Question

A uniform steel beam has a mass of 930 kg. On it's left resting half a identical beam. At the ends of the full beam are two supports.

What is the vertical support force at each end?

Answer 1

You know that the two forces must equal the weight of the twopieces so

   L + R = Mg + (1/2) Mg  

    L + R =  9.8 *(3/2) *930 =   13671

To find R, consider the torques acting on the slabs, using theleft end as the point of reference. Then each torque (weight ofhalf slab, weight of full slab, force of R) is just the force timesthe distance from the left end:

    (1/2) Mg * (1/4) L  +   Mg * (1/2) L   =   R * L

solve for R:      R = (5/8)Mg = (5/8) *930*9.8 = 5696.2 N

and so   L = 13671 - 5696.2 =  7974.7 N