Figure 1(c) shows a roller or a wheel support with lateral constraint. A lateral force P exerted by the We extend our methods for two-dimensional equi- guide on the wheel can exist in addition to the nor-librium to the case of three-dimensional equilib- mal force N.
rium. The general conditions for the equilibrium A ball-and-socket joint is shown in Fig. 1(d).
of a body require that the resultant force and re- The joint free to pivot about the center of the ballsultant couple on a body in equilibrium be zero. can support a force R with all three components.
These two vector equations of equilibrium and their Figure 1(e) shows a fixed connection. In addition to three components of force, a fixed connection can support a couple M represented by its three first three scalar equations state that there is no Figure 1(f) illustrates a thrust-bearing support.
resultant force acting on a body in equilibrium in Thrust bearing is capable of supporting axial forceany of the three coordinate directions. The second Ry as well as radial forces Rx and Rz. Couples Mx three scalar equations express the further equilib- and Mz must be assumed zero in order to provide rium requirement that there be no resultant mo- statical determinacy.
ment acting on the body about any of the coordi-nate axes or about axes parallel to the coordinateaxes. These six equations are both necessary and sufficient conditions for complete equilibrium. Thereference axes may be chosen arbitrarily as a mat- Application falls into four categories. These cate- ter of convenience, the only restriction being that gories differ in the number and type of independenta right-handed coordinate system should be chosen equilibrium equations required to solve the prob-when vector notation is used. The six scalar rela- tionships are independent conditions because any • Category 1, equilibrium of forces all concurrent of them can be valid without the others.
at point O, requires all three force equations,but no moment equations because the moment of the forces about any axis through O is zero.
• Category 2, equilibrium of forces which are The summations include the effects of all forces on concurrent with a line, requires all equations the body. Figure 1 describes the most common sit- except the moment equation about that line.
uations of force transmission in three-dimensionalproblems.
• Category 3, equilibrium of parallel forces, re- Figure 1(a) shows a member in contact with a quires only one force equation, the one in the smooth surface or a ball-supported member. Force direction of the forces and two moment equa- must be normal to the surface and directed toward tions about the axes which are normal to the A member in contact with a rough surface is • Category 4, equilibrium of a general system of shown in Fig. 1(b). The possibility exists for a forces, requires all three force equations and force F tangent to the surface to act on the mem- A force of magnitude P = 180 N is applied to the stationary machine handle as shown in Fig. 2.
Write the force and moment reactions at O as vec-tors. Neglect the weight of the handle assembly.
Solution. With unit vectors i, j and k in the x-, y- , and z-directions, the zero summation of forces andmoments for equilibrium yields the vector equa-tions 180(cos 15◦i+sin 15◦k)+(Oxi+Oyj +Ozk) =0 and (0.2j + 0.125k) × 180(cos 15◦i + sin 15◦k) +(MOxi + MOyj + MOzk) = 0 where (Ox, Oy, Oz) and (MOx, MOy, MOz) are the force and moment components in the three directions. Equating the coefficients of the i-, j- and k-terms to zero gives Ox = −173.9, Oy = 0 and Oz = −46.6 N, andMOx = −9.32, MOy = −21.7 and MOz = 34.8Nm. Thus, O = −173.9i − 46.6k N and M [1] J. L. Meriam and L. G. Kraige, (2011), Engi- neering Mechanics, Volume 1, Statics, 7th edi- Figure 1: Examples of free body diagrams.

Source: http://www.robotics.it-chiba.ac.jp/okubo/dynamics/files/equilibrium_in_three_dimensions.pdf

20100619 kisc early detection weekly report 6

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