Question

PLEASE SHOW ALL WORK STEP-BY-STEP AS MUCH AS POSSIBLE

0.75 in 0.75 in. 8.71 A wedge A of negligible weight is to be driven between two 100- ib blocks B and C resting on a horizontal surface. Knowing that the coeffi- cient of static friction at all surfaces of contact is 0.35, determine the smallest force P required to start moving the wedge (a) if the blocks are equally free to move, (b) if block C is securely bolted to the horizontal surface. 8.72. A wedge A of negligible weight is to be driven between two 100- lb blocks B and C. Knowing that the coefficient of static friction is 0.35 between the blocks and the horizontal surface and zero between the wedge and each of the blocks, determine the smallest force P required to start moving the wedge (a) if the blocks are equally free to move, (b) iſ block C is securely bolted to the horizontal surface. Fig. P8.71 and P8.72

Answer 1

solution: coledge) FBD Wedge: O= tans (0.75) = 10.620 +0.75in x 0.75 in. For block Bi sin 100 lb fz/ºs. 100lb Ps= tantus - fan 0.35 s = 19.29° By geometry: El. = 100 sinds sin40.8 (force triangle) =) fi= 100 x sin 19.290 40.8° floop sin 40.8 fr (Block B fi= 50.56 lb Due to symmetry (f = Fg) 3 -. By geometry calculate P: Parisin Corps) +F3 sin (0+%s) = 2f, Cords) p=2x50:56 sin (10.62°+ 19.290) ca IP=50:42 lb | Aus (6) When Block e is securely bolted: FBD will be same canchanged) .: smallest force required to schood start moving the wedge if cis bolted to horizontal P = 50.42 db Aus M

Plz Rate!