Physics Help! Student is standing on a beam

A 3.0-m-long rigid beam with a mass of 125 kg is supported at each end. An 60 kg student stands 2.0 m from support 1. How much upward force does each support exert on the beam?

right support :


left support:

Help I'm stuck!

call the support forces L and R representing the left and right supports

the equation of vertical equilibrium is

L+R = (125kg + 60kg) x 9.8m/s/s = 1813N

now we consider rotational equilibrium; let's sum torques around the left pivot; the weight of the board and person will generate torques in one direction; the torque from R will be in the other, therefore our condition for rotational equilibrium will be

torque due to R = torque due to weight of board plus torque due to weight of student

these torques are:

R : R x 3m
student: weight of student x 2m = 60 g x 2
board: weight of board x 1.5 m (assuming board is uniform, its weight acts as if it is at the center of mass)

we have:

3R = 1.5mx125kg x 9.8m/s/s + 2m x 60kg x 9.8m/s/s

R=1004.5 N

therefore, the left end provides support of L = total weight - R = 808.5N

Start with the beam. 62.5 kg on each support. As the student is close to support 2 by 2/3 take the 60 kg and divide by 3. Ad 40 kg to support 2 and 20 kg to support 1. The result is 102.5 kg by support 2 and 82.5 kg by support 1. Easy???