Question

PROBLEM 5 The weights of block A and cylinder B are W4 = 250 lb and W3 = 100 lb. If the coefficient of friction is 0.35 for all surfaces, determine the maximum force P required for equilibrium. 30°

Answer 1

solution: for maximum load I force Load / force P : • free body diagram of the block and cylinder: ! FAB WB Caso WACOSO NAB FAB FSB a NAB ISA NA No Wasino Wo sinot 닉 force balance in balance in vertical direction to the plan for A: =0 NA-WA CosO+ PAB where, NA-950 (330 + 0.35 NAB =0 friction force b/w A and B surfaces is LAB = - 4 NAB FAB = 0.35 NAB NA = 216.51 +0:35 NAB NA = 216.51 -0.35 NAB CS$canned with CamScanner

(where, NAL Su NAB: Normal Reaction force blue A and B surface force balance in (or) along the horizontal direction to the plan Plan ferr A : P - WA sino - NAB-ISA 0 P - 250 x sin 30 - NAB – 0.35NA-0 fs, a friction force bfw Block A and inclined surface fs, A = M NA P-125- NAB -0.358/216.51 -0.35 NAB)=0 P = 200.78 +0.878 NAB 3-0 fs, A = 0.35X NA - - force balance in perpendicular to the Plan for B: NB - WB cos OfAB =0 FAB = NAB NB - 10x (0530 10 (030 - MANAB 0 LAB 20:35 NAB [No 86.6 + 0.35 NAB] where, NAI NB : Normal Reaction force How blocks and inclined surface blacks } - force balance along the plane for B: 0 fs, B = 4 NB 20:35 *NB NAB - WB sino + fs, B NAB - 1 ro XSin 30 +0.35% (NB)=0 NAB 50 +0.35% ( 86.6 +0.35 NAB) –0 17.54 16 IN AB = CSScanned with CamScanner

→ NAB value in eq O: - pa 200.78 + 0.878x (17.54) p = 216.18 la. Ang Maximum force for equlibrium CSScanned with CamScanner