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1、1,Chapter 5 Force System in Space,2,第五章 空间力系,3,4,5,51 Concurrent force system in space51 空间汇交力系52 System of force couples in space52 空间力偶系53 The moment of a force about a point or an axis53 力对点的矩与力对轴的矩54 Reduction of a force system in space to a givenpoint 54 空间一般力系向一点的简化,6,55 Discussion of the resu
2、lt of the reduction of a force system in space55 空间一般力系简化结果的讨论56 Equilibrium equations of a force system inspace and their applications56 空间一般力系的平衡方程及应用57 The center of a system of parallel forces and the center of gravity of an object57 平行力系的中心与物体的重心 Class of exercises 习题课,7,1. The projection and d
3、ecomposition of a force on the axes in space: 一、力在空间轴上的投影与分解：(1)The representation of force inspace:1.力在空间的表示：a force is given by itsmagnitude, its direction and its point (line) of application 力的三要素：大小、方向、作用点(线),8,magnitude:大小： point of application: the point at which the force acts onto the object
4、.作用点：在物体的哪点就是哪点direction: it can be determined bythe three angles , , g or by the angle of inclination and the angleof depression .方向：由、g三个方向角确定由仰角 与俯角 来确定。,9,10,11,12,then:,13,(1) The graphical method: It is same as in the case of the composition of a coplanar system of concurrent forces. It can be
5、 solved by the force polygon.The resultant force is equal to the geometrical sum of the components.,1、几何法：与平面汇交力系的合成方法相同，也可用力多 边形方法求合力。即：合力等于各分力的矢量和,14,15,16,The resultant force is given by,17,18,19,20,21,22, Composing F1 and R we getF2 at the point d composing F1and R we get F2 at point c , R-F1=F2
6、 , R- F1= F2 . 由反向平行力合成得： F1与R合成得F2，作用在d点F1与R合成得F2，作用在c点且R-F1=F2 ，R- F1= F2 The force couple (F1，F1) in plane I is replaced by (F2， F2) in plane II. 在I内的力偶（F1，F1）等效变成II内的（ F2， F2 ）,23,24,25,26,27,In the plane it is an algebraic quantity. In space, it is a vector. 在平面中：力对点的矩是代数量。在空间中：力对点的矩是矢量。,28,29,
7、O,30,O,31,32,结论：力对/它的轴的矩为零。即力F与轴共面时，力对轴之矩为零。,Conclusion: the moment of a force about an axis which is parallel to the force is zero. In other words, if the force F and the axis are coplanar the moment is zero.,33,34,35,36,即：,37,38,39,40,41,According to the Theorem of Translation of a Force we transl
8、ate every force to the point O. We get a concurrent force system in space and an additional force couple system . notice are the moments of the forces about point O. 根据力线平移定理，将各力平行搬到O点得到一空间汇交力系： 和附加力偶系注意 分别是各力对O点的矩。,42,Because the force couple in space is a free vector it can be always translated to
9、 the point O. 由于空间力偶是自由矢量，总可汇交于O点。,43,Composing , we get the principle vector . In other words, (the principle vector passes the point O, and it does not depend on the position of the point O)合成 得主矢 即 （主矢 过简化中心O，且与O点的选择无关）,44,Composing , we get the principle moment ,(the principle moment depends on
10、the position of point O) 合成 得主矩 即： （主矩 与简化中心O有关）,45,If the given center O is the origin of the coordinate system, then the magnitude of the principle vector is 若取简化中心O点为坐标原点，则：主矢大小The direction is 主矢方向,46,According to the relationship between the moment of the force about a point and the moment of t
11、he force about an axis through the point 根据力对点之矩与力对轴之矩的关系:The magnitude of the principle moment is则主矩大小为： and the direction is: 主矩方向：,47,48,49,50,51,52,53,54,55,notice 注意 The invariant quantities after reduction are Rand M/ (they do not depend on the given center). 力系简化中的不变量（不随简化中心改变）有：R， M/ If the
12、given center is the point O we get M. 简化中心为O时：为M,56,But if the given center is the point O it is changed to M. M/ is an invariant quantity (it does not depend on the given center and the force couple of the force system) 当简化中心为O时，为M 但M/总是不变的（它是原力系中的力偶与简化中心无关）,57,58,59,60,所以空间任意力系的平衡方程为：还有四矩式，五矩式和六矩式
13、，同时各有一定限制条件。,So the equilibrium equations of a force system in space areThey can also be expressed by 4 moment equations, 5 momentequations or 6 moment equations,but each case has some limiting conditions.,61,62,空间平行力系的平衡方程，设各力线都 / z 轴。因为均成为了恒等式。,The equilibrium equations of a parallel force system
14、in space, supposing all the action lines are parallel to the axis z arethey are all identities.,63,64,65,66,67,68,69,70,71,72,73,74,75,76,77,78,79,80,81,82,物体分割的越多，每一小部分体积越小，求得的重心位置就越准确。在极限情况下，(n- )，常用积分法求物体的重心位置。,If the body is divided into smaller and smaller parts the determination of the center
15、of gravity becomes more accurate. In the limiting case, (n- ), we used to use an integration to determine the position of the center of gravity.,83,If I denote the weight of part No. I and Vi is the volume of it, then . Substitution into the equations above gives in the limiting case:Where: this is the accurate formula to determine the center of gravity.,
