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在学习杠杆的过程中,很多学生都会遇到“画杠杆中最小动力的方向”的问题.其实,基本思路都一样:根据杠杆的平衡条件:F1l1=F2l2,要使F1最小,必须使l1最大,也就是使动力臂最长,因为此时F2l2的乘积是一定的,所以就转化为求最长力臂的问题.根据实际情况,在初中可以把此类问题分成三类.一、动力作用点固定时的最小力例1如图1所示,在弯曲杠杆A端吊一重物G,要想在B
In the process of learning leverage, many students will encounter "In fact, the basic idea is the same: According to the lever balance conditions: F1l1 = F2l2, to make F1 the smallest, you must make l1 maximum, that is, the power arm longest, because F2l2 at this time the product is a certain degree, it is converted to the problem of seeking the longest arm. According to the actual situation, in junior high school can be divided into three types of problems. Dynamic point of the fixed force minimum force 1 As shown in Figure 1, hanging in the bending lever A hanging a heavy object G, in order to B