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|dw:1350335809666:dw|

|dw:1350336122902:dw|

where is fn then

What is fn

normal force

and how did u get t cos 65 because that is y coordinate right so it would be t sin 65 ?

can u plz help me with this

what do u mean by what angle

yes i know that cos =... and sin=.. but ..

|dw:1350337034801:dw|

X = T cos(65), right?

yuppppp

thx

i was doing other way but same

glad to help. I'm pretty out of practice on these, but the diagram helped me remember a little.

|dw:1350337344014:dw|

yes, that's what I get. I have to take off, just signed on for a minute... good luck!

Tcos 65=mg

Yes, you can find T then.

then how to find the velocity

-T sin 65= mv^2/r

Correct :)

got it thx u

hold on what is the answer for tension

5102

but book gave 5.1*10^3N

\[T=\frac{mg}{\cos65}=5101N=5.1\times10^3N\]

and the last one velocity |dw:1350338528028:dw|

sorry,\[v=\sqrt{\frac{Tr\sin65}{m}}\]

how can i omit the -