M3: COM

bump

Your diagram is not quite correct.

You get $\lambda$ from taking moments about $B$. Hence we have:

$(4+k)mg \cdot d_1 = \lambda m g \cdot d_2$,

where you know $k$ and can determine $d_1, d_2$ fairly easy.

For $d_2$ I suggest you have a look at the second image and notice that $d_1 + d_2 = 2r \sin 60$.

thanks.

so we need to find the perpendicular distance from the COM to the line of action and not the other way around?

but the mass is a point and not a line....?

Very odd questions...


Obviously the perp. distance will be the same between the two things regardless from which you begin measuring...?



Yes mass is concentrated onto a single point in these models... where you are getting the impression that mass is 'a line' ... ?