Plank and two supports using torque

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A 100kg man stands on a 60m uniform plank weighing 50kg, supported at both ends. The left support is at the plank's left end, while the right support is 2m from the right end. Using torque equations, the force exerted by the left support is calculated to be 122.5N, while the right support exerts 1347.5N. If the man's mass doubles, the force from the left support will increase due to the additional weight. The calculations confirm that equilibrium conditions were met throughout the analysis.
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A man with a mass of 100kg is standing on a uniform plank of mass 50kg. The man is standing directly above the right support and the left support is located at the left end of the plank. The right support is 2m from the right end. the plank is 60m long. What is the magnitude of force the left support applies to the plank, the magnitude of force the right support applies, and if the man's mass was doubled the force by the left support would remain the same, increase or decreas?
Equations: Tnet=0
Fnet=0

I tried solving the force by the left support...I set the pivot point on the right support. I got the distance of the man from the pivot point is 0. the beam is 6m long so the center mass is mg(1/2* 6) my equation is tnet=0 I get 100(9.8)(0m)+50(9.8)(3m)-Force left(4m)=0. I know this isn't right but can somone point me in the right direction?
 
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The weight resultant force acts at the cg of the plank, which is not 3 m from the right support.
 
the 3m is half the length of the board. so should I move my pivot point to the far left support? so the center mass of the plank would be 3m? This is the equation I get: 50(9.8)(3m)-Force left(0m)+100(9.8)(4m)=0 there's nothing to solve with that though
 
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Ahhhh...after struggling for 2 hours and wearing down my eraser...I think I may have solved this. take a look at what I did and let me know if its finally right.

these are the equations I used. I had 2 unknown so I set the pivot point on the left support to =0.
1.) 50kg(9.8)(3m)+100(9.8)(4m) - Force right support(4m) = 0 solve this to get Fr=1347.5N.

2.) now that I have the force the right support gives I can use it in the equation fnet=0 to solve for the left support force. FL+FR-mplank(g) - mman(g)=0
FL+1347.5 - 490 - 980 = 0. solve for FL to get 122.5N

Correct?
 
Yes, for a system in equilibrium, you can sum moments about any point=0, and use also F_net =0, to get the result, which you did.
If you summed moments about the right support, you get the same answer , but you mistakenly took the lever arm of the weight force as 3 m when you tried it that way, when you should have used 1 m as that distance. Then , summing moments about the right support, your equation should have read

100(9.8)(0 m)+50(9.8)(1 m [/color]) - Force_left(4 m)=0, from which
Force_left = 122.5 N, same result. It is always a good idea, however, to sum moments = 0 about both supports independently, and use F_net = 0, to check your work to be sure you didn't make a math or other error.

Now you should answer the last question, if you so choose, "...if the man's mass was doubled the force by the left support would remain the same, increase or decrease"?
 
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