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GEOMETRIC PROOF OF THE LAW OF THE LEVER

A lever consists of a rigid bar which is free to turn about a fixed point called the fulcrum.

The law of the lever states that the equilibrium condition of balance exists when the effort force Fe multiplied by the length de of the effort arm is equal to the load force Fl multiplied by the length dl of the load arm i.e. when

Fe de = Fl dl .

See Fig. 1. This law can be derived by a geometric argument as follows:

Proof of the law of the lever. Consider a uniform bar ABCD suspended by its middle point M so that it balances. See Fig. 2. Imagine that it is divided at EF into two parts AEFD and EBCF whose middle points are at K and L. Let WAEFB be the weight of AEFD and WEBCF be the weight of EBCF. Then the weights of AEFD and EBCF are proportional to GI and IH i.e.

and may be supposed to be concentrated at K and L. Referring to Fig. 2, the law of the lever then states that

or

We will prove that this assertion 3) is true by showing that

Proof. Denote the length GH of the bar by and the distance MI by i.e.

We then find the distances GI, IH, ML, and MK in terms of and

Thus

and so

Hence,

Fig. 3 shows that the weights may be hung at any depth below the bar, and that any equal weights may be substituted for the parts of the bar.

With some modifications this proof was given by Stevinus of Bruges, Galileo and Lagrange.

Reference.

Cox. Mechanics.