# Torque - Where is the normal force in this FBD?

Silly question I know, but I want to just clarify this.

The actual question:
A 0.45kg bug is crawling to the right along a metre-stick which is resting on the edge of a desk. If the mass of the uniform metre-stick is 0.37kg, how far from the edge of the desk can the bug crawl before the metre-stick flips off the desk?

[PLAIN]http://img842.imageshack.us/img842/9154/58811064.jpg [Broken]

My question:
Eventually, I worked out that FN is the green arrow. But why is it there instead of being at where the red arrow is? I realize the entire table is the support. So should it not be in the middle of where the support touches the metre-stick?

I also know found no way to solve this without putting FN in the middle. I got the answer, but I want to clarify this and double-check.

[PLAIN]http://img194.imageshack.us/img194/9067/24168499.jpg [Broken]

In case I am not correct by putting FN in the middle, the answer I got is 0.1644m. Is that right?

Many thanks!

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the normal force should be where the center of mass is.

Hi MHrtz, thanks for the reply.

Do you mean this then?

[PLAIN]http://img153.imageshack.us/img153/2333/95048564.jpg [Broken]

I can't solve the question if it is this, because any weight on the right side will tip the ruler - which isn't suppose to happen.

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Okay, I after further pondering, I realized that FN could range from mg to 0. So that means the bug could only have as much torque as mg.

So the answer would be 0.16m. Am I right or still totally wrong?

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PhanthomJay
Homework Helper
Gold Member
You don't indicate how far from the edge of the table that the meter stick is projecting. If initially, with the bug at the edge of the table, the stick was delicately balanced half on the table, and half off, the bug could not move an iota of a mm from the edge of the table, without tipping the stick over. What's the overhang distance?

My apologies...I forgot to include that 0.70m of the stick is in contact with the table. Sorry! I am not missing any other information though, I double-checked to make sure.

PhanthomJay
Homework Helper
Gold Member
Okay, I after further pondering, I realized that FN could range from mg to 0. So that means the bug could only have as much torque as mg.

So the answer would be 0.16m. Am I right or still totally wrong?
You are right. But the normal force is always equal to the stick's plus bugs weight...its resultant location varies as the bug moves to the right, ultimately ending up as a concentrated force at the at the edge of the table at the point of pending tipover.

You are right. But the normal force is always equal to the stick's plus bugs weight...its resultant location varies as the bug moves to the right, ultimately ending up as a concentrated force at the at the edge of the table at the point of pending tipover.

Ah! I get it now. Thank you very much :)

normal always acts through the center of mass
ie the center of the rod in this case

PhanthomJay
Homework Helper
Gold Member
normal always acts through the center of mass
ie the center of the rod in this case
No. The resultant weight always acts through the center of mass...the normal (perpendicular contact force) does not always act through the center of mass.

No. The resultant weight always acts through the center of mass...the normal (perpendicular contact force) does not always act through the center of mass.
hm...you are right
normal acts at the point of contact