Question

letter b

DETERMINE THE FRICTION FORCE BETWEEN THE Book SND THE ROUGH FLOOR. PREPORE 1 UNIFORon THE REQUIRED PRO'S. 700 N E 20° 70.9 To.sm UN mach=100kg F.H.N 1.6m * 0.33 Mica 0.21 b) Hock iP THE mock tipsovER.

Answer 1

TOON=F - Giver : - - Ioam. F = 700N mass-m=lookg 4s=stallo friction co-eftfcient Ys = 0.33 44 kinelfe friction coefficient A • 2 a M = 0.21 y=23141N 700N Take g = 9.810/22 Ff = 4 N = Force of trickson for equilibrum; {fy= 0 (Atre) u. Fig &N=W=0 20 - 657.79N 4 ... 23141 +-981=0 W = mg= 9814 Freed Body Diagram TooN = 749.59] Is N= 0.33 * 749.59 =247.37N But (Fix) > (FI)s i. Block will move in the direction of Net Fx C Also frictional to Force, acting on block will be Kinetfe füctional force (F) k. . 0.21 * 749.59 = 157.414N . Fly - 231.414700 N=F1 . Required FBD is Note: line of action of Wand N is some. Fra= 657.79N 109 m A +0.8m 157.41 N-749 59N. V The acceleration, an W=981N - Free Body Dragram In Equilibrium; Efx = max ( the)

o Fre -(F) k=mae. 657.79 – 157.41 = 1000x oo ax= 5.0038 m/s2 6 To check if o for point A the block tipsover. - 6 tve) Moment about A MA=(fy) #106+ (Fix) * 0.9 +W+0.8-N#08 2. A -231-41*1.6 + 657.79 of +981703-7 MA = 406.88 Nm het moment about A is los clockwise directson. So block with not tipsover cet point A. for point B :- @the) e oment about pt. B, MB= 657.79*0.9 + 749.59+08 - 981*0.8 2.MB- 406.88 Nm. Since, Net moment about B is in clockwise din 406.88 Nm Block B will tipsover at point B.
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