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d0minicz · « **on:** October 26, 2009, 11:13:49 pm »

Hey for VCAA 06 ! exam 1; Question 4.b)

im a bit confused with how to work without the angle given; i mean i get the working out... just T being slanted without an angle throws me off
how do we do these if none is given?
is that just the diagram? and T is the same despite the angle ??

thanks

/0 · « **Reply #1 on:** October 26, 2009, 11:25:33 pm »

You can choose the angle for yourself. It can be the angle with the horizontal or the vertical.

If $\theta$ is the angle with the horizontal then
$F_x=\frac{g}{2}-T\cos{\theta}=0$
$F_y = mg - T\sin{\theta}=0$

Hence,
$T\cos{\theta} = \frac{g}{2}$
$T\sin{\theta}=mg$

$\Rightarrow \left(T\cos{\theta}\right)^2+\left(T\sin{\theta}\right)^2 = \left(\frac{g}{2}\right)^2+(mg)^2$

$\Rightarrow T^2(\cos^2\theta+\sin^2\theta) = g^2\left(\frac{1}{4}+m^2\right)$

$\therefore T^2 = g^2\left(\frac{1}{4}+m^2\right)$

If $\theta$ is the angle with the vertical then
$F_x=\frac{g}{2}-T\sin{\theta}=0$
$F_y = mg - T\cos{\theta}=0$

And the working to find T is the same.

anonuser0511 · « **Reply #2 on:** October 26, 2009, 11:40:27 pm »

lol you didn't say which year this exam was from - 06 but i guess you kids know them all huh.

i think using Pythagoras would be an easier way to solve

=> (g/2)^2 + (4g)^2 = T^2
T = sqrt(g^2(1/4 + 16))
=g*sqrt(65)/2

/0 · « **Reply #3 on:** October 26, 2009, 11:49:13 pm »

Ah yes it is... I've never been a fan of vector triangles, but I guess it works well

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