&;&0('>+3#$E>36&$'1&$0+.&$><$2(<<;&<<$ is its acceleration. Since the rope is massless F = T is =1&0& # # the tension in the rope. The rope pulls upward on the package with a force of magnitude F, so "$D$&$><$'1&$'&3<>+3$>3$'1&$0+.&#$:1&$0+.&$.-;;<$-.=(05$+3$'1&$.(6)(7&$=>'1$($/+06&$+/$ Newton’s second law for the package is 2(73>'-5&$"B$<+$@&='+3A<$<&6+35$;(=$/+0$'1&$.(6)(7&$>< F +$C$# FN –$$D$# mpg =% mB pap, "$*$" ' ( ( ( where # mp$ is the mass of the package, ap is its acceleration, and FN is the normal force exerted by =1&0& ( ><$ '1&$ 2(<<$ +/$ '1&$ .(6)(7&B$ %($ ><$ >'<$ (66&;&0('>+3B$ (35$ "'$ ><$ '1&$ 3+02(;$ /+06&$ the ground on it. Now, if F is the minimum force required to lift the package, then FN = 0 and ap &F&0'&5$,4$'1&$70+-35$+3$>'#$@+=B$>/$"$><$'1&$2>3>2-2$/+06&$0&G->0&5$'+$;>/'$'1&$.(6)(7&B$ = 0. According to %&%'*%+,,-.)/$0%!-%!"#%1#,-$)%2(3%#45(!/-$%6-.%!"#%7(,8(0#9%!"/1%:#($1% the second law equation for the package, this means F = mpg. Substituting mpg !"#$ !"%&%'%($)%# $ for F in&*%;5<1!/!5!/$0%% the equation for &%6-.%!%/$%!"#%#45(!/-$%6-.%!"#%:-$8#=9%3#%1-2>#%6-.%# the monkey, we solve for am: !%&%% $ $ %? @ ! − %% & ( % $ − %% ) & (AB 80 − A' 80 ) ( C*D :E1 ) = = = F*C :E1 @ * #% = %% %% A' 80 (b) +1% As )/1,511#)9% discussed,I#3!-$J1% Newton’s1#,-$)% second leads for7(,8(0#% the package G<H% 2(3%law 2#()1% !-% !to− % $ & = % $ # $ %6-.% !"#% ($)% and for!"#% the:-$8#=*% monkey.K6%If!"#% the(,,#2#.(!/-$% acceleration-6%of package is downward, then the ! − %% & = %% #% %6-.% !"#%the 7(,8(0#% /1% )-3$3(.)9% !"#$% acceleration of the monkey is upward, so am = %–a the first equation for F p. Solving %&%L# !"#%(,,#2#.(!/-$%-6%!"#%:-$8#=%/1%573(.)9%1-%# $*%;-2>/$0%!"#%6/.1!%#45(!/-$%6-.%! ! " # ! = %$ & + # $ = %$ & − #% $ ($)%15<1!/!5!/$0%!"/1%.#152!%/$!-%!"#%1#,-$)%#45(!/-$9%3#%1-2>#%6-.%#%? # (% = $ − %% ) & = (AB 80 − A' 80 ) ( C*D :E1@ ) = @*' :E1 @ * and substituting this result into the second equation, we solve for am: (c) The result is positive, indicating that the acceleration of the monkey is upward. (d) Solving the second law equation for the package, we obtain 60. The motion of the man-and-chair is positive if upward. (a) When the man is grasping the rope, pulling with a force equal to the tension T in the rope, the total upward force on the man-and-chair due its two contact points with the rope is 2T. Thus, Newton’s second law leads to so that when a = 0, the tension is T = 466 N. (b) When a = +1.30 m/s2 the equation in part (a) predicts that the tension will be . (c) When the man is not holding the rope (instead, the co-worker attached to the ground is pulling on the rope with a force equal to the tension T in it), there is only one contact point between the rope and the man-and-chair, and Newton’s second law now leads to so that when a = 0, the tension is T = 931 N. (d) When a = +1.30 m/s2, the equation in (c) yields T = 1.05 × 103 N. (e) The rope comes into contact (pulling down in each case) at the left edge and the right edge of the pulley, producing a total downward force of magnitude 2T on the ceiling. Thus, in part (a) this gives 2T = 931 N. (f) In part (b) the downward force on the ceiling has magnitude 2T = 1.05 × 103 N. (g) In part (c) the downward force on the ceiling has magnitude 2T = 1.86 × 103 N. (h) In part (d) the downward force on the ceiling has magnitude 2T = 2.11 × 103 N. 61. The forces on the balloon are the force of gravity (down) and the force of the air (up). We take the +y to be up, and use a to mean the magnitude of the acceleration (which is not G !"#$%&'$()*+',$)-$.&'$/011))-$0*'$.&'$()*+'$)($2*034.5$ !" $67)8-9$0-7$.&'$()*+'$)($.&'$04*$ G #$ $6:;9#$<'$.0='$.&'$>%$.)$/'$:;?$0-7$:,'$ $$.)$@'0-$.&'$ !$"&'()*+$)($.&'$0++'1'*0.4)-$ its 68&4+&$ usual use in 4.,$ this:,:01$ chapter). When the mass<&'-$ is M .&'$ (before is .&'$ thrown out) 4,$ -).$ :,'$ 4-$ .&4,$ +&0;.'*9#$ @0,,$the 4,$ ,ballast $ 6/'()*'$ /0110,.$ 4,$ the acceleration is downward and Newton’s second law is .&*)8-$):.9$.&'$0++'1'*0.4)-$4,$7)8-80*7$0-7$A'8.)-B,$,'+)-7$108$4,$$ Mg$D$C = –Ma. #F$a$C$–," ,$# After the ballast is thrown out, the mass is M,–$C$m!(where is the mass of the ballast) and the E(.'*$.&'$/0110,.$4,$.&*)8-$):.?$.&'$@0,,$4,$ $68&'*'$m!$4,$.&'$@0,,$)($.&'$/0110,.9$0-7$ acceleration is upward. Newton’s second law leads to .&'$0++'1'*0.4)-$4,$:;80*7#$A'8.)-B,$,'+)-7$108$1'07,$.) – (M = (M #F$a$C$6 ,$C$–!m)g 9"$D$6 ,$C$–!m)a. 9$# The%&'$;*'34):,$'F:0.4)-$243',$ previous equation gives Fa = this plugs into the new equation to give #$M(g $D$,–6"a), $C$$and 9?$0-7$.&4,$;1:2,$4-.)$.&'$-'8$'F:0.4)-$.)$243'$ ! " ! " ! " , " −$ − , −! " = ,−!$ != G ,$ # "+$