Question

please explain even small steps and im having trouble with the moments so please explain that too thanks

(6) The forces and couples shown are applied to two screws as a piece sheet metal is fastened to a block of wood. Reduce the forces and the couples to an equivalent wrench, and determine (a) the resultant R. (b) the pitch of the wrench, (c) the point where the axis of the wrench intersects the xz plane. 6 Nm 375 mm 500 mm 10 N

Answer 1

- A As shown in the diagram Two forces are working at poin . I are working at poinat A & B (in -y dięection) to the resultant R = FA +FB . = -10 9 - 11 = -21 Ĵ R = 21n and acting towards -y-direction The positions vectors are in tz direction. So The positions vectors from Center O is. K = Co0 mm & RR = -325 mm to the Resultant of moment at point o is ño = RAX(-10 î. + RB X (-11 h ) + (67)+(6 Î) (Nimi soox 10-3 Ê XC-10) ſ +6325/310-3 FX(-119) 6 + 12. Î

Pitch CP) Mo - 2 - 4 12 1 + 1 2 3 = (0.875 ^ +12 Î , a.m. Now the pitch of the wrench is a . (Resultant force). (momentato) . ( Resultant force) 2 Ř. Ñ. p = TR12 (-21 Î). (0.875 î+ 129 ) (-2172 0 - 21*12. 21X2] pa-0.571 m Ô The couple Vector 1. M = (Pitch ) x (Resultant force) = (-0.571)*(-21 ſnim = 11,991 Î (elim) Suppose the position of the point vector where it intersects with x-z plave. = neî + z So Mi + x (R) = mo --> 11iggiſ + (x +2Ã )*(-21 Î) = 0.8757+025?

2112 Î +11.991 2138 = 0.875 +12 ſ so by comparing both sides. by 212 = 0.875 1212= 0.875 pe= 0 2= 21 0041667 m z = 41.667 mm