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

In summary, a 600 kg homogeneous board on two supports must have a weight of 45 kgf evenly distributed between the supports.
  • #1
stackptr
10
3

Homework Statement


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

?temp_hash=4375eeb1df1bf1d1e2a87797b0447c2d.jpg


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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  • #2
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.
 
  • #3
Your textbook is wrong. The sum of the forces must be 600 kgf (the weight of the board) not 60 kgf.
 
  • #4
stackptr said:
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.
 
  • #5
kuruman said:
Your textbook is wrong. The sum of the forces must be 600 kgf (the weight of the board) not 60 kgf.
Doc Al said:
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
 
  • #6
stackptr said:
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.
 
  • #7
Doc Al said:
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
 

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

1. What is torque?

Torque is a measure of the force that can cause an object to rotate around an axis. It is calculated by multiplying the force applied to the object by the distance from the axis to the point where the force is applied.

2. How is torque related to the board problem?

In the board problem, torque is used to determine the stability of a board resting on a pivot or fulcrum. The weight of the board and the distance of its center of gravity from the pivot point determine the torque acting on the board, which in turn affects its stability.

3. What factors affect the torque of the board in the problem?

The torque of the board in the problem is affected by the weight of the board, the distance of its center of gravity from the pivot point, and the angle at which the board is resting on the pivot.

4. How is torque different from force?

Force is a measure of the push or pull on an object, while torque is a measure of the turning force on an object. While force can be applied in any direction, torque is specific to rotational motion around an axis.

5. How can torque be calculated in the board problem?

To calculate the torque in the board problem, you must first determine the weight of the board and the distance of its center of gravity from the pivot point. Then, multiply these values together to get the torque acting on the board. The direction of the torque can also be determined by the direction of the force and the angle at which the board is resting on the pivot.

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