PHYS 00 Worked Eaple Week 05: ewton nd Law Poor Man Acceleroeter A drver hang an ar frehener fro ther rearvew rror wth a trng. When acceleratng onto the hghwa, the drver notce that the ar frehener ake an angle of about 5 degree wth repect to the horzontal. What the acceleraton of the car? ( Coprehend the Proble We ve got a hangng ar frehener (of unknown a acceleratng forward. It hangng fro a trng at 5 degree fro the vertcal. ro th angle we re to fnd the object acceleraton. Snce the ar frehener appear at ret to the drver, t ut be acceleratng at the ae rate a the car. Snce the car acceleraton all horzontal, the ar frehener acceleraton hould be entrel horzontal a well. a 5 0 ( Epre the Proble n oral Ter (Decrbe the Phc We re lookng for the acceleraton of the ar frehener, o we probabl need to fnd all the force actng on the ar frehener that caue th acceleraton. The free bod dagra hown at the rght. T There are two force on the ar frehener, a tenon up and to the left and a weght traght down. The acceleraton to the left, o the u of thee two force ut pont to the left. We ll chooe to the left a the + drecton o the acceleraton purel (th hould ake the ath pler later. The relatonhp between force and acceleraton gven b ewton Second Law,. To fnd the force, we ll need to know the drecton of all the force n the proble. We know the weght pont traght down (all -drecton, and the tenon n the trng ha a potve -coponent and a potve -coponent (.e. up and to the left. 5 0 W (3 Plan a Soluton We ll wrte ewton Second Law a two coponent equaton, one n the -drecton and the other n the -drecton. We ll have two unknown, the agntude of the tenon T and the acceleraton n the -drecton a. We can olve the two equaton for the two unknown T and a.
PHYS 00 Worked Eaple Week 05: ewton nd Law (4 Eecute the Plan T 5 0 W Wrte ewton Second Law a two coponent equaton, one n the - drecton and the other n the - drecton. = = = -drecton:,,,,, ( T 5 ( 0 T 5 T + W + Epre the force a a u of all the force actng on the object., -drecton: ( T co 5 + ( g = ( 0 T co 5 T + W = g g T = co 5 T 5 g We can nert the tenon value fro 5 5 the -drecton equaton nto the - co drecton equaton. a = g tan 5 = ( 9. 8 ( 0. 466 a = 4. 6 The ar frehener (and therefore the car accelerate at 4.6 /.
PHYS 00 Worked Eaple Week 05: ewton nd Law Reel It In A crate of a 00 kg connected to a wnch on the top of a tower b a cable. The reel upple a contant force of 3000 to the cable, ldng the crate acro a ver lck floor. A the crate lde toward the tower acro the ver lck floor, the angle the cable ake wth the horzontal get larger. Anwer the followng two queton. a If the ntal angle the cable ake wth the horzontal 5 degree, what the ntal acceleraton of the crate? b At what cable angle doe the crate frt get lfted off of the floor? ( Coprehend the proble Th tuaton nvolve a crate beng pulled acro the floor. We know the tenon n the cable rean a contant 3000 and the a of the crate 00 kg. We re alo gven the ntal angle the crate ake wth the horzontal 5 degree. We need to fgure out the acceleraton that all the force on the crate produce at th tartng pont. After th, we are to fnd the frt angle the cable ake wth the horzontal at whch the crate accelerate upward (.e. get lfted off of the floor. ( Epre the Proble n oral Ter (Decrbe the Phc There are two queton we ut addre. The frt nvolve fndng the ntal acceleraton of the crate. The econd nvolve fndng the frt angle at whch the crate accelerate upward, rather than jut horzontall. Snce the acceleraton of the crate caued b the force on the crate, let begn b dentfng all the force on the crate. There are three force on the crate. the tenon fro the cable on the crate up and to the rght: the weght fro the earth attractng the crate downward: 3 the noral force fro the floor on the crate upward: So the cable pull the crate up and to the rght, the crate weght pull t traght down, and the noral force fro the floor puhe traght up to prevent the crate fro peratng the floor. Snce we re aked about oton along the floor and perpendcular to the floor, let chooe our ae for th proble to be horzontal (.e. parallel to the floor for and vertcal (.e. perpendcular to the floor for. 3
PHYS 00 Worked Eaple Week 05: ewton nd Law or both queton a and b, we can dentf the force on the object and are aked about the acceleraton. The relatonhp between force and acceleraton eboded n ewton Second Law: where the force the u of all the force actng on the crate. Th vector equaton tand for two calar (.e. jut nuber equaton, one n the -drecton and one n the -drecton: where each equaton deal onl wth - or -coponent, repectvel. We know the followng paraeter: (3 Plan a Soluton a We know the agntude of the tenon n the cable and we can fnd the agntude of the crate weght ug the relatonhp between weght and the acceleraton due to gravt near earth,. We alo know the angle the tenon ake wth the horzontal, ce t pont along the cable at 5 degree. To fnd the ntal acceleraton of the crate, we can fnd the vector u of all the force on the crate and dvde b the a. Th wll gve u the ntal acceleraton. b Here we are aked to fnd the frt cable angle that lft the crate off of the ground. Th happen jut after the noral force goe to zero. How do we know th? The noral force needed ntall becaue the vertcal coponent of the cable tenon not a bg a the vertcal coponent of the crate weght downward. (ee the fgure on the page Th ean that f there were no noral force, there would be a force downward on the crate. The crate would therefore accelerate downward and perate the floor. Snce th doen t happen, the floor ut be puhng upward wth a noral force jut bg enough to ake up the dfference between the -coponent of the tenon and the weght. A the crate approache the tower, the cable pont n a ore vertcal drecton. Th ncreae the -coponent of the tenon. Eventuall the tenon -coponent wll be large enough to cancel the crate weght on t own, and the noral force won t be needed (.e. t wll be zero. If the -coponent of the tenon grow pat th value, there wll be a force upward on the crate (the noral force can t pull downward unle the floor tck. A upward force ean that the crate wll accelerate upward. 4
PHYS 00 Worked Eaple Week 05: ewton nd Law Intal Stuaton: A noral force needed to keep the block fro acceleratng downward Later: The tenon -coponent jut cancel the crate weght Even Later: The tenon -coponent larger than the crate weght W T T T ree Bod Dagra W T coponent W T T ree Bod Dagra T W T coponent T T T W ree Bod Dagra W T coponent (no -acceleraton (no -acceleraton (accelerate upward We hould therefore wrte the equaton for the force on the crate n the -drecton. We wll then et the noral force agntude equal to zero and olve for the angle that ake th happen. That wll be the lat angle at whch the crate rean on the floor. An larger angle wll accelerate the crate upward. (4 Eecute the Plan a nd the agntude of the crate ntal acceleraton fro ewton Second Law. Break the general tateent of ewton Second Law nto - and -coponent equaton. We are told that the crate ntall lde along the floor, o t -veloct rean zero. Th ple that t -acceleraton zero. All that left to fnd the -acceleraton. Wrte the force on the crate a a u of all the force actng on the crate (ee the free bod dagra drawn earler. We ve choen horzontal and vertcal a our - and -ae. Up and rght are the potve drecton. We ll tck wth the - drecton ce we alread know that the - acceleraton zero. Solve for the -coponent of the acceleraton and nert the nuber fro the proble. T co,, = ( 0, + W + T ( T co( θ ( 0 + (0 + 0 ( θ 0 a T co = = ( θ 0 kg ( 3000 ( co5 00 kg = 9.0 5
PHYS 00 Worked Eaple Week 05: ewton nd Law b nd the cable angle at whch the crate lfted off of the floor. We ll do th b fndng the angle that ake the noral force zero and keep the acceleraton n the -drecton equal to zero. An angle bgger than th wll lft the crate off of the ground. Break the general tateent of ewton Second Law nto - and -coponent equaton. We re keepng the -acceleraton zero becaue we re lookng for the angle jut before the crate lft off the ground. Snce we re ntereted n -oton, let look at the -drecton frt. We wrte the force a a u of all the force on the crate (ee the free bod dagra drawn n tep ( We et the noral force equal to zero and wrte the other -coponent, reeberng that we choe up a potve. Solve for the lat angle wth zero upward acceleraton and nert the nuber fro the proble. Once the cable reache 9 0 the crate wll lft off of the ground.,, = + W ( 0, + T ( 0 + ( g + ( T ( θ ( θ g + T lat ( θ lat θ = g T lat lat = = θ g T ( 00 kg ( 9. 8 ( 3000 ( lat =. = 0 37 9 6
PHYS 00 Worked Eaple Week 05: ewton nd Law Groovn A phere of a M retng n the groove created b two nclned plane eetng at ther tp. The plane have known dfferent angle of nclne wth repect to the horzontal. nd the agntude of the noral force uppled b each plane. ( Coprehend the Proble We have a phere retng on two flat urface, each wth t own known nclne wth the horzontal. A ketch of the tuaton look lke th: θ θ Ν Ν We know that the phere retng on the nclne. We know the a of the ball, a well. We are alo aued to know the two angle of the nclne, θ and θ. We know that there a gravtatonal force (the weght pullng the phere traght down. Therefore, there ut alo be a noral force on the ball fro each plane to keep the ball fro peratng the urface. We are to fnd the agntude of each of thee noral force. Snce we know the ball rean at ret, t acceleraton ut be zero. We therefore know t acceleraton ut be zero (veloct never change fro zero. Snce we know the acceleraton and are aked about force, we ll probabl ue ewton Second Law to deterne the noral force. ( Repreent the Proble n oral Ter (Decrbe the Phc We can organze our qualtatve dea about what gong on wth a free bod dagra of the ball. Th ean that we draw the ball b telf (.e. free of the ret of the object wth onl the force O the ball. θ W θ Ν Ν We know we have three force to deal wth. We ll probabl need to know ther drecton. We can chooe our - and -ae an wa we lke, but ce none of thee force are ether perpendcular or parallel to each other we ght a well jut chooe up a + and rght a +. Th at leat ake the weght force pont entrel n the -drecton. W To fnd the drecton of the noral force, we ll need to ue oe geoetr. Let look at frt: Ν The noral force alwa drected perpendcular (.e. noral to the urface that provde t. If we etend drecton back to the urface that provde t, we eet that urface at a rght angle, a hown. θ 90 ο θ θ Etendng th lne all the wa to the -a ake a rght trangle underneath the left-hand urface. All the angle of a trangle add up to 80 o, o the angle we re ntereted n hould be: 80 o 90 o θ = (90 o θ Slarl, the angle between and the horzontal (90 o θ. ote that pont n the negatve -drecton whle pont n the potve -drecton. 7
PHYS 00 Worked Eaple Week 05: ewton nd Law We therefore have the followng force: agntude = noral force fro the left urface = angle ( 90 θ fro the horztonal agntude = noral force fro the rght urface = angle ( 90 θ fro the horztonal agntude g W = weght of the phere = angle - 90 fro the horztonal We can relate thee force b ug ewton Second Law: =. (3 Plan a Soluton We wll break ewton Second Law nto - and -coponent equaton. Snce we know the acceleraton zero (phere rean at ret, we can olve thee two equaton for the two unknown and. (4 Eecute the Plan rt, we break ewton Second Law down nto two coponent equaton. We know that the phere rean at ret, o the acceleraton along both the - and -drecton zero. ow we wrte the force n each drecton a the u of each force coponent. (ee the free bod dagra n tep ( We are alo ug the trgonoetrc dentte 90 θ = co θ co ( ( ( 90 θ = ( θ = = =,,,, (, =, = 0, =, = ( 0 -drecton, + + W ( co( θ co( θ ( ( 90 + 90 + 0 ( θ ( θ -drecton, + + W ( ( θ ( θ ( ( 90 + 90 + g co ( θ co( θ + 8
PHYS 00 Worked Eaple Week 05: ewton nd Law We now have a te of two equaton (one fro each denon wth two unknown and ( θ ( θ. co( θ + co( θ g We can olve the frt equaton for and ubttute t nto the econd. ow we can deterne Here we have ultpled the top and botto b (θ to get rd of the annong fracton n the denonator. Before we plug th back nto the frt equaton to get, let plf t ug the trgonoetrc addton dentt: θ co θ + θ co θ = θ + θ ( ( ( ( ( ow we can plug th pler epreon for nto the frt equaton (.e. the equaton fro the -drecton. We now have each noral force agntude n ter of the known a, acceleraton due to gravt, and angle of the nclne. ( θ ( θ ( θ ( θ ( θ = ( θ ( θ co( θ ( θ ( θ ( θ co( θ + co( θ ( θ g ( θ ( θ ( θ + ( θ ( θ co + g co( θ + co( θ g co( θ + co( θ = g g = co co = = g ( θ ( θ ( θ ( g ( θ ( θ + θ ( θ ( θ + θ ( θ ( + = θ θ θ = g = g = g ( θ ( θ + θ ( θ ( θ + θ = (5 Interpret and Evaluate the Soluton Let ee f thee two epreon ake ene b checkng oe lt. If the left nclne were flat, then θ would be zero. The ball at ret would then ret entrel on the flat nclne one. Pluggng θ =0 nto the epreon gve =g and =0. Th check, becaue the left (flat nclne cancel the entre weght g and the rght nclne doen t uppl an noral force at all. 9