MP HW 6 1 FBD of block: 2- P T 1-5 WE FBD of man: -P Wat F FB Nf : Mp = 0.3 EFx= P -F1 = 0 EFy = Nf - W = 0 E (Mo)} = 450XR = 1.5P EFx= FB - P = 0 EFy = WB - Ng = 0 [ (Mo) = 200xm= 1-5P We also know, Ff = MANA : F1 = 0.3 NF Fridge Boy No : MB = 0.6 N= = 450 Pp= 135 Ng = 200 PB = 120 => F = 0.3 (450) = 135 N FB= MB NB FB = 0.6 NB => FB = 0.6 (200) = 120 N (Fm) max= 120 lb F> (Fm)max force that moves man to the right > Frictional force 135 > 120 the man would slip.
2 Az F 1 Ay -3 W ÃŽ NB > FB No 3NB = 1.5 W 1 P 3Ng = 1.5 (200x 9.81) + : No = 981 N 0.75 L T Wp Fc Nc=117.2-0.6P 961.21 No : FB = 0.342P = 0.342 (360) = 123.12 N P= 2.1875Fc : Fc = 164.6 N FB = UNB >MB = 0126 Fc = 1 Nc => Mc = 0.171
3 FBD of B: FBD of A: P A B P 10" uNa UND Na 1 1 Nb k = 15 KN/m Block A : ØNB = 10° EFy : Fs = Na : Na = 1425 N F =Kx 95 = 15 KN/m × 1000 = 1425 N Block B: EFy: 1425 + 0.35 No sin10 - No COS10 = 0 => No [0:35 sin10 - cos10] = - 1425 : No = 1542.2 N EFx: P - NosinIO - UNa - UND cos10 =0 .P= 1298.097 6 Fc NC