# Homework Help: Torque. I thought i understood it. Obv I don't.

1. Nov 7, 2008

### DDRchick

Two people are holding up the ends of a 4.70-kg wooden board that is l = 1.80 m long. A m = 8.00-kg box sits on the board, 0.50 m from one end, as shown below. What forces do the two people exert?
Fright person = ?
Fleft person =?

Okay so this is what i did:
I drew a diagram and found out that each half of the board is 0.9 m. The box is 0.5 m away from the end of one side, so i subtracted 0.5 m from 0.9, which left me 0.4 m from the box to the center of gravity.
4.7 + 8 = 12.7 kg <---Total
1.8-0.5=1.3
0.9(4.7)+1.3(8)=14.63
14.63=1.8x
=8.13 N

To find one side I did:(I think it's left, but i put in the answer for both sides anyway)
12.7-8.13=4.57

It was of course, marked wrong. I redid it to see if i might've typed something in my calculator wrong...everything was "right"...meaning I did the whole process completely wrong.

2. Nov 7, 2008

### mgb_phys

The centre of gravity isn't important.
Imagine that it's pivoted at one end, workout the torque at the far end, then do the same for the other end. This givesyou the ratio of the forces at each end.
You know the total downward force.

3. Nov 7, 2008

### DDRchick

12.7 is the total but i still dunno how to find the torques at each side...
I tried 8(1.3)=0.9x
which gave me 11.6.
It wasn't right.
This is problem is probably ridiculously easy too...

4. Nov 7, 2008

### mgb_phys

Assume a pivot at end 1, there is a torque down for each of the loads which is just distance from end 1 * weight (the weight of theboard is at it's centre )
To hold these up you would need a torque up acting at the other end of the board = F * length of board.

So you have F2 * 1.8 = 0.5 * 8g + 0.9 * 4.7g
And similairly from the other end to work out the upward force F1 at end 1.
So you have a ratio of F1/F2 and a total force F1+f2 which you know.
it's then simple to get F1 and F2

5. Nov 8, 2008

### DDRchick

Okay so you put the pivot on the side with the box right?
You multiplied 4.7*0.9...why? I thought you always did distance away from the pivot...?
You also multiplied 0.5*8 because that would be the box...this is where (i think) i found that you put the pivot on that end..(I think)
Sorry I'm causing you trouble, I just wanna make sure I understand this. Thanks again :)

6. Nov 8, 2008

### mgb_phys

You have to do the calc for a pivot at both ends.

Put a pivot at the very end so it's distance is zero.
Then consider all the other forces.
The box is 0.5m from one end and has a force 8g down.
The board acts at it's centre, 0.9m (ie 1.8m/2) from the end and a force of 4.7g
We then work out what force F you would need at the other end at a distance of 1.8m

So from end 1, we have: 0.5m*8g + 0.9m*4.7g = 1.8m * F(other end)

From the other end the calculation is the same except that thebox is at a distance of 1.8-0.5=1.3m. so 1.3m*8g + 0.9m*4,7g = 1.8m * F(first end)

7. Nov 8, 2008

### DDRchick

Oh. Hmm so that would be...
0.5*8+0.9*4.7=1.8*F
=24.57
1.3*8+0.9*4.7=1.8*F
=8.13
It's still saying they're wrong. :/
I tried this again earlier this morning and i added in a whole thing of F=mg and finding the downward force/upward force...it was still wrong lol. Ugh. stupid webassign. >:/

8. Nov 8, 2008

### mgb_phys

Thats' only the first step to get the ratios F1/F2 you then need to separate them.
F1/F2 = (0.5*8g + 0.9*4.7g) / ( 1.3*8g + 0.9*4.7g) = 8.23g/14.63g = 0.56
(depending on which you labelled one and two of course)

You also know that F1+F2 = 8g + 4.7g = 12.7g so it's simple to get F1 and F2

BUT before you hit the calcualtor - you need to know what answer you expect.
If the box was in the middle then the two forces would be equal, and since the total force is 12.7g it would just be 12.7g/2 = 6.35g each.
But the box is about half way between the middle and one end, so one force is going to be about 50% more than 6.35g and one will be about 50% less.

9. Nov 8, 2008

### DDRchick

Ohh okay. That's easy then! C: Thanks so much! Sorry for putting you through a lot :/