see here for a situation of the problem http://i40.tinypic.com/6fbblk.jpg
The Attempt at a Solution
Ok well I've solved this problem but one thing about the solution process troubles me.
Namely when trying to find the force in the bar...
Well what you're trying to do is find the total distance to the second train station.... so you're told the train travels at this acceleration you've found for another 135s, you need to find the change in distance with this acceleration.
2pi to 5 decimal points is 6.283185.
and the period is 6.283185, so the period is 2pi!
I'm looking at the problem statement and cannot see why the max allowable range wouldn't be 0-90, given that we are looking for the max change in distance x.
the restriction on the possible angle would i assume range from 0-90, or 0 - pi/2?
I don't get what you're saying when you say divide by 2n.
I divide theta ≈ 0.74559 + 6.28319n by 2n? I don't see how that would help me see where the angle lies.
Ok well I had a real nasty equation in which i could not solve.
I used wolfram alpha to get out this answer "theta ≈ 0.74559 + 6.28319n for integer n".
can someone please tell me what this statement means.
do i assume theta in degrees equals 0.74559(180/pi)...
Ok well I've given up on trying to solve this analytically. Just one last question - i'm using wolfram alpha to get the final answer which they give as, "theta ≈ 0.74559 + 6.28319n for integer n".
can someone tell me what this statement means.
do i assume theta in degrees equals...
Ok so this problem is still giving me some trouble. My current attempt reads as follows:
do a force balance at position c
2T + (Mc)*gcos(∅)*Us - (Mc)*gsin(∅) = 0
do a force balance at A,
sub back into equation 1,
then using a/√(a2+b2) =...
mmm still getting confused with this question. So i have x as a function of theta, but i'm getting 44 degrees to give me higher distance then 45. Am i meant to take the derivative of this function and set it to zero then solving for theta?
Ok now I've assumed we are talking about static friction my final attempt is as follows:
and T =1226.25
but solving the first equation i can't see how to go about this because the minus sign doesn't allow me to use that trig identity you kindly suggested earlier.
Ok i'm getting confused when trying to solve for time t.
Δy =-16, so
Vosin(∅)t + (1/2)(g)t^(2) +16
but when solving for t how do i know which one to use resulting from the quadratic equation?
if i use (-Vosin(∅)/g + (Vosin(∅))^(2) -4(1/2)(g)(16))^(1/2)/g i get the max angle at 45...