How to calculate a width of a crane, so it won't tip over

[Ghetto Cowboy]

Homework Statement

Hey all, I'm stuck on some homework and any help would be awesome. this is the question

The crane shown in the diagram must carry a load of 600 lbs, at a distance, shown in the figure, of 20 feet from the center of gravity of the crane. If the crane weighs 1,500 lbs, how wide must the base (d in the figure) of the crane be to ensure that the crane does not tip?

Homework Equations

I feel like i have to sum some moments and forces in the x and y direction

The Attempt at a Solution

I think i have to sum the moment around the point where the cranes right "foot" is and it might go something like this

\sumM = -1500(d) + By(d) - 600(20+d)

to make things clear, the point of the crane's right foot is A, and the left foot is B..

anyhelp would be great!

 

[spartandfm18]

You're on the right track. You need a couple moment of (20ft)*(600lb) at the base of the crane to keep it in moment equilibrium. To get that couple moment you're going to need to use the reaction force and the distance between the legs. Don't forget to do a force balance too.