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\fdO'^ - Old Forge Coal Mines

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ELEMENTARY MECHANICS. 185<br />

(446) 900 X .18 X 2 = 324 lb. = force required to overcome<br />

the friction. 900 + 324 = 1,224 lb. = total force.<br />

IS;^ = --«-• ^-<br />

.<br />

(446)<br />

18 -T- 88 = .2045. Ans.<br />

(447) See Art. 962. D = f^= ^^^^<br />

(448) See Fig. 39. 125<br />

-47.5 = 77.5 lb. = downward<br />

pressure.<br />

77.5^4 = 19.375 lb.<br />

= pressure on each support.<br />

Ans.<br />

(449) See Fig. 40.<br />

Fig. 40.<br />

(450) See Fig. 41. 4.5 ^ 2 = 2.25.<br />

12<br />

2.25 X 6 X 30 = 9G0 lb. Ans.<br />

(451) (a) 960^30 = 32. Ans.<br />

{d) 790 -^ 960 = .8229 = 82. 29^. An§,<br />

= 12.438.<br />

Ans.

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