Question

Any notes about the steps helps me learn. also the formulas posted before the numbers are plugged in will be greatly appreciated

Question 8 --/10 View Policies Current Attempt in Progress Block B in the figure weighs 701 N. The coefficient of static friction between block and table is 0.22; angle is 29°; assume that the cord between B and the knot is horizontal. Find the maximum weight of block A for which the system will be stationary. Knot- Number Units

Answer 1

Weight of block B = 701N, "=0.22 free body diagram for stationary condition of B NWB=701N tUN, = 0.22*701 t=154.22 N free body diagram of A and To= f = 154.22 N TA = WA for knot Tsino so=29 for stationary condition Toso = To and Tsino = Ta dividing Teose - TB > to the = WA = MWB coto plug in the values UNB ll WB WA WA We = 0.22x701 cotat = TWA = 85.48N