Statics Example a road-rollerruns over a barrier, the weight of the roller is P=20kN, its radius is r60cm and the height of the barrier h=8cm Determine the magnitude of the horizontal force F acting on the center of the roller and the force with which the roller acts on the barrier Solution: ( 1choose the roller as the body to be studied B isolate the body, draw the force diagram h When the roller just left the ground N=0, Fis maximum F, gravitation and reaction force NB form of a balanced force system According to the graphical condition of equilibrium B F=Ptga A B cosa by the geometrical relation /2-(r-h) tga r-h -=0.577
11 [Example] A road-roller runs over a barrier, the weight of the roller is P=20kN, its radius is r=60cm and the height of the barrier h=8cm. Determine the magnitude of the horizontal force F acting on the center of the roller and the force with which the roller acts on the barrier. 0.577 ( ) tg 2 2 = − − − = r h r r h By the geometrical relation ①choose the roller as the body to be studied; ②isolate the body, draw the force diagram. Solution: ∵When the roller just left the ground NA=0,F is maximum. F, gravitation and reaction force NB form of a balanced force system. According to the graphical condition of equilibrium F=Ptg cos N P B = , .
学 「例]已知压路机碾子重P=20kN,r=60cm,欲拉过h=8cm的障碍 物。求:在中心作用的水平力F的大小和碾子对障碍物的压力。 解:①选碾子为研究对象 B ②取分离体画受力图 ∴当碾子刚离地面时N=0,拉力F最大,这时 拉力F和自重及支反力N构成一平衡力系 O 换况何条件,刘多封闭,故 又由几何关系:gC2(=h2037 r-h 12
12 [例] 已知压路机碾子重P=20kN, r=60cm, 欲拉过h=8cm的障碍 物。求：在中心作用的水平力F的大小和碾子对障碍物的压力。 0.577 ( ) tg 2 2 = − − − = r h r r h 又由几何关系： ①选碾子为研究对象 ②取分离体画受力图 解： ∵当碾子刚离地面时NA=0,拉力F最大,这时 拉力F和自重及支反力NB构成一平衡力系。 由平衡的几何条件，力多边形封闭，故 F=Ptg cos N P B =
Statics Therefore F=11. 5KN N =23.IkN by the relation between action force and reaction force, the force with which the roller acts on the barrier is 23. 1kN This problem can also be solved by the method of the force polygon, measure the magnitude of the force by a scale. The steps of the graphical method are: Choose a body to study; @draw the force diagram draw the force polygon with a proper scale; determine the unknown quantities The shortages of the graphical method are: Othe precision is not enough, the error is large @drawing the diagram need higher precision it can not show the functional relations among the forces. We shall study the other method of the composition and equilibrium of coplanar systems of concurrent forces The analytical method
13 By the relation between action force and reaction force, the force with which the roller acts on the barrier is 23.1kN. This problem can also be solved by the method of the force polygon , measure the magnitude of the force by a scale. Therefore F=11.5kN , NB=23.1kN. The steps of the graphical method are: ①choose a body to study; ②draw the force diagram; ③draw the force polygon with a proper scale; ④determine the unknown quantities. The shortages of the graphical method are: ①the precision is not enough, the error is large; ②drawing the diagram need higher precision; ③it can not show the functional relations among the forces. We shall study the other method of the composition and equilibrium of coplanar systems of concurrent forces: The analytical method
学 所以F=115kN,N2=23kN 由作用力和反作用力的关系,碾子对障碍物的压力等于231kN。 此题也可用力多边形方法用比例尺去量。 几何法解题步骤:①选研究对象;②作出受力图; ③作力多边形,选择适当的比例尺; ④求出未知数 几何法解题不足:①精度不够,误差大②作图要求精度高; ③不能表达各个量之间的函数关系 下面我们研究平面汇交力系合成与平衡的另一种方法: 解析法
14 由作用力和反作用力的关系，碾子对障碍物的压力等于23.1kN。 此题也可用力多边形方法用比例尺去量。 所以 F=11.5kN , NB=23.1kN 几何法解题步骤：①选研究对象；②作出受力图； ③作力多边形，选择适当的比例尺； ④求出未知数 几何法解题不足：①精度不够，误差大 ②作图要求精度高； ③不能表达各个量之间的函数关系。 下面我们研究平面汇交力系合成与平衡的另一种方法： 解析法
Statics 82-2 The analytical method of composition and the equilibrium of a coplanar system of concurrent forces I Projection of a force on an axis X-F=Fcosa Y FE Y=F=F'Sina=F coSB A F +F F X X F Y COSCo= FF cosA=F 15
15 F F F X x cos= = F F F Y y cos = = 2 2 F = Fx +Fy §2-2 The analytical method of composition and the equilibrium of a coplanar system of concurrent forces Ⅰ Projection of a force on an axis X=Fx=F·cos ： Y=Fy=F·sin=F ·cos