PHYSICS 151 Notes for Online Lecture #11



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PHYSICS 151 ote for Online Lecture #11 A free-bod diagra i a wa to repreent all of the force that act on a bod. A free-bod diagra ake olving ewton econd law for a given ituation eaier, becaue ou re odeling the te a oething ipler than it actuall i. o draw a free-bod diagra: 1. Draw a eparate diagra for each bod in the proble. If ou have to deal with two ae, for exaple, ou need two diagra.. Repreent the bod a a point. or now, we don t know how to deal with force that caue rotation. We re onl dealing with force that caue linear otion. Regardle of what bod we re dealing with, we re going to repreent it a a point.. Draw each of the force that act directl on the bod. here are two iportant idea here: act directl and on. Recall that onl force that act directl on a bod are counted when ou u up the net force. he econd operative word i on : ake ure that ou have included onl thoe force acting on the bod. Don t include force that the bod exert on other thing, or force that aren t exerted on the bod that ou re analzing.. Draw all of the force o that their tail are at the dot and their head point in the direction of the force. 5. Label all of the force Ex. 1-1: Draw a free-bod diagra for the exaple of a hand holding a block againt the wall that i alo upended b a tring. Solution: he block i repreented a a point. We have four force: tenion due to the tring, the applied force fro the hand, the weight of the block and the noral force of the wall on the block. he free-bod diagra look like: A A g g 1

Ex. 1-: A peron i oving a 5 kg file cabinet b puhing it downward at an angle of 0 below the horizontal with an applied force of 5. What i the acceleration of the file cabinet? If ou draw a free-bod diagra for thi ituation, it will look like thi: θ = 0 A W Part of the applied force act to the right and part act downward. he cabinet i accelerating to the right not up and down, o the net force on the cabinet ut be zero. hat ean that part of the applied force add with the weight. he noral force ha to upport both the weight of the box and the force downward that the peron i appling. Regardle of the exact aount the applied force contribution downward, the noral force will alwa be greater than the weight. he equation for ewton econd law in the -direction i = g Ain 0 = 0 which allow one to olve for the noral force a = g+ A in 0. he iportance of the noral force will oon becoe apparent a we talk about friction. he equation for ewton econd law in the x-direction i x = Aco0 A co0 = ax for which one can olve for the acceleration ax =. Σ x = Acoθ = ax A coθ ax = = = kg ( 5 ) co0 0.87 ( 5 ) You r How would the proble change if the force were applied 0 above the horizontal. (It wouldn t change our anwer for acceleration at all in the preent proble. It would change it quite a bit if we were conidering friction becaue it would greatl affect the noral force.)

Ex. 1-: An elevator i being planned for a building. If the elevator ha a a of 1.50 x 10 what tenion ut the upporting cable be able to withtand? Draw the force acting on the elevator. kg, + W = 0 = W = g = a = (1.50 x10 kg)(9.80 ) = 1.7 x10 he elevator deigner forgot that the plan pecified that the elevator would experience a axiu upward acceleration of.50 / and intalled a cable that could upport.00 x 10 without breaking. I the elevator afe? o deterine whether the elevator i afe, we need to find out what tenion the cable will experience when accelerating. If we re-draw our diagra, thi tie including the acceleration in the picture, we find that when we appl ewton econd law, we have a light change. = a W = a = W + a = g ( + a) = (1.5 x10 kg) g a =.5 / ( 9.80.50 ) + =.00 x10 +x 1.50 x 10 kg g + +x You r What acceleration would the elevator have to experience in order for the cable to break? Recall that the cable can handle.00 x 10. Anwer: 10. / You r What tenion would the cable experience if the elevator were being accelerated at.50 / in the negative direction (i.e. down?) Hint: draw the diagra and ake ure ou get the right ign on the acceleration) Anwer: 9.5 x 10

You r A hopper puhe a 7.5-kg hopping cart up a 1 incline, a hown below. ind the agnitude of the horizontal force needed to give the cart an acceleration of 1.1 /? x = coθ ginθ = a (7.5 kg) 1.1 + ( 9.81 ) in1 a ( + gin θ ) = = = coθ co1 8

5