Question

Chapter 6, Problem 6/101 Multistep Determine the range of mass m2 for which the system is in equilibrium. The coefficient of static friction between the block and the incline is H1 = 0.37 and that between the cord and the fixed disk on the block is u2 = 0.20. M2 mi 15 Mi m2

Answer 1

we know that if the system is in equilibrium it won't that mone. we can determine what forces (mass ma) is equal to move the block up the inclined and down the incline and the body will be in the equilibrium over the range of these two Nalues. Let's first draw the FBD of the block in movement down the lincline. Sime the block more down the incline, we can conclude that tension Ti is grates than Ti = Temo initiating Te B X. V IN MN we can see from sketch above that angle ß is B =90-15=75 = 1.3089 rad. M2=0.20 " = C, ? e-2017-308 7 T, = 1.299 m. g By equation equating, sum of fancer im y direction to zero will determine noormal reaction farce N. e Fy=0 N-cocosis-ta Cosis=0 N = co, costret Ta Cosir Replace Ty by wr W=(wirus) Cosis - (mitm) og los 15

By equating sum of forces inxdirection to reo will determine mass me required to mone the block downwards. & fge=0 MN-T2 sim 15-w. Simis+T, = 0 0.37 com tag Cosis - megsimis-migsimis +1.29 amaz lets divide both side by g 0.37 (m, tm.) cos 15-mesimis-m, simit + 1.299m2 = 0 0.357 m, +0.357 m2-0.2588 m₂ - 0-2588m, +1.299 m2=0 0.0982 m. +1.3972 M = 0 m2 - -0.0982 m, = -0.0702m, 1.3972 Next Lots first draw the FBD of the block initiating movement up the incline. Since the block move down the incline, we can conclude that hension Ti is less then Tz, so write extression fas T T- emo Ti V N MN T2

We can see from the sketch above that angle & is P = 90-15 = 75 = 1.3089 M₂ = 0.20 Taong By equating Ti= mag 0.20x1.3089 T1=0.7696 mag E fy=0 N- ww, Cosis-T2 Cos=0 N = wiCosit +T2 Cos 15. N= (wit W₂) Cosir Cometnam) a Cosis Sum of farces inx direction to zero will determine mass me reg to the block upwards. Efx=0 - M,N-Tesimet - WisinistTi =0 0.37 (mptoma)g Cosis-magsimis -migsinir + -0.78 96 mg -0.35m 0.357m2 0.25 88 m₂ - 0.2988 m to.78 96m=0 0-6158 m +0.15 88 m2 = 0 0.6158 mi = 4.0039 mi mone 0.1538 Range 0.07025 m2 a 4.003am,