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stackptr

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## Homework Statement

A homogeneous board weighing 60 kgf lies on two supports as shown. Find the forces acting on the supports.

## Homework Equations

$$\frac{F_A}{F_B} = \frac{OB}{OA}$$

where ##F_A## is the force acting on the support on the left, ##F_B## is the force acting on the support on the right, and point ##O## is the center of rotation (which I assumed to be the center of mass, 2.5 m; correct me if I'm wrong). Points ##A## and ##B## are the points where the supports are located.

$$\Sigma\tau = (L/2)F_O = \tau_A + \tau_B$$

##\Sigma\tau## is the net torque of the entire system; ##L## is the length of the entire board; ##F_O## is the weight of the board acting at point O; ##\tau_A## and ##\tau_B## are the torques acting at points A and B, respectively.

## The Attempt at a Solution

$$\frac{F_A}{F_B} = \frac{OB}{OA} = \frac{2.5m - 2m}{2.5m - 1m} = \frac{1}{3}$$

Therefore, ##F_B = 3F_A##.

$$\Sigma\tau = (L/2)F_O = \tau_A + \tau_B$$

$$(L/2)F_O = (OA)F_A + (OB)F_B$$

$$(L/2)F_O = (OA)F_A + (OB)3F_A$$

$$(L/2)F_O = F_A[(OA)+ (OB)3]$$

$$\frac{(L/2)F_O}{OA + 3OB} = F_A = \frac{\frac{5 m}{2} * 60 kgf}{(\frac{5 m}{2} - 1 m) + 3(\frac{5 m}{2} - 2 m)} = 50 kgf$$

If ##F_A## is 50 kgf then using the other equation we get ##F_B## is 150 kgf. But my textbook says the right answers are 15 kgf and 45 kgf. I suspect that my mistake is in the net torque equation, since the ratios of my answers are consistent with the ratios of the correct answers.

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