Weights and Friction on Inclined Rods
Statics of loaded beams, a pulley-and-counterweight, and friction reckoned in quarters and eighths
The sheet works through problems in statics: how the weight carried at the end of an inclined rod or beam is distributed, and how friction ('confregazione') is reckoned as fractions — quarters and eighths — of the load. One diagram raises a supported beam with a pulley and counterweight, while another arrays a horizontal rod against five inclined ones to compare the friction in each, Leonardo noting 'try these out' beside one figure. The lower fragment of the leaf is blank.
On this page
Half the obliquity is not half the weight
The line a b is not half the obliquity, but it is indeed half the weight that falls upon one of the ends of the rods. The figure is a double quadrant of a circle lettered b – a.
Weight at the end of an inclined rod, reckoned in quarters
If 4/4 at a comes from 4 pounds of the rod a b, then 3/4 at a — needed to resist 3/4 of friction at b — comes from 3 pounds of a c. Divide the pounds, therefore, into quarters. The rod is lettered a 4/4 – c – b 12/4.
A right-angled support carries none of the load at o
The right-angled support n m discharges at o no part of the weight o p; therefore at o there are the 3/4 of the weight o p, which is 3 pounds, having 3/4 of a pound for friction. The inclined beam is lettered p 4/4 – m – o.
A support raised by a pulley and counterweight
n S is a spherical right angle; because at e there is one and at n there are 3, the rope a e of right-angled conjunction at e sustains it. Reckoning the friction of 3 at n against the one at g, the mean of 4/4 with 3/4 gives 7/4, and halved, 7/8. The figure is lettered d c b a – 1 g – e – S – n 3.
Comparing friction across five inclined rods
A horizontal rod is set against inclined ones, each labelled with its share of load and friction — 4/4, 3/4, 2/4, 1/4, each against 3/4 of friction — resolving to 6/8. Beside a related support marked 3 6 1 Leonardo writes 'try these out'.