FzzXWM3XVersion 3.0.3: 15Nov95IE80 MSWindows'WRLDLINK _[ T<?JHome@@FHome@@RLINK V ??Iڴ!??vHvb)@???Iڴ???IڴmassLINK U'poinLINK .b ASystem Center Of Mass'LINK pun_Mkg??un_Dm??unRt@.?5Ȯun_CA??un_KK??unsmmole??un_Lcd??un_Ts??unFFN??unENJ??unWWW??unCHC??unRs??unVlV??unHz??unRV??unSp??D /s@.?5Ȯm/s^2??m/s??N/m??N-s/m??kg-m^2??N-m??kg/m^2?? N-m^2/kg^2?? N-m^2/kg^2??'LINK 'LINK   'LINK  'CSTRLINK @ ԕ*?ݒ@o3yq?'%`- self.mass #* (9.80664999999999942)???b A'LINK  'CSTRLINK  @ q>?'b A'LINK  'CSTRLINK @@qp`angle(self.v#)????'b A'LINK  'CSTRLINK !t`??`??`??q>?'b A'LINK Lk??Q?Q'LINK 6eTexMDTEXTttxtDTAexDC6eT98exVXMoVV avi .eTexDXTEXTDCGTDXF'LINK 'LINK 'LINK &LINK  &LINK &LINK massLINK @ٙ??@ٙ&ּ_{?DƮ}uXw'%`?zGy?ז Չ????ѷb A RectanglepoinLINK `(0.0)??`{body[3].height}/2.0???b A Base PointCSTRLINK k<?Q@qv?b AForcemassLINK ?ʠ@@?ʠ&ffffffwwwwwwT??\(C@????ѷb A RectanglepoinLINK . b AAnchorpoinLINK .ٙ?uXw'%`b APointpoinLINK ."6?)^z`b APointCSTRLINK nqz?uXw'%`?b A Pin Joint~ poinLINK .?pm?Ta+?uXw'%`b APoint poinLINK .@b APoint CSTRLINK @buSkq`?uXw'%`? @buSkb ARope LINK ],ud`vt?|F|?u<vb A Tension of Rope 11@`time#???`- constraintforce(11).x#??? LINK Y'ZTo draw to scale, multiply by 4. a=4*2.5 c=6.075*4 a=10 b=24.3 Next, to find the length of b we must use Pythagorean's Theorem. b=sqrt(24.300^2- 10^2). b=22.147 b ATextLINK YIZNow to find the angle for theta, we must use the inverse tangent function. Theta = tan^-1 (10/22.147). Theta = 24.3 degrees.nb ATextLINK Y"hZ Theta b ATextLINK Y/hZ bb ATextLINK Y"hZabl4b ATextLINK Y"hZcbFb ATextLINK ],d`t?Fx?Fy?k|F|?u<v<}b ATotal Force on Rectangle 3@N`time#???`Body[3].mass #* Body[3].a.x#????`Body[3].mass #* Body[3].a.y#????`Body[3].mass #*| Body[3].a#|????