2—6.If the handle of a jackscrew isturned 16 complete revolutions toraise the jack 2 inches, the pitchof the screw is1.1/32 in2.1/16 in3.1/8 in4.1/4 in2–7.You are pulling a 21–inch lever toturn a jackscrew having a pitch of3/16 inch.The theoreticalmechanical advantage of thejackscrew is about1.1,0002.7003.4004.1002–8.A jackscrew has a handle 35 incheslong and a pitch of 7/32 inch. Ifa pull of 15 pounds is required atthe end of the handle to lift a3,000–pound load, the forceexpended in overcoming friction is1.12 lb2.9 lb3.3 lb4.5 lb2-9.Refer to textbook figure 5–3.Howmany complete turns of the thimbleare required to increase theopening of the micrometer by 1/4inch?1. 252. 103. 54. 4When answering items 2–10 and 2–11,orefer to textbook figure 5–4.2–10.If the micrometer’s thimble isturned exactly five completerevolutions,the new reading is2–11.Assume that the graduation mark 5on the thimble is opposite point X.How much farther do you open themicrometer in turning the thimbleuntil the graduation mark 15 isopposite the point for the firsttime?1.0.125 in2.0.010 in3.0.0125 in4.0.0010 in2–12.How do you find the actualmechanical advantage that ajackscrew provides in lifting aload?1.Multiply the length of the jackhandle by the radius of thescrew and then divide by thepitch of the screw2.Divide the load by the amountof effort required to lift theload3.Multiply the length of the jackhandle by 2 and then divide bythe pitch of the screw4.Divide the distance the screwtravels by the number of turnsit makes and then subtract theamount of frictional resistance2–13.If a jackscrew has a pitch of 5/32inch, the length of the handlerequired to obtain a theoreticalmechanical advantage of 800 isabout1.30 in2.25 in3.20 in4.15 in2–14.If a jackscrew requires a force of15 pounds at the end of the handleto lift a 3,000 pound load, itsactual mechanical advantage is1.4,5002.2,0003.4504.2001.0.753 in2.0.703 in3.0.628 in4.0.517 in13