To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
[tex]\sum \tau = F*d[/tex]
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
[tex]F*(27-6)= 6*600[/tex]
[tex]F = \frac{6*600}{21}[/tex]
[tex]F= 171.42 lb[/tex]
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
