# Search results

1. ### Problem involving Newton's second law

Ah i see now - its stated in the problem that it is to the horizontal! thanks for that.
2. ### Problem involving Newton's second law

thanks for the prompt reply, however i don't understand what you mean stated to the horizontal can you please elaborate for a slow poke like me. to me it looks like B is hanging from the vertical.
3. ### Problem involving Newton's second law

Homework Statement see here for a situation of the problem http://i40.tinypic.com/6fbblk.jpg Homework Equations F=ma 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...
4. ### Help Laws of Motion and question help

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.
5. ### Help Laws of Motion and question help

Yes that is correct however you know that u=0 right? because as you said earlier the train starts from rest, and u represents the initial velocity of the train.
6. ### Help Laws of Motion and question help

as i said before use this equation..... s= ut + 1/2at^2 you know what s, u, and t are can you not solve the equation for a?
7. ### Help Laws of Motion and question help

that is correct.
8. ### Help Laws of Motion and question help

use the equation you stated, s= ut + 1/2at^2
9. ### Help Laws of Motion and question help

well the question states that the train travels 225m in 30s - try using these.
10. ### Help Laws of Motion and question help

Can you find the acceleration over the 225m distance given you have time?
11. ### Work-kinetic energy theorem

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.
12. ### Work-kinetic energy theorem

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 thanks for clearing that up guys.

So if i was solving an equation say something like x'(theta) =0 would n be zero when solving for theta?

Homework Statement 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)...
16. ### Work-kinetic energy theorem

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...
17. ### Newton's Second law problem

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, T=1226.25 sub back into equation 1, -3924sin(∅)+1962cos(∅)=-2452.2 then using a/√(a2+b2) =...
18. ### Work-kinetic energy theorem

I'll give it a go, thanks Simon for the help!
19. ### Work-kinetic energy theorem

Are you saying i need to take the derivative of the function? because even if i do, i do not have the required math knowledge to solve it at x'(theta) =0.
20. ### Newton's Second law problem

haruspex don't you mean c/√(a2+b2) = cos (θ+ψ)? nvm i'm confusing myself.
21. ### Work-kinetic energy theorem

I used wolfram alpha to graph the x v theta graph and its giving me the maximum at 41.79. how do i find this value analytically though, thanks!
22. ### Work-kinetic energy theorem

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?
23. ### Newton's Second law problem

Ok now I've assumed we are talking about static friction my final attempt is as follows: 2T+Mgcos(a)-Mgsin(a)=0 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. Any...
24. ### Constrained motion problem

yep, i understand now haruspex, thanks!
25. ### Constrained motion problem

How did you know that x was a function of time?
26. ### Work-kinetic energy theorem

ah i see, i was thinking that if i change the angle then h changes. Thanks for clearing that up Simon.
27. ### Work-kinetic energy theorem

That was for the case where the angle was 30 i.e the angle was given.
28. ### Work-kinetic energy theorem

How can i calculate the distance though, without already assuming the angle's value, which also leads me to the question of how can i tell which t is nonsense, without knowing Vo and the angle theta?
29. ### Work-kinetic energy theorem

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...
30. ### Polar coordinates question

that it should be set to radians lol, my bad its late here.