Dynamics 4600:203 Homework 08 Due: March 28, Solution: We identify the displacements of the blocks A and B with the coordinates x and y,

Transcription

1 Dynamcs 46:23 Homework 8 Due: March 28, 28 Name: Please denote your answers clearly,.e., box n, star, etc., and wrte neatly. There are no ponts for small, messy, unreadable work... please use lots of paper. Problem : Hbbeler, 3 36 Determne the acceleraton of block A when the system s released. The coeffcent of knetc frcton and the weht of each block are ndcated. Nelect the mass of the pulleys and cord. We dentfy the dsplacements of the blocks A and B wth the coordnates x and y, toether wth the drectons (î,ĵ) and (ê,ê 2 ). These drectons are related as î = cos ψ ê sn ψ ê 2, ĵ = sn ψ ê cos ψ ê 2. The acceleraton of each block s 2 T ê T ĵ a A = ẍ ê, a B = ÿ ĵ, f r ê and these coordnates are related as 2x = y. Wth the tenson n the cable defned as T, the free-body darams for each block are shown to the rht. W A ĵ N ê 2 W B ĵ Therefore, lnear momentum balance, = m a, on each block yelds WA 2T ê + f r ê + N ê 2 W A ĵ = ẍ ê, WB T ĵ W B ĵ = ÿ ĵ. Elmnatn the unknown tenson T and ÿ from these equatons yelds WA + 4W B 2W B W A snψ + f r = ẍ, N = W A cos ψ. If the system s released from rest, block A tends to slp down down the plane, so that f r = +µ k N, and fnally the acceleraton of the block s ẍ = W A (sn ψ µ k cos ψ) 2W B W A + 4W B.

2 Wth the parameters ven n the problem, ẍ = 4.28ft/s 2 so that the acceleraton of block A s a A = ẍê = W A (sn ψ µ k cos ψ) 2W B W A + 4W B ê = ( 4.28ft/s 2) ê Problem 2: Hbbeler, 3 4 A parachutst havn a mass m opens hs parachute from an at-rest poston at a very hh alttude. If the atmospherc dra resstance s D = k v 2, where k s a constant, determne hs velocty when he has fallen for a tme t. What s hs velocty when he lands on the round? Ths velocty s referred to as the termnal velocty, whch s found by lettn the tme of fall t. Wth the acceleraton of the parachutst ven as a P = v ĵ, lnear momentum balance leads to = k v2 ĵ m ĵ = m v ĵ = m a P. The resultn dfferental equaton that overns the moton of the parachutst s smply v = k m v2, whch can be nterated as m 2 m 2 k v(t) [ ln [ m + k v + v(t) v() dv k m v2 = t ] dv = t, m k v ( m ) ( m )] + k v ln k v = t. dt, Ths may be solved for v as m v(t) = k e (2 k/m) t. e (2 k/m) t + As t, the termnal velocty s found to be v = m/k. Problem 3: Hbbeler, 3 48 Block B has a mass m and s hosted usn the cord and pulley system shown. Determne he mantude of force as a functon of the block s vertcal poston y so that when s appled the block rses wth a constant acceleraton a B. Nelect the mass of the cord and the pulleys. 2

3 Problem 4: Hbbeler, 4 9 When the drver apples the brakes of a lht truck traveln 4km/h, t skds 3m before stoppn. How far wll the truck skd f t s traveln 8km/h when the brakes are appled? If the truck s skddn to the rht wth velocty v T = ẋî, the resultn sldn frcton force s ven as r = µm sn(ẋ)î, the work done by ths force throuh a dstance d s W = d µm dx = µm d. Therefore f t skds d = 3m when traveln wth ntal speed v = 4km/h, work-enery can be wrtten as W = µm d = m 2 v2 = T. Therefore the unknown coeffcent of frcton can be solved as µ = v2 2 d. nally, f the truck s traveln at speed v, wth the above coeffcent of frcton t sldes a dstance d wth d = v2 2µ = dv2 v 2 or v = 8km/h, the truck sldes a dstance d = 2m. When you double your speed, you ncreased the dstance requred to skd to a stop by a factor of four. Drve safe. Problem 5: Hbbeler, 4 4 Determne the velocty of the 6lb block A f the two blocks are released from rest and the 4lb block B moves 2ft up the nclne. The coeffcent of knetc frcton between both blocks and the nclned plane s µ k =.. We dentfy the dsplacements of the blocks A and B relatve to the corner as x and y respectvely, toether wth the drectons (î,ĵ) and (ê,ê 2 ). 3

4 These drectons, Wth ψ = 3, are related as 2 T ê 2 W B ĵ î = cos ψ ê + snψ ê 2, T ê ĵ = sn ψ ê + cos ψ ê 2. The acceleraton of each block s f A ê 2 a A = ẍê 2, a B = ÿ ê, and these coordnates are related as 2x + y =. Wth the tenson n the cable defned as T, the free-body darams for each block are shown to the rht. W A ĵ N A ê N B ê 2 f B ê Therefore, the total force actn on each block s A = ( N A )ê + (2T f A )ê 2 W A ĵ, = ( N A + W A snψ) ê + (2T f A W A cos ψ) ê 2, B = ( T + f B )ê + (N B )ê 2 W B ĵ, = ( T + f B + W B snψ) ê + (N B W B cos ψ) ê 2. If block B moves up the plane, then block A moves down the plane so that ẋ > and ẏ <, and the frcton forces may be wrtten as f A = µ k N A = µ k W A sn ψ, f B = µ k N B = µ k W B cos ψ. If the dsplacement of B s y f y = d = 2ft then the dsplacmeent of A s x f x = d/2 = ft, and the total work s W f = = ( A v A + ) B v B dt, ( 2T + µ k W A snψ W A cos ψ) dx + ( T + µ k W B cos ψ + W B sn ψ) dy, = d 2 (2T + W A(cos ψ µ k sn ψ)) + d ( T + W B (sn ψ + µ k cos ψ)), ( = d W B (sn ψ + µ k cos ψ) W ) A 2 (cos ψ µ k snψ). The knetc enery of the system, n terms of the velocty of A, s ven as T = W A 2 v A 2 + W B 2 v B 2 = W A 2 ẋ2 + W B 2 ẏ2, = (W A + 4W B ) ẋ2 2. nally, solvn for ẏ usn work-enery, we fnd that, when the system s released from rest ẏ = d 2W B (sn ψ + µ k cos ψ) W A (cos ψ µ k snψ). 4W A + W B 4

5 Wth the ven parameters and dsplacements, the work s W = 2.3lbft, and the knetc enery s T = 3.4sl ẋ 2, so that the speed of block A s ẋ =.77ft/s. The velocty of ths block s therefore v A = ẋê 2 = (.77ft/s)ê 2. Problem 6: Hbbeler, 4 8 The collar has a mass of 2k and rests on the smooth rod. Two sprns are attached to t and the ends of the rod as shown. Each sprn has an uncompressed lenth of m. If the collar s dsplaced s =.5m and released from rest, determne ts velocty at the nstant t returns to pont s =. 5