What forces act upon this system and where?

[nickdwaters]

Given the following description: A 10 kg board 1 meter long rests on a table with 0.3 meters extending beyond the edge. I expect two forces: The normal force on the board from the table, and the weight of the board upon the table. The board is in equilibrium.

I think these forces occur at the center of gravity which is -0.2 meters from the edge of the table (in the middle of the board at 0.5 m). Assuming even weight distribution within the board, I expect 5 kg of board mass to occur on either side of the center of gravity.

Continuing with the description, a 3 kg box is placed on the overhanging portion of the board. At what distance from the edge of the table will the box cause the board to fall?

This is where I get confused. The pivot point will be the edge of the table. But the center of gravity is at -0.2 m from it. Is a normal force present at the pivot point? Do I calculate a separate center of gravity for the portion of the board not in contact with the table?

I've been staring at this problem for hours and reading the text and notes. There is something fundamental I do not understand.

 

[azizlwl]

You can also look at the possibilities of rotation.

 

[PhanthomJay]

Given the following description: A 10 kg board 1 meter long rests on a table with 0.3 meters extending beyond the edge. I expect two forces: The normal force on the board from the table, and the weight of the board upon the table. The board is in equilibrium.

yes

I think the resultant of [/color] these forces occur at the center of gravity which is -0.2 meters from the edge of the table (in the middle of the board at 0.5 m).

OK.

Assuming even weight distribution within the board, I expect 5 kg of board mass to occur on either side of the center of gravity.

yes

Continuing with the description, a 3 kg box is placed on the overhanging portion of the board. At what distance from the edge of the table will the box cause the board to fall?

This is where I get confused. The pivot point will be the edge of the table. But the center of gravity is at -0.2 m from it.

yes

Is a normal force present at the pivot point?

yes

Do I calculate a separate center of gravity for the portion of the board not in contact with the table?

Not necessary. When the board is about to tip, all the normal force occurs at the pivot point at the edge of the table, because the entire rest of the board is about to lose complete contact with it. Try summing moments about that pivot point.

I've been staring at this problem for hours and reading the text and notes. There is something fundamental I do not understand.

At the tipover point, the normal force occurs at the pivot.
Note: You have good communication skills.