Hi please help with solving the equation to two unknowns here for some reason I'm not getting it. I know the answer I just need the working to understand it please
I have \[\sum_{}^{} F _{x}=0; 0 = WC \cos 30^{o} - 1.375 Cos \theta \]
\[\sum_{}^{}F _{y}=0; 0=WC \sin30^{o} + 1.375 \sin \theta - 1.5\]
Im stuck at solving for the angle and the WC, my answer is always wrong :(
can u specify which problem is it? 3-12 or 3-13?
sorry 3.13
its been a while since i did this so my memory is vague, i will need ur help if ok? what's WC?
weight at C :) ty
let me google a bit
u need to have to sum of forces, one over the X axis and the other over the Y axis, let me think more
two sums not 'to'
@miszery can we say over the X-axis the following? 1.375 cos theta = Wc cos 30
yes
so that's ur 2nd equation then Wc = (1.375/cos30) . cosTheta and can u substitue Wc above in ur first equation to find Theta?
i try to but I get sin theta and cos theta and Im confused how to solve equation with two different trig functions :(
isn't sin = 1/cos or am I wrong?
I forgot the relation but there was one between the 2 I think one is sin2 + cos2 = 1 ??
i dont know sorry I can't remember the trig identities :S
you may be right i'll have to chech that out
@ganeshie8 hi, do u know much about forces in physics ? kind of stuck on this :(
horizontal components : \(1.375*\cos(\theta) - C*\cos(30) =0\) looks good to me
solving the 2 equations over X and Y axis to find Theta and Wc is the hard bit
vertical components : \(-1.5+ 1.375*\sin(\theta) + C*\sin(30) = 0 \)
do u remember the triangle method for equilibrium ?
that's correct, both @miszery and I got that but can progress any further
let me look it up
me too...
equation I Wc + 2.75 sinTheta = 3
1.588 cos T + 2.75 sin T = 3 , where T = Theta what is Theta (T) ?
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