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HW22FrictionSoln - Problem 6/5 A 10.2—kg block is place...

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Unformatted text preview: Problem 6/5 A 10.2—kg block is place on the inclined plane. The coefficient of static friction is 0.9. Find the magnitude of the total force exerted by the inclined plane onto the block. Answer: 100.1 N (up). IZOOmm 9‘53qu 6.3m librium. ._ 4. o as: tan ‘ “fir 13.4—3 {FM-:0: ~F+ £0.2(fi.‘3l)sin 9‘«'10) F“: 3L0 N Zita-=0: Ne mafia) case-x M} N“ ‘VH N Fma #5“ 2 cm ($440: 85.4 N > F=3LL N Problem 6/6 The uniform block weighs 100 N. The block is twice as wide as it is tall. The coefficient of static friction between the block and the floor is 0.5. A force is applied to the corner as shown. What is the magnitude of this force that puts the block on the verge of moving? Will the block tip or slide? Answer: 44.8 N :Fkflo: “F3“Fc+?c°$3°° ”“0 M 2:53 “5‘0" NB+NL+?$&1 30°wmq==o (2) With FawfisNa é Fem/16M TNM‘Q‘ 0) it». +0 attain 1" == am N355 30° 1" 0.0530" th‘x jug-‘05, “Pr: "PS a 0.4mm Ii {9 ', m % 1° 2.: ~-a---=3—————— = osaamj-at’» Cos 30°+Zsin 30° ' t'P For these. conditéns) 511%?53 WW“ char “first. Problem 6/7 The light bar is used to support the 50 kg block in its vertical guides. The coefficient of static friction is 0.3 at the upper end of the bar and 0.4 at the lower end of the bar. (a) Find the friction force acting at each end for x = 75 mm. (b) Find the maximum value of x for which the block will not slip. Answers: FA = FB = 126.6 N, xmax = 86.2 mm :5er 96: 75" mm; 6: 5x5£= [4.5" Fr/é‘fzzin any/6 96: kfl"}‘¢5 ,6.» A ,3 g = fin“’0.40= 21:5” u a ,, 9643 == fien'J0Ja: 16.7“ 515:: a *4: a 5! d5 : (50" «faves 1147‘ 5/1,} a‘ gzga» Wfané. a Jb/fiQI)/an /4.5a :: 126.6 N ‘W’ For Increased )C beer 54,95 fzks/ at 3 M i m - c? v. f . .70 wa‘h a... 9’3” 2’6. 7 . 77m!) xmax" 300 3m /6 3 86.12 Mm Problem 6/14 A window sash weighting 10 pounds is normally supported by I 36 m. I two 5-lb sash weights. Knowing that the window remains I open after one sash cord has broken, determine the smallest possible value of the coefficient of static friction. (Assume that the sash is slightly smaller than the frame and will bind only at points A and D.) Answer: us = 0.750 FBD window: ' T = 5 1b ”214; =0 NA ~ND =0 NA =ND Impending motion: FA = ,uSNA FD =flsND CZMD =0: (18 in.)W—(27 in.)]\£1 —(36in.)FA =0 W=101b W=%NA +2,uSNA 2W 3 + 4/15 A = ‘ 121+; =0: FA=W+T+FD =0 NOW FA+FD=flS(NA+ND) =2flsNA W 2W _._—.2 Then 2 'uSB+4,ua or ,us =0.750 4 ...
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