A force of 182.6 lb should be applied to lift 700 lb object.
<h3>How to calculate the force?</h3>
Let the force required to lift be F
Total length = 29 in.
Distance of short end from pivot = in.
Distance from long end = 29-6 = 23 in.
Now Equating moments
F × 23 = 700x6
23F = 4200
F = 4200/23
= 182.6 lb
A force of 182.6 lb should be applied to lift 700 lb object.
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You've answered this yourself hong would wake up at 8:35
By the Pythagoras theorem:-
h^2 = x^2 + x^2
(where h = hypotenuse)
h^2 = 2x^2
h = sqrt2 x
let's firstly convert the mixed fraction to improper fraction, and then divide.
![\bf \stackrel{mixed}{11\frac{1}{2}}\implies \cfrac{11\cdot 2+1}{2}\implies \stackrel{improper}{\cfrac{23}{2}} \\\\[-0.35em] ~\dotfill\\\\ \cfrac{23}{2}\div \cfrac{3}{4}\implies \cfrac{23}{2}\cdot \cfrac{4}{3}\implies \cfrac{23}{3}\cdot \cfrac{4}{2}\implies \cfrac{23}{3}\cdot 2\implies \cfrac{46}{3}\implies 15\frac{1}{3}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7Bmixed%7D%7B11%5Cfrac%7B1%7D%7B2%7D%7D%5Cimplies%20%5Ccfrac%7B11%5Ccdot%202%2B1%7D%7B2%7D%5Cimplies%20%5Cstackrel%7Bimproper%7D%7B%5Ccfrac%7B23%7D%7B2%7D%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Ccfrac%7B23%7D%7B2%7D%5Cdiv%20%5Ccfrac%7B3%7D%7B4%7D%5Cimplies%20%5Ccfrac%7B23%7D%7B2%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B3%7D%5Cimplies%20%5Ccfrac%7B23%7D%7B3%7D%5Ccdot%20%5Ccfrac%7B4%7D%7B2%7D%5Cimplies%20%5Ccfrac%7B23%7D%7B3%7D%5Ccdot%202%5Cimplies%20%5Ccfrac%7B46%7D%7B3%7D%5Cimplies%2015%5Cfrac%7B1%7D%7B3%7D)
