Question

2013 Michael Swanbom CCOO abcd hi ha 2.44 in 4.6 in 0.74 in 14.7 in 4.11 in 9.4 in A certain type of piping system uses deformable rings to clamp pipes onto fittings. A crimping tool that is used to make these connections is shown here. Find the forces indicated in the table below when a force of F = 52 lb is applied to the handles as shown. Assume that the crimp ring exerts purely vertical forces on the tool's jaws. Forces: (Denote rightward or upward forces with positive values; leftward or downward forces with negative values.) This body exerts forces of -> Fhorizontal Fvertical on this body crimp ring AB AB AD BE CD be sure to include units with your answers

Answer 1

2.44 in 4.6 in 0.74 in 14 7 in | 4'll in | 914 in Tb Given F=5216. The force exerted by the crimp ring is purely vertical as given For EM,=0 - Fox hy - Foyc - Fd =0. - For hi/2 + Foy C + F2 = 0 . ---(1). Foy Fox L Fox =-FER Foy = - FEY > - Eme=0 Fer h1/2 - FEY C + F2 =0 - Foxhila + Foy C + Fd=0 --- (2) Eq. (1) - Eq (2) gives Fox=0 So, Foy = - Ed = – 526147 - 0.74 = -1032.97 lb

Rese= 0 cy Rey= F - Ed = F(c-d) =_52×(14.700.74). 0.74 = -980.97 16. 77 TAN EMAFO >-f, a + Ed b=0 52 x 14.7X46 2.44 x 0.74 1947•4 I lb. - 4, + Fay + F d = 0 - Fay = -f, - Ed - Fdb - Ed - Red (67) 52x1417X(4.6+2.44) 16 2.448 0.74 - - 2980.38 (b. 1 on this body AD This body exerts Fharrison vertical forces erimp ring 0 lb 1947.4116 AB - 2980:38 L1 AB 12980.38 lb AD 1-1032.97 lb 1 1032.97 lb 0 Lb. 1980.9716 mo BE CD CE