UCSD Phys 4A Intro Mechanics Winter 2016 Ch 4 Solutions

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1 USD Phys 4 Intro Mechnics Winter 06 h 4 Solutions 0. () he 0.0 k box restin on the tble hs the free-body dir shown. Its weiht 0.0 k 9.80 s 96 N. Since the box is t rest, the net force on is the box ust be 0, nd so the norl force ust lso be 96 N. (b) ree-body dirs re shown for both boxes. is the force on box (the top box) due to box (the botto box), nd is the norl force on box. is the force on box due to box, nd hs the se nitude s by Newton s third lw. is the force of the tble on box. ht is the norl N force on box. Since both boxes re t rest, the net force on ech box ust be 0. Write Newton s second lw in the verticl direction for ech box, tkin the upwrd direction to be positive. 0 N 0.0 k 9.80 s 98.0 N N 0 N N 98.0 N 0.0 k 9.80 s 94 N 3. hoose up to be the positive direction. Write Newton s second lw for the verticl direction, nd solve for the ccelertion. 63 N 4.0 k 9.80 s 4.0 k.8 s Since the ccelertion is positive, the bucket hs n upwrd ccelertion. N N op box (#) otto (#). () Since the rocket is exertin downwrd force on the ses, the ses will exert n upwrd force on the rocket, typiclly clled the thrust. he free-body dir for the rocket shows two forces the thrust nd the weiht. Newton s second lw cn be used to find the ccelertion of the rocket k N.750 k 9.80 s s 3. s (b) he velocity cn be found fro Eq. -. v v t s 8.0s 4.87 s 5 s (c) he tie to rech displceent of 9500 cn be found fro Eq. -b. x x x x v t t t 78s s N box

2 7. ree-body dirs for the box nd the weiht re shown below. he tension exerts the se nitude of force on both objects. () If the weiht of the hnin weiht is less thn the weiht of the box, the objects will not ove, nd the tension will be the se s the weiht of the hnin weiht. he ccelertion of the box will lso be zero, nd so the su of the forces on it will be zero. or the box, N 30.0 N 47.0 N N N (b) he se nlysis s for prt () pplies here N 60.0 N 7.0 N N (c) Since the hnin weiht hs ore weiht thn the box on the tble, the box on the tble will be lifted up off the tble, nd norl force of the tble on the box will be 0 N. 33. We drw free-body dirs for ech bucket. () Since the buckets re t rest, their ccelertion is 0. Write Newton s second lw for ech bucket, cllin UP the positive direction k 9.80 s 3N 0 3.k 9.80 s 63N N op (# ) otto (# ) (b) Now repet the nlysis, but with non-zero ccelertion. he free-body dirs re unchned. 3. k 9.80 s.5 s N 35 N 7N 37. he net force in ech cse is found by vector ddition with coponents. 0. N 6.0 N () x y N tn he ctul nle fro the x-xis is then hus the net force is 9.0 N t N.03 s t k net (b) o o cos N sin N x y N 0.9 N 4.03N 4.0 N N tn s t k net 30 o

3 44. or ech object, we hve the free-body dir shown, ssuin tht the strin doesn t brek. Newton s second lw is used to et n expression for the tension. Since the strin broke for the.0 k ss, we know tht the required tension to ccelerte tht ss ws ore thn. N. Likewise, since the strin didn t brek for the.05 k ss, we know tht the required tension to ccelerte tht ss ws less thn. N. hese reltionships cn be used to et the rne of ccelertions. x x.0 ;.05 ; x x N. N 9.80 s 9.80 s.0 k.05 k x x s.03 s 0.8 s.0 s 54. We drw free-body dir for ech ss. We choose UP to be the positive direction. he tension force in the cord is found fro nlyzin the two hnin sses. Notice tht the se tension force is pplied to ech ss. Write Newton s second lw for ech of the sses. Since the sses re joined toether by the cord, their ccelertions will hve the se nitude but opposite directions. hus. Substitute this into the force expressions nd solve for the tension force. pply Newton s second lw to the sttionry pulley k.k 9.80 s 4.4 k 0 34N. k 3. k

4 56. ecuse the pulleys re ssless, the net force on the ust be 0. ecuse the cords re ssless, the tension will be the se t both ends of the cords. Use the free-body dirs to write Newton s second lw for ech ss. We re usin the se pproch tken in proble 47, where we tke the direction of ccelertion to be positive in the direction of otion of the object. We ssue tht is fllin, is fllin reltive to its pulley, nd is risin reltive to its pulley. lso note tht if the ccelertion of reltive to the pulley bove it is R, then. hen, the R ccelertion of is, since is in the opposite R direction of. : R : R : pulley: 0 Re-write this syste s three equtions in three unknowns,,. R R R R R his syste now needs to be solved. One ethod to solve syste of liner equtions is by deterinnts. We show tht for Siilr nipultions ive the followin results.

5 R 4 4 ; 4 () he ccelertions of the three sses re found below. 4 R R (b) he tensions re shown below. 4 8 ; () free-body dir is shown for ech block. We define the positive x-direction for to be up its incline, nd the positive x-direction for to be down its incline. With tht definition the sses will both hve the se ccelertion. Write Newton s second lw for ech body in the x direction, nd cobine those equtions to find the ccelertion. : sin x : sin dd these two equtions x sin sin sin sin (b) or the syste to be t rest, the ccelertion ust be 0. sin sin 0 sin sin sin sin k 6.8k sin sin 3 he tension cn be found fro one of the Newton s second lw expression fro prt (). : sin 0 sin 5.0k 9.80 s sin3 6N (c) s in prt (b), the ccelertion will be 0 for constnt velocity in either direction. sin sin 0 sin sin sin sin sin sin 3 y N- x N- y x

6 79. () We ssue tht the xiu horizontl force occurs when the trin is ovin very slowly, nd so the ir resistnce is neliible. hus the xiu ccelertion is iven by the followin N x x s 0.6 s k (b) t top speed, we ssue tht the trin is ovin t constnt velocity. herefore the net force on the trin is 0, nd so the ir resistnce nd friction forces toether ust be of the se nitude s the horizontl pushin force, which is N.