Dekoyl's Questions

[kamil9876]

lol TT great minds think alike. That's how I did it back in the days. However I think there are quicker ways. Dekoyl's way of v(t).k would work to, this is basically finding the angle between the two vectors and , which can be done using dot product or TT's way of finding the magnitude of the two vectors, and and using trigonometry.

[Mao]

Hmm. Okay, the same question but...
what if it didn't say "with the north direction"? What if it just wants you to find the angle to the ground at which it crashes at? How would you approach that?

Would you dot v(t).k ? And then 90 minus the angle you find from v(t).k?

Thanks again.

That is correct.

Both ways are equally as valid, but I find this way more.. 'fool-proof'

[dekoyl]

I don't understand how the k is obtained below.

"The numerical solution of when , given when , is

Why is the k not within the integral.. or why is there even a +k at all?

Thanks!

Thanks TT. I understand now

[TrueTears]

That's a rule you should remember, it pops up a lot in exams.

But to see the deviation check your essentials book pg 346

[dekoyl]

I used the cosine rule for this question but is there another way (without using sine/cosine rules)?

"A particle is acted on by a force of magnitude acting in a south-westerly direction and by a second force of 7N acting due south. Find the magnitude of the resulting force."

Thanks!

[TrueTears]

I'd draw the cartesian axes and then componentise the forces.

Consider the N force.

Let positive x direction, positive y direction be i and j respective.

Thus in the i direction:

N

Consider the 7 N force.

Thus in the i direction:

0 N

Now consider the N force in the j direction:



Consider the 7 N force in the j direction:



Total force in the i direction:

Total force in the j direction:

Total force: 

To find the magnitude just let

Thus magnitude

[dekoyl]

Awesome Thanks TT

[kamil9876]

Quite nice how there are two different ways to solve it. Often in mathematics if you find two different ways of solving the same problem, or counting the same collection of things, you can find a cool relationship between the two ways. I first noticed in some thread a month or so ago that by "equating" these two methods you can find a simple proof of the cosine rule .

[TrueTears]

Quite nice how there are two different ways to solve it. Often in mathematics if you find two different ways of solving the same problem, or counting the same collection of things, you can find a cool relationship between the two ways. I first noticed in some thread a month or so ago that by "equating" these two methods you can find a simple proof of the cosine rule .

Just like the way you showed me that "imaginary" parabola question.

[Your way owns though.]

[dekoyl]

When you've got a system like this:

and you want to find the amount of force required so that the crates are just about to move, why don't you need to consider crate 2?

Thanks

[TrueTears]

If you are talking about the WHOLE system, then yes you need to consider it.

If say just Crate 1 or just Crate 2, then you don't.

[dekoyl]

If you are talking about the WHOLE system, then yes you need to consider it.

If say just Crate 1 or just Crate 2, then you don't.

Well this is the question:
"Whats the minimum pulling force P (still at 30 degrees to the horizontal) that you would need to apply so that the crates are just on the point of sliding?"  So I'm guessing that's the WHOLE system?

[TrueTears]

Yeap, notice how it says crates

meaning the whole system XD

[dekoyl]

Ah I thought the answers were wrong.

I thought the answered only resolved around crate 1 (but the solutions are loooong) so I didnt see they resolved around crate 2, too.

[dekoyl]

In a question, I'm told to solve . But I turned it to when solving it.

So, Instead of getting as the answer wanted, I ended up with .

I guess the problem was at one point, I had (note modulus).
How would I overcome this problem if I had decided to put the negative into the ?

Thanks.