Structures Q - Safety factors

[anonuser0511]

A uniform beam of mass 1500kg is used as a cantilever as shown below. If the supports are made of wood and there is to be a safety factor of 6, what is the minimum cross-sectional area of each support?

ANS: 5.6E-4m^-2 (left) 6.1E.3m^2 to 1.9E-3 (right)
^wtf m^-2?

earlier in the book they have a table of Young's modulus for wood 1E9 - 1E10 Nm^-2 and then they have theses "strengths"  wood has a tensile strength of 4E7 Nm^-2 and compressive strength of 1E7 - 3.5E7 Nm^-2

so i'm guessing answer is wrong but how would you work out a safety factor Q - by multiplying the F in the stress by 6?

[TrueTears]

http://vcenotes.com/forum/index.php/topic,9668.msg153486.html#msg153486

Kamil explained safety factors really well

[simplicity123]

Correct me if im wrong, F1 is in tension and F2 is in compression. To apply the safety factors, u divide the tensile and compressive strengths by 6. That is the maximum allowable stress allowed on each support respectively. Then u go on to use ur stress formula for each support to find the cross sectional area.

[anonuser0511]

Awesome so i rearranged the formula and i got

     Safety Factor * Force
A =___________________
               MAX

So the answer i got for the above question was

LEFT: 5.625E-4m^2
RIGHT: 1.125E-2m^2 to 3.214E-3m^2

well answers agree with one of them eh, but then again they have m^-2, damn you Nelson! (3rd edition)