# Homework Help: Statics table problem

1. Aug 27, 2014

### toothpaste666

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

How close to the edge of the 26.0-kg table shown in the figure (Figure 1) can a 64.0-kg person sit without tipping it over?

2. Relevant equations

∑F = 0 ∑T = 0

3. The attempt at a solution

forces A (left leg) and B (right leg) are the forces from the two legs of the table and x is the distance of the person from the right edge of the table (what i am trying to find)

∑F = A +B - 26g - 64g = 0
= A + B = 90g
= A + B = 882

for the torques I chose leg A as the pivot point

∑T = -(.6)26g + (1.2)B -[(1.7-x)64g] = 0
= -15.6g + 1.2B - (108.8g-x64g) = 0
= -15.6g + 1.2B -108.8g + x64g = 0
= -152.88 +1.2B -1066.24 +627.2x = 0
= 1.2B + 627.2x = 1219.12

I have 2 equations but 3 unknowns. What am I missing here? please help :(

#### Attached Files:

• ###### tablestatics.jpg
File size:
7.9 KB
Views:
535
2. Aug 27, 2014

### BvU

Choosing A as pivot point indicates to me that you are after a value of x in the interval [1.7, 2.2] . At the point of tipping, what can you say about FB ?

3. Aug 27, 2014

### toothpaste666

The point of tipping? When it lifts off the ground? It would become 0 right? because the normal force is no longer pushing back up from the ground? But I thought I was not allowed to set it to 0 because I DONT want the table to tip?

4. Aug 28, 2014

### BvU

If you don't like the 0, you set it to 10-20 or something, and then take the limit to zero.
As long as it's > 0 no tipping.
How realistic do you want this to be ? The legs don't rest on spike tips, etc, etc.

5. Aug 30, 2014

### toothpaste666

I wasnt trying to be difficult. I thought about it more and I understand what you are saying now. Thanks.

6. Aug 31, 2014

### BvU

I know. Glad to be able to help.

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted