Homework#4 Solutions

Homework#4 Solutions - Problem] 824 Suppose that you want...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Problem] 824 Suppose that you want the bar of length L to act as a simple brake that will allow the workpiece A to slide to the left but will not allow it to slide to the right no matter how large a horizontal force is applied to it. The weight of the bar is H", and the coefficient of static friction between it and the workpiece A is #5. You can neglect friction between the workpiece and the surface it rests on. Ca] What is the largest angle a for which the bar will prevent the workpiece from moving to the right? (b) If cr has the value determined in (a), what horizontal force is required to start the workpiece A toward the left at a constant rate? Solution: [a] To resist motion to the right no matter how large the horizontal force requires a ver}r large friction force. For impending motion to the right. the sum of the moments about the bar hinge is EM 2 +%Lsino — FNLsmo + choer = 0, where FN is the normal force and f is the friction force resisting the impending motion. Noting that f = grgFN. the sum of moments yields W sin Ct FN= manor—its coscq If the denominator vanishes. the normal force and hence the fric- tion force become as large as required. Thus sin a —fit, one a = 0. from which g; = 12.31th.01' cs = tan—lute) [bl For impending motion to the left. the sum of moments about the bar hinge is EM 2 +w [3)5ino —FNLsino— choscr:U__ where the friction force opposes the impending motion. Noting FN = then W sino _ W sinct _ gs}? 2(sinct+msa) _ 4005c? _ 4 5'1: f: is the force required to start motion to the left. Problem 8.46 To obtain a preliminary evaluation of the stahilir)r ol‘a turning car, imagine subjecting the sta- tionary car to an increasing lateral force F at the height of its center of mass, and determine whether the car will slip {skid} laterally before it tips over. Show that this will be the case if blr'h :=- 2:45. l[Notice the importance of the height of the center of mass relative to the width of the car. This reflects on recent discussions ot'the stability of sport utility,r vehicles and vans that have relatively high centers of mass.) Solution: EQUILIBRIUM Eons: 2F}: F—fL—fR=D 2F”: Elli}: Assume skid and tip simultaneously. N1, + NR — mg 2 U —hF + 6N3 — 3mg 2 0 fit =I—‘5NL~ ffl Zfilsll‘l'fliskidll and My :0 (tip). fL = 3- fa =irsm9. The equilibrium eqns become F = fa =irsNR =irsm9 and the moment eqn. uses 6: —la|:}.tsmg] + bfimg) — 51019 = U or f = Etta For TE- I::v 2H5 . slip before tip -::2 2st“. tip before slip big N low cm. relative to track width Julo-IJ'10-IJ'10-I small N high cm. relative to track width Problem 8.74 The carjack is operated by turning the threaded shaft at A. The threaded shaft fits into a mating groove in the collar at B, causing the collar to move relative to the shaft as the shaft turns. As a result, points 3 and D move closer together or farther apart, causing point C“ [where the jack is in contact with the car} to move up or down. The pitch of the threaded shaft is p = 5 mm, the mean radius of the thread is r = ll] mm, and the coefficient of kinetic friction between the thread and the mating groove is 0.15. What couple is necessaryr to turn the shaft at a constant rate and raise thejack when it is in the position shown ifF = 6.13 kN‘? Solution: Isolate members BC." and ED. Assume that half the car load is carried by these members. The equilibrium conditions for member BC are: Zr: =cI—BI :0. F 2M3 = 0.3 (E) — 0.150, = 0. These equations are solved for E zfizaasm 2 2 to obtain B: = 6.5 ML which is the force on the collar to be balanced by the rotating threaded shaft. The pitch angle is 5 _ —1 _ o ct—Lan (2fi10)_4.55. The angle of kinetic friction is 9, = tail—1(U.15)= 3.530. The moment required to rotate the shaft at a constant rate is M = (0.01)(6.5)tan(8.53° + 4.55“) = 0.0151kN m = 15.1 N m 300 mm -—3 DD mat—— Problem 8.84 The mass of the suspended object is 4 kg. The pttlle}r has a IOU—mm radius and is rigidly,r attached to a horizontal shaft supported by journal hear- 25?" ings. The radius ofthe horizontal shaft is 10 mm and the coefficient of kinetic friction between the shaft and the ‘5 hearings is 0.26. What tension must the person exert on 100 mm the hope to raise the load at a constant rate? Solution: H : 0.1m fill, = 0.26 ShaftradjusOD] in grkfshaftj = [3.26 tanlfik= 2m. 9;, =14.5?° .11, = rFsinB;r m=4k3 R:C|.lm NK=U26 To Find F. we nrust find the forces acting on the shaft. ZFI: OJ. —Tc0525°=O (I) Shaft radius 0.0] m K (shaft) : 0.26 Zr}: 09 —Tsmas°—mg=n(2;t F = fog + 03 (3;. 2 M0: at" — ngl — M, = o (4;. M, firms. (5;. Unknowns: 0,: 03,. T. £11,, F Solving. we get T = 40.914 Also. F =8'FGN. M, 20.1’TON-m O: =3?.1N. o, = 56.5 N Problem 8.91) The shaft is supported by thrust bear— ings that subject it to an axial load of 800 N. The coef— ficients of kinetic friction between the shaft and the left and right bearings are 0.20 and 0.26, respectively. lWhat couple is required to rotate the shaft at a constant rate? Solution: The Iq‘rbenring: The parameters are r0 = 38 rum. 1”; =0. e: 245°. in, 20.2. and F =SDCIN. The moment required to sustain a constant rate of rotation is 2}.th J‘g — 3"? 3COSQ r3 —r-l2 Mk”, 2 ) 25.73Nm. The right hearing: This is a fiat-end hearing. The parameters are pig 2 0.26. r = 15 mm. and F = 800 N. The moment required to sustain a constant rate of rotation is .trfight = 2mg“ = new m. The sum ofthe moments: AI = 5.?3 + ECG 2 ?.81 N rn Problem 8.99 The couple required to turn the wheel of the exercise bicycle is adjusted by changing the weight H". The coefficient oi'kinetic friction hetween the Wheel and the belt is Ink. Assume the wheel turns clockwise. Ca] Show that the couple :‘lf required to turn the wheel is M = n-‘Rn _ 9-3-4“). {b} If H" = 40 lb and Ink 2 0.2, what iorce will the scale S indicate when the bicycle is in use? Solution: Let :3 he the angle in radians of the belt contact with wheel. The tension in the top belt when the belt slips is Tapper = We—“k'g. The tensioninthelowerbeit isle“ = TV. The moment applied to the wheel is M = REED,” —Tuppfl.) = RWU — 9—"k’9j. This is the moment required to turn the wheel at aoonstant rate. The angle .6 in radians is 7r 180 from which M = Hl’lr’il — e‘a'it‘k 1]. [hi The upper belt tension is :3 = 7r + (30 —15;. 1: ) =3_4o radians. Twp" 2409'3'4m'2l = 20.28113. This is also the reading of the scale 5. ...
View Full Document

This note was uploaded on 11/29/2010 for the course MECH&AE 101 taught by Professor Mal during the Fall '08 term at UCLA.

Page1 / 5

Homework#4 Solutions - Problem] 824 Suppose that you want...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online