How o Solve Sye Dynaic Proble A ye dynaic proble involve wo or ore bodie (objec) under he influence of everal exernal force. The objec ay uliaely re, ove wih conan velociy, conan acceleraion or oe cobinaion of hee, bu all objec of he ye have he ae oion. Anoher hing of iporance i he diincion beween inernal and exernal force. 1. Read he proble carefully! A hi poin you need o deerine wha objec coprie he ye. An objec uually belong o he ye if i i a i known and / or i i known o be conneced o anoher objec. Generally, urface and he Earh ielf are no conidered a par of he ye Read he following pracice exercie and deerine wha objec ake up he ye. i. Pracice 1: A caerpillar racor puhe a couple of car wreck on a level junk yard. One car ha a a of 1500 kg. Thi car ouche a ruck wih a a of 2300 kg. The racor applie 20 kn of force o he car. The coefficien of kineic fricion beween boh wreck and he ground i 0.60. Deerine he ye acceleraion and he force of he car on he ruck and he ruck on he car. The car and he ruck coprie hi ye. We exclude he racor a we do no know i a. We exclude he urface for eenially he ae reaon. ii. Pracice 2: Uing he diagra provided deerine (a) which way he ye ove, (b) if he ye ponaneouly ove, (c) if i were in oion i acceleraion and (d) if i were in oion he enion in all connecing ring. M1 and M2 are he ye, a we know heir ae and hey are conneced by a coon ring. The urface of he incline and he Earh will be ouide hi ye.
2. Nex you need o deerine he direcion of ravel for he ye and he locaion of he x-axi. You ay need o conider hi objec by objec epecially if each objec i on a differen urface. However, in general er all objec will ove eiher o he lef or righ. i. Pracice 1: A caerpillar racor puhe a couple of car wreck on a level junk yard. One car ha a a of 1500 kg. Thi car ouche a ruck wih a a of 2300 kg. The racor applie 20 kn of force o he car. The coefficien of kineic fricion beween boh wreck and he ground i 0.60. Deerine he ye acceleraion and he force of he car on he ruck and he ruck on he car. We are old he yard i level o he x-axi i horizonal. In hi proble i i enirely up o you o have he racor puh righ or lef bu i u ouch he car. I have arbirarily choen righ. ii. Pracice 2: Uing he diagra provided deerine (a) which way he ye ove, (b) if he ye ponaneouly ove, (c) if i were in oion i acceleraion and (d) if i were in oion he enion in all connecing ring. Boh ae can only ove along he incline plane, herefore he x-axi will be iled for each a. A for direcion of ravel we have wo opion eiher i will be obviou or no. I uually no obviou if one or ore objec are on an inclined plane. So wha you do when i i no obviou i calculae he exernal force on he objec ha could ake he ove. Ignore all fricion a hey will no caue oion. Ignore force or coponen perpendicular o he poenial direcion of
ravel. Ignore connecing force a hey are inernal. In hi exaple, i i only he x-coponen of graviy ha can caue oion. You now u deerine he ize of all of hee and add up he one poining righ, and hen add up he one poining lef. Which ever ide i bigger deerine he direcion of ravel. Since F g2x > F g1x, hen M2 ove down hill while M1 ove up hill. 3. In pracice 2 we have an addiional challenge of deciding if an iniially reing ye will ponaneouly ove. Thi require olving for he ye acceleraion. However, if a fricion will be ued in he ne force equaion i will only be aic fricion. You hould now calculae he oher exernal force uch a noral force and aic fricion. Once you know he value of hee exernal force olve for he ye acceleraion wih he ne force equaion. Again only ue force along he x-axi o calculae acceleraion. The value of he acceleraion deerine he anwer. If a i negaive or zero, hen he ye ay a re. If a i poiive, i will ponaneouly ove. i. Pracice 2: Uing he diagra provided deerine (a) which way he ye ove, (b) if he ye ponaneouly ove, (c) if i were in oion i acceleraion and (d) if i were in oion he enion in all connecing ring. F = a=σf ΣF ne wih again ΣF wih ΣF again F F F F g2x g1x 1 2 (72.3 50.2 13.8 5.06) N = 0.135 24.0kg Thi ye ponaneouly accelerae. 2
4. In boh pracice, we u find he acceleraion auing ha oehow hey go in oion. If olving for he ye acceleraion and fricion are involved only be kineic fricion will appear in he ne force equaion. NEVER USE STATIC AND KINETIC FRICTION AT THE SAME TIME. Only ue force on he x-axi o olve for he acceleraion. The value of he acceleraion deerine he naure of he oion. If a i negaive hen he ye low down and op. I o likely will reain opped unle he aic fricion i really low. If a i zero, hen he ye ove wih conan velociy in he deired direcion of ravel. If a i poiive, he ye will accelerae. i. Pracice 1: A caerpillar racor puhe a couple of car wreck on a level junk yard. One car ha a a of 1500 kg. Thi car ouche a ruck wih a a of 2300 kg. The racor applie 20 kn of force o he car. The coefficien of kineic fricion beween boh wreck and he ground i 0.60. Deerine he ye acceleraion and he force of he car on he ruck and he ruck on he car. F = ΣF ΣF ne wih again ΣF wih ΣF F F F again 1 k1 k2 (20000 13538 8829) N = 0.635 3800kg Thi ye i lowing down o a hal. 2
ii. Pracice 2: Uing he diagra provided deerine (a) which way he ye ove, (b) if he ye ponaneouly ove, (c) if i were in oion i acceleraion and (d) if i were in oion he enion in all connecing ring. F = a=σf ΣF ne wih again ΣF wih ΣF again F F F F g2x g1x k1 k2 (72.3 50.2 2.8 1.0) N = 0.758 24.0kg Thi ye i peeding up. 2
5. Finding connecing force in any proble require elecing one objec and olving he ne force equaion for i. Ue he ye acceleraion calculaed in he la ep. i. Pracice 1: A caerpillar racor puhe a couple of car wreck on a level junk yard. One car ha a a of 1500 kg. Thi car ouche a ruck wih a a of 2300 kg. The racor applie 20 kn of force o he car. The coefficien of kineic fricion beween boh wreck and he ground i 0.60. Deerine he ye acceleraion and he force of he car on he ruck and he ruck on he car. Uing he ruck a he arge objec: F = ΣF ΣF ne 2 wih again 2300( 0.635 2) rd car on ruck k 2 F = ( 1461+ 13538) N = 12077N car on ruck = F F By Newon' 3 Law, he force of he ruck on he car ha he ae value in he oppoie direcion.
ii. Pracice 2: Uing he diagra provided deerine (a) which way he ye ove, (b) if he ye ponaneouly ove, (c) if i were in oion i acceleraion and (d) if i were in oion he enion in all connecing ring. Uing M1 a he arge objec: F = ΣF ΣF ne 1 wih again 15(0.758) = F - F - F T g1x k1 F = 11.37 + F + F T g1x k1 F = (11.37 + 50.2 + 2.8) N = 64.4N T