How Do Forces Distribute on a Board Supported at Two Points?

[stackptr]

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.

 

[Doc Al]

Some of your equations are confusing. If you take torques around the board's center (which is perfectly fine, but any point would do just as well), then the torque due to the board's weight is zero. What you end up with is the torque from Fa being equal and opposite to the torque from Fb. And, indeed, Fb = 3Fa.

Combine that with an equation for the sum of the forces and you'll get the textbook answer.

 

[kuruman]

Your textbook is wrong. The sum of the forces must be 600 kgf (the weight of the board) not 60 kgf.

 

[Doc Al]

A homogeneous board weighing 600 kgf

Given your textbook answer, I assumed that this was a typo and you meant to write 60 kgf. (Silly to make such assumptions!)

Either way, the two forces from the supports must add up to equal the weight of the board.

 

[stackptr]

Your textbook is wrong. The sum of the forces must be 600 kgf (the weight of the board) not 60 kgf.

Given your textbook answer, I assumed that this was a typo and you meant to write 60 kgf. (Silly to make such assumptions!)

Either way, the two forces from the supports must add up to equal the weight of the board.

My bad, I meant to write 60 kgf and not 600 kgf. I've edited the question appropriately

 

[Doc Al]

My bad, I meant to write 60 kgf and not 600 kgf. I've edited the question appropriately

Good. Now review my response above and take another crack at the problem.

 

[stackptr]

Good. Now review my response above and take another crack at the problem.

The problem is a lot simpler than what I had made it out to be.

The weight of the board, 60 kgf must be balanced by the two supports. Therefore

$$ 60 kgf = F_A + F_B$$

We know $F_B$ is 3 times the value of $F_A$, so

$$60 kgf = 4F_A$$,
giving us 15 kgf; the other force being triple that, is 45 kgf