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26-10-2005, 02:56 AM
| #21 | |
| Guardian Angel | Quote:
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27-10-2005, 09:26 PM
| #22 |
| ANBU |
Would anyone mind explaining how to do this? I'm completely stumped by it. You're pushing a 122 kg crate at constant speed up a frictionless inclined plane of q = 36 degrees with the horizontal. Assuming you push with a horizontal force, what is the magnitude of the force F required to move the crate? I found out the answer is 870N. But I have no idea how to get it...any help would be appreciated |
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27-10-2005, 10:46 PM
| #23 |
| Kazekage |
I got the answer. Let me draw you a diagram. The green represents the coordinate system that I chose for the diagram. The arrows show the positive directions. With these types of problems, you sum the forces in the x and y directions and assume equilibrium. However for this question, you're only concerned about the force in the x-direction. Summing the forces in the x-direction: -W*sin(36) + Fx*cos(-36) = 0 You already know that W = m*g = (144 kg)*(9.81 m/s^2), so just plug that in, and solve for Fx.
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27-10-2005, 11:23 PM
| #24 |
| ANBU |
Woo, thanks. But one more question >.>How do you know when to use sin or cos in each situation? I really don't see an "opposite" to the 36 degree angle on W >.> I'm so confused >.< |
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27-10-2005, 11:24 PM
| #25 |
| ANBU Join Date: Nov 2004 Location: where the alcohol is cheap
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Nicely done. You could have also shown the resultant force of gravity against F which would be g x tan 36
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28-10-2005, 01:35 AM
| #26 | |
| Kazekage | Quote:
Like below, I've shown in pink the "imaginary" components of your forces. All forces have an x and y component, and that is represented by trigonometry. The pink below the y-axis illustrates the x-component of the weight. You can tell it's the x-component because it is parallel to the x-axis. This is one way to imagine the x-component of forces, and it answers your question about seeing the side "opposite" to the 36 degree angle. You can move the pink line on top of the x-axis and use the cos(90-36), and that will produce the same answer. The "opposite" of 36 degree angles won't always give you x-components. In this same problem, the "opposite" of another 36 degree angle gives a force in the vertical direction. Make sure that your free-body diagrams are clear and that you are familiar with trigonometry. Also, choose your coordinate system that's convenient for your situation. There is no "right" or "wrong" coordinate system; just make sure that you are consistent with your sign conventions. That will make these statics problems easy to solve.
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28-10-2005, 01:46 AM
| #27 |
| ANBU |
Bleh, ok...now I see...I think it was just the way you drew the diagram. I'll put the way I usually see them up when I get a chance I suppose. But thanks for the explination. It really helped |
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28-10-2005, 05:04 PM
| #29 |
| Hunter-nin |
how come?
__________________ -^^is from Farre ty ty lots |
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28-10-2005, 11:50 PM
| #31 | |
| A://Stocks | Quote:
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30-10-2005, 12:55 AM
| #32 |
| Hunter-nin Join Date: Apr 2005 Location: in the mountains
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I want to write a very very short program in C++. It will read input from a parallel port and print out the status onto the screen. As practice, I've got output working, I use outportb(0x378,N); and can turn on 8 LEDs that are each connected to a separate output pin, 2-9. I tried using inportb(0x378); to get the input pin status but that just gives me the status of the output pins. Can anyone give advice as to how to connect input to the port connector (which pins?) and how to read the status of the input pins, whichever ones they are? EDIT: I suppose no one has a clue :( Last edited by RagingVortex; 30-10-2005 at 11:22 PM. |
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31-10-2005, 06:27 AM
| #33 |
| Administrator |
i've done C++ last year , but it was Borland so quite different + we never had to do anything like that. sorry peep.
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31-10-2005, 05:45 PM
| #34 |
| Hunter-nin |
i am not good with computers..sorry=(
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03-11-2005, 06:27 AM
| #35 |
| Kazekage |
how do u write this formula out? cobalt (III) permanganate?
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03-11-2005, 07:01 AM
| #36 |
| Life = Sex + Guns Join Date: Oct 2004 Location: Red Light District
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Co(MnO4)2 i think that is it |
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03-11-2005, 01:55 PM
| #37 |
| Elite Jounin |
Help!!! --> Java urgent! i need to write 2 programs and i have no idea how to do them. one has to take a user input of an integer and print a right justified pyramid made of "[]" with as many rows as the user entered has to print this: [] [][] [][][] [][][][] [][][][][] the other is the same but has to print: [] [][] [][][] [][][][] [][][][][]
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03-11-2005, 11:05 PM
| #38 |
| Banned |
when do ya need em by? i can ask my frind
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04-11-2005, 12:28 AM
| #39 |
| ANBU |
Well, here's an old lab I did. Code: public class Nested3
{
public static void main(String[] args)
{
for(int r=0;r<5;r++)
{
for(int c=3;c>r-1;c--)
{
System.out.print(" ");
}
for(int b=0;b<r+1;b++)
{
System.out.print("[]");
}
System.out.println();
}
}
}
EDIT: Sorry I'm drawing a blank =\ If I think of something I'll add it It's a bit simple since I did it in my first year so hope this helps. Last edited by Raug; 04-11-2005 at 12:38 AM. |
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04-11-2005, 04:25 PM
| #40 |
| Elite Jounin |
thanks that helps, i modified it to how i liked it, thx mucho
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