Next we will knock a tintack into any point _B_, and tie a string on to _B_. Then if I pull at the string in any direction _B C_ the board tends to twist round the screw at _A_. What will the strength of the twisting force be? It will depend on the strength of the pull, and on the “leverage,” or distance of the line _C B_ from _A_. We might imagine the string, instead of being attached at _B_, to be attached at _D_; then, if I put _P_ as the strength of the pull, the twisting power would be represented by _P_ × _A D_. This is called the “moment” of the force _P_ round the centre _A_. It would be the same as if I had simply an arm _A D_, and pulled upon it with the force _P_. It is an experimental truth, known to the old Greek philosophers, that moments, or twisting powers, are equal when in each case the result of multiplying the arm by the power acting at right angles to it is equal. Now suppose _A B_ is a pendulum, with a bob _B_ of 10 lbs. weight, and suppose it has been drawn aside out of the vertical so that the bob is in the position _B_. Then the weight of the bob will act vertically downwards along the line _B C_. The moment, or twisting power, of the weight will be equal to 10 lbs. multiplied by _A D_, _A D_ being a line perpendicular to _B C_. [Illustration: FIG. 36.] Now suppose that another string were tied to the bob _B_, and pulled in a direction at right angles to _A B_, with a force _P_ just enough to hold the bob back in the position _B_. The pull along _D B_ × _A B_ would be the moment of that pull round the point _A_. But, because this moment just holds the pendulum up, it follows that the moment of the weight of the pendulum round _A_ is equal to the moment of the pull of the string _B D_ round _A_. Whence P × A B = 10 lbs. × A D. Whence P = 10 lbs. × (A D)/(A B).