Author Topic: Dekoyl's Questions (Read 25878 times) Tweet Share

kamil9876 · « **Reply #60 on:** July 24, 2009, 10:23:45 pm »

Yeah Archimedes tells me your one would float it better

dekoyl · « **Reply #61 on:** July 26, 2009, 02:58:28 pm »

Okay... Another one =(

A block of mass 2kg lies on a horizontal table, the coefficient of friction being 0.5. Find the magnitude of the force on the block which, acting upwards at 45º to the plane of the table, produces in the block a horizontal acceleration of 2.45 $m/s^2$ .

Why (by fluke/chance) does $mgsin(\theta) = 13.859 N$ yield the answer? I know this method is most definitely wrong, and it wasn't my method - just part of the working out I did and I spotted the answer in this part.

How would I have approached the question, considering all components of the question? I'm getting extreeeemely slow for some reason.

/0 · « **Reply #62 on:** July 26, 2009, 03:15:40 pm »

$F = ma$

I would componentize in the x and y directions.

The normal force is $2g - F\sin{45^{\circ}}$

So the horizontal frictional force is $0.5(2g-F\sin{45^{\circ}})$

The horizontal pulling force is $F\cos{45^{\circ}}$

$\Rightarrow F\cos{45^{\circ}} -0.5(2g-F\sin{45^{\circ}}) = 2(2.45)$

Solving gives $F = 13.86N$

As for $mg\sin{\theta} = 13.859N$ , I think that was probably a fluke, since it does not depend on either the coefficient of friction or the acceleration.

dekoyl · « **Reply #63 on:** July 30, 2009, 06:37:02 pm »

Q: A locomotive of mass $M kg$ pulls a convoy of cars loaded on trucks where the mass of each truck is $\frac{M}{12} kg$ . The pulling force of the engine is $P$ and the greatest tension that a truck coupling can withstand is $\frac{2P}{3}$ . Find the maximum number of trucks that can be safely pulled.

Thanks.

kamil9876 · « **Reply #64 on:** July 30, 2009, 07:02:05 pm »

|=====|---->-------<---|=========|--------->
T T P

Rope doesn't stretch hence the two vehicles have the same motion => same acceleration, a:

P-T=Ma
T=ma (m is mass of second vehicle).

We wish to get an expression for T in terms of m and M. One way is to divide both equations:

$\frac{P-T}{T}=\frac{M}{m}$
$<br />P=\frac{M}{m}T + T<br />$
$<br />T=\frac{P}{\frac{M}{m} + 1}$

However $T\leq\frac{2P}{3}$ and so:
$<br />\frac{P}{\frac{M}{m} + 1}\leq\frac{2P}{3}$

$1\leq\frac{2}{3}(\frac{M}{m}+1)$

Now $m=\frac{nM}{12}$ if there are $n$ trucks

$1\leq\frac{2}{3}(\frac{12}{n}+1)$

$\frac{3}{2}\leq\frac{12}{n}+1$
$\frac{1}{2}\leq\frac{12}{n}$
$2\geq\frac{n}{12}$
$24\geq n$

So 24 trucks maximum.

dekoyl · « **Reply #65 on:** August 05, 2009, 05:41:56 pm »

=(

A body of mass $m$ falls from rest under constant gravitational force against air resistance which is equal to $kv$ when the speed is $v$ , where $k$ is a positive constant.
Find its speed at any time. Calculate the time taken to reach the speed $v_1$ where $4kv_1=mg$ .

I should be okay with this question.. it's just I'm not sure when to use the last piece of given information. =\
Thanks!

Mao · « **Reply #66 on:** August 06, 2009, 02:11:17 pm »

$a = \frac{dv}{dt} = \frac{mg - kv}{m}$

$\implies t = \int \frac{m}{mg - kv}\; dv = -\frac{m}{k}\log_{e} |mg - kv| + C$

falling from rest, :. t = 0, v = 0

$\implies C = \frac{m}{k}\log_{e} (mg)$

$\implies t = \frac{m}{k}\log_{e}\left| \frac{mg}{mg - kv}\right|$ .... (1)

$\implies v = mg - mg\cdot e^{-\frac{k}{m}t}$ .... (2)

(2) describes the velocity at any time t, however, to calculate time taken to reach v1, (1) will be used:

$v_1 = \frac{mg}{4k}$

$\implies t_1 = \frac{m}{k} \log_{e} \left| \frac{mg}{mg - mg/4}\right| = \frac{m}{k}\log_{e} \left(\frac{4}{3}\right)$

dekoyl · « **Reply #67 on:** September 15, 2009, 02:41:12 pm »

Urgh, still having problems with crashing angles.

The position vector $r(t)=sin\left(\frac{t}{10}\right)$ i $+\left(1-cos\left(\frac{t}{10}\right)\right)$ j $-log_e\left(\frac{t+1}{20}\right)$ k, $t \geq 0$ .

i is a unit vector in the east direction
j " " north direction
k" " vertically up

Q: At what angle does the path of the feather make with the north direction as it reaches the ground?

Thanks!

TrueTears · « **Reply #68 on:** September 15, 2009, 02:46:29 pm »

$v(t) = \frac{1}{10}cos(\frac{t}{10})i + \frac{1}{10}sin(\frac{t}{10})j - \frac{1}{t+1}k$

Now to find the angle between the north direction you need to dot v(t).j

Notice it is the velocity vector because it gives the direction, not the position vector.

The feather touches the ground when t = 19

So sub in t = 19 into the velocity then dot.

[Notice when it says the "path" of the feather, it is referring to the direction in which it is travelling in.]

dekoyl · « **Reply #69 on:** September 15, 2009, 02:48:24 pm »

Thanks TT but why the north direction and not the vertical direction?

Heff 05 for anyone interested

TrueTears · « **Reply #70 on:** September 15, 2009, 02:51:56 pm »

Well the question defined "j" to be north, so you dot with the j. If it defined "k" as north, then yes you'd dot with the vertical component.

dekoyl · « **Reply #71 on:** September 15, 2009, 02:54:00 pm »

Gah, I'm getting borderline retarded. First I thought you remembered the question because you mentioned feather (when I stated that it was a feather) and now I finally see that the question wants the north direction after the 15th try.

Thank you.

dekoyl · « **Reply #72 on:** September 15, 2009, 02:56:07 pm »

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.

TrueTears · « **Reply #73 on:** September 15, 2009, 03:14:47 pm »

Ahh, a similar question pop up in my SAC recently.

If it said find the angle it makes with the ground, I'd do this.

It's kinda like physics, you have a vertical component and a horizontal component of velocity.

Lets change the our velocity vector into another notation for convenience sake.

Let $v(t) = x(t)i + y(t)j + z(t)k$

The vertical component can be worked out from $|z(t)|$

The horizontal component (ie the component on the "flat" ground) is $\sqrt{x(t)^2 + y(t)^2}$ .

Thus the angle it makes with the flat ground, (let $\theta$ be the angle it makes): $tan(\theta) = \frac{|z(t)|}{\sqrt{x(t)^2 + y(t)^2}}$

Again notice how you use velocity for everything because that gives the direction in which it is traveling and thus the angle.

ryley · « **Reply #74 on:** September 15, 2009, 04:30:15 pm »

I think there was a question similar to this on the 07 VCAA exam 2 (if you want another example)

EDIT: Yep, Q4 in the extended response.

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