MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.001 Mechanics of Materials I Spring,2003 Solutions for Quiz 1 Problem I mI m1 putley① F 03 707777刀77777 a) DYaw FBD f ench mais aud wile down ey.of equilibrim m,9 2F3:N=m,9 =7fe,=Am,9 T← 2=7 T=um,g -0 N 艺可 N2 Nit mg ↓m =(m,+mz】9 fF不N 7斤.：μ(m,tm,)9 2：T2-T++F, we① N =μm,9Ψm9+m4m)9 T2=3Am9+rm,9一@ 7T2 25: N3=N2+m9=(m,+m,+m3)月 F,=(m,+m.+m)9 2n:F=T2+6,1F;
= F=3Hm,9寸Am29+(m,+m?)9 +H(m,+m2+m3)9 F=5Hm,9+3m:9+Am39 (b) feezes H, T2=T32 -7万=万e Redh.aw im时Le FBD ms 业F ZR 9T3 F T3 FF+FFs 个F -,下 N3 F=Tze ulmitm:)9 十H(m+m2tm3)9 F=(3umg+um,g)e +2((m,+m2)9+m39
PROBLEM 2 la)Driw the FBp g th mimber AcD (M 2BM M-R.d=o R=yA、—0 25 E^+F-em日二。 FicFt M/d 3x: A-FBc Co2 o =o RA=Fec Coe ，F+g.me 5i78 RA F+MA raw FBD4F中pntB Fx: Rt FBc Cose R 乙5：R8+f6eSmB=0 字R=-(F+Mh)
(b) 长一X州M Fxx 返：m:-R对 ↑g Fxy 5则=-对 2M:M=× 、4 AFD g SFD BMD
Problem 3 uilbrim9了oint D ：-。6,-62g00 2:-6o红8n6f局 o Defwmahimn Eyurhing 5w·段75：k。一@ 50”e75·t68。—④ w :7=大5w一⊙ epa'6r的Lmnditon5 Sn·(a,°-u,)an0+(u°-uy)ho =u°-⑥ 8o-(u-u)a8,)+(u-uy)a, 。ukCn8,+4y8,一⑦ c。÷(u”-u,)CfB2)+(uy°-4m6. :山C8,+uP2⑧
(b, d=1 8,÷3 ”6,:60 eg.⑥，D (8 heerme ShD-Up loe these w +uy e4·③，④audE Sep FA=KP +) 0 Fco (2,”+9g) Putriug 0 复生)9-(5u”+厚)片= D -+u-里吲-要y-0 -2u以-=0 A From es② -k,要+以生-k(生以+复)g-F=0 4 4 F =7 05 Puk
-2uw-2(父-長）=o -2以”+3 +B 2 反=a F -0 4 2 up B×号 3 Sot 5 42= -(华6)-发 二3 sop