Recent content by Whotto

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

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.

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?

Hi MHrtz, thanks for the reply. Do you mean this then? [PLAIN]http://img153.imageshack.us/img153/2333/95048564.jpg 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.

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...

I got the answer now. I have thought about why we would change the axis for the inclined planes, but that is due to the fact we knew the Normal force is perpendicular to the surface. The confusion that I had with this problem is that there is no indication that the angle between the string...

All right, seems fairly odd to me that they would go against what teachers have taught. Thanks!

Thanks for your reply! But the diagram does not tell me the angle is 90. Wouldn't that mean I cannot assume it is 90?

Homework Statement A mass suspended by a string is held 24o from vertical by a force of 13.8N as shown. Find the mass. http://img525.imageshack.us/img525/3421/problem32.jpg 2. The attempt at a solution [PLAIN][PLAIN]http://img205.imageshack.us/img205/6100/attemptv.jpg I've split it up...

Ah, I get it now! Much appreciated and thank you for taking the time, rockfreak, collinsmark, and bhat! and yes - I will definitely keep better track of my signs.

Hi Collinsmark, It is much clearer to me now. I got 0.474 seconds with -4m/s instead. So the equation 0 = \frac {1}{2}at^{2} + v_{0}t - d would include the height that the diver has gained while taken off, even though d = 3m anyways? Thank you for the good explanation! Thanks for the...

I get what you've said so far, but I don't understand how it works into my problem. I will try ask my question in another way: Previous: http://img638.imageshack.us/img638/1486/diagram2b.jpg New situation: [PLAIN][PLAIN]http://img205.imageshack.us/img205/1645/diagram3.jpg If I were given...

Thanks for the quick reply rock freak! So then d = displacement I take it. If the diving board was right on top of the water then d = 0?

This is a projectile motion problem at a Grade 12 level in Canada. My main problem here is conceptual, not the math itself. 1. The given question A diver takes off with a speed of 8.0m/s from a 3.0-m high diving board at 30o above the horizontal. How much later does she strike the water? My...

Thanks! I got #2 with the formula T = ( 2 pi r3/2 ) / sqrt(GMe). But I still don't get #1. Can I get some more pointers?