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9.1:.24' Uh 4.1 12 Given: Cellar door prepped open as shown; Weight of door = 40 lb
Find: Force in the supporting stick Solution: Free body diagram: There is a force directed along QP exerted on the door at P; the weight of the door downward at W, and then
couples and forces at each of the hinges as shown. We have six possible equilibrium conditions to use for ﬁnding F8. However, note if we choose to sum the
moments with respect to O, we can eliminate many of the unknowns: 2M0 =Zrmi xF, +20, =O21.5ﬂix(Fy,j+Fnk)—1.5ﬁix(Fy1j+Fz1k)+(0.9j+1.2k)?x(— 40m)
+(~ 3i+1.8j+2.4k)xFs((— 3i —2.2j+2.4k)/ 32 +2.22 +2.42 )+ Cy1j+Cz1k+ Cy2j+szk = o This is particularly true if we look just at the xcomponent for the moments with respect to O, that is, if we not
the moment with respect to thexaxis = 0: (2mm: (Zrm xF,)i+(ZC,)oi =0 2 (1.5m X(Fy1j+ Fz1k))oi — (1 .5ﬂ‘i X(Fy1j+ Fz1k))i +((0.9j + 1 21¢}: x (— 40tbk))oi + k— 3i+1.8j+2.4k)>< th((— 3i — 2.2j+2.4k)/J32 + 2.22 +2.42 )1. i +(Cy1j+ 621k + 6,2] + szk)oi = o The ﬁrst two terms can be seen to be zero, either by calculating the cross product or by noting the lines of action of the forces run through the x—axis, so the moment about the xaxis should be zero. The last four terms can be
simply seen to be zero when the dot products are calculated. When the other dot products are calculated, we get 0 + o + (ask40):? o lb +Fsk(1.8)(2.4)+(2.4)(2.2))/ 32 + 2.22 +2.42 J+ o = o r0? : o; I "IQ—3 cm 3'53) *"Q‘Ysrkﬁgb w: 553} ,a‘f‘éoscﬁ S«n55'jj t A . + w 3 ~40 k
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 Fall '05
 Downing

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