1. Jul 27, 2015

### Orex

1. The problem statement, all variables and given/known data
Hello, I attempted solving a problem from my practice sheet, but got stuck on part B. If my part A is incorrect, please let me know!

2. Relevant equations
Not sure which equations to use!

3. The attempt at a solution

I don't think the solution is too difficult for part B, but I'm stuck on not knowing if part A is correct, so i cannot move on. Any help, and an attempt at the solution would be appreciated! :)

2. Jul 27, 2015

### SteamKing

Staff Emeritus
The ∑Fy equation looks OK.

For writing the torque equation, you should pick a reference point to use to calculate torque (or moment).

Since the problem wants you to find how far Bob can walk toward the right end of the plank before it starts to tilt, a good choice of reference would be to use the location of the right sawhorse.

When writing moment (torque) equations, it's understood that T = F × d, so go ahead and substitute any actual forces or distances into your moment equation. Terms like "TA" can be confusing, to you and others.

3. Jul 27, 2015

### Orex

Here is my second attempt. Is my torque equation right?
I know 8.333...m cannot be the right answer, it does not make sense. What am I doing wrong? Any help would be appreciated!
Also Note: I believe the mass of A and B are given, 50kg and 90kg respectively.

4. Jul 27, 2015

### andrewkirk

You are not calculating the torque from the plank's weight correctly. In your equation, you calculate it as though the weight of the 5m section of plank to the left of the right pivot is all applied at A. It is not. You need to work out the torque $\tau_{left}^{plank}$ of the left-side plank's own weight by integrating:

$$\tau_{left}^{plank}=\int_0^5 l\cdot g\cdot \frac{200kg}{9m}dl$$

You then need to do a similar integration to get the opposite torque of the weight of the right-hand section of plank.

Then add in the torques of Anna and Bob.

5. Jul 27, 2015

### SteamKing

Staff Emeritus
You calculated the moment due to mass A correctly.

Why did you choose the moment arm of the plank's mass as 5 m from the second sawhorse?

Why did you assume the direction of the moment due to the plank's mass is opposite that of mass A?

Why did you assume the direction of the moment due to mass B was the same direction due to the moment due to mass A? Aren't they acting in opposite directions?

For future problems, assume moments acting counter-clockwise are positive.

6. Jul 27, 2015

### SteamKing

Staff Emeritus
There's no need to get this complicated.

The plank is uniform, and its center of mass should be easy to find without resorting to integral calculus.

7. Jul 27, 2015

### andrewkirk

Good point. I hadn't noticed that.

8. Jul 28, 2015

### Jobrag

Without knowing the mass of A and B or at least the ratio of their masses I don't see how no you can resolve this, sorry I hadn't read steamking's comment before posting

9. Jul 28, 2015

### SteamKing

Staff Emeritus
Apparently, the OP neglected to mention that MA = 50 kg and MB = 90 kg in the original problem statement. These figures appear in Post #3 in his calculations.