# Homework Help: Moments physics problem

1. Dec 4, 2013

### Apothem

1. The problem statement, all variables and given/known data

"A uniform plank 3.0m long and a weight of 200N is supported horizontally and symmetrically across two trestles, A and B, placed 2.0m apart. Calculate the reaction force of each trestle. What is the maximum weight that can be placed at one end without the plank tipping over?"

2. Relevant equations

Unsure

3. The attempt at a solution

I have drawn a diagram of the situation, with the weight of 200N acting vertically downwards at 1.5m, with each trestle being 0.5m from each end of the plank, as it is symmetrical. From that I am unsure how to calculate the reaction forces. On regards to the second part where you have to calculate the maximum weight which can be placed at one end without the plank tipping, I feel as if I should be able to calculate that, one I have guidance on working out the reaction forces

2. Dec 4, 2013

### PhanthomJay

For calculating the support reaction forces from the weight of the beam only, that is correct procedure to place the resultant 200 N load at the beam centerline. Now just use the equilibrium equations , with which hopefully you are familiar, to solve for them. Then chck your result based on the symmetry of the system.....would you expect one support to support more than the other? For the second part, again use the equilibrium equations for the new load case (200 N load at beam center and unknown load at beam end), but to solve for the unknown load, you first need to consider what the far reaction force must be when the beam is about to tip over. Please show your work.

3. Dec 4, 2013

### nasu

You need to draw all the forces and then write two equations:
1. equilibrium of forces
2. equilibrium of toques.

4. Dec 4, 2013

### cepheid

Staff Emeritus
Hello Apothem,

Calculating the reaction forces should be as simple as employing Newton's Second Law, which states that for the plank, Fnet = ma. Since the plank is in equilibrium, both its vertical and its horizontal accelerations are zero. So the vector sum of forces in each of these directions must be 0:

ƩFx = 0
ƩFy = 0

This will probably only constrain the sum of the two upward reaction forces. To solve for each of them individually, you will also require the rotational equivalent of Newton's second law for torques:

Ʃτ = 0

which will probably end up telling you that the two reaction forces are the same, since they have the same lever arm.

My advice would be to start with a free body diagram (FBD) for the plank. It shows only the plank (no other objects) and the forces acting on it. This avoids confusion, and ensures you have a complete inventory of the forces for your sum.

5. Dec 4, 2013

### vela

Staff Emeritus
The board in in equilibrium, so the forces and moments have to sum to 0:
\begin{align*}
\sum_i \vec{F}_i &= 0 \\
\sum_i \vec{\tau}_i &= \sum (\vec{r}\times\vec{F})_i = 0
\end{align*}

6. Dec 5, 2013

### Apothem

From the diagram I used I took the moments about trestle B
1.0m*200=2*FtrestleA
From this I deduced that the reaction force of trestle A was 100N, due to the fact that the trestles are symmetrical based upon the centre of gravity, then the reaction force of trestle B is 100N, is that right?. From this to work out the maximum weight I am in the process of saying that the weight is at the end with trestle B, so I am taking moments from trestle B, however I know that there will be no force at trestle A, as it is the weight just before it tips. Is this the right way?

Last edited: Dec 5, 2013
7. Dec 5, 2013

### vela

Staff Emeritus
Yes, your work sounds good so far.

8. Dec 7, 2013

### Apothem

So in order to calculate the maximum weight that can be placed at one end without the plank tipping over, I have taken moments about trestle B:
F*0.5=200*1 (where F is the maximum weight)
Thus after rearranging, I have found that F=400N, is this correct?

9. Dec 7, 2013

### vela

Staff Emeritus
Yes, that's correct.