Physics questions...

[Zeeky Boogy Doog]

Well, I was working on the second problem, and I think I had it, and my teacher made it due two hours earlier that it's supposed to be, and when he said it was due, which sucks for me, and everyone else in the class I immagine :laugh:
bet he'll be catching a lot of crap over this one.

1. A van accelerates down a hill, going from rest to 27.0 m/s in 6.00 s. During the acceleration, a toy (m = 0.700 kg) hangs by a string from the van's ceiling. The acceleration is such that the string remains perpendicular to the ceiling. Determine the angle theta and find the tension in the string.

so, i've got that the acceleration of the car is 4.5 m/s^2, the tension in the verticle is 6.86N (equal to force of gravity on the toy) and thats it. I thought that the tension in the horizontal was m*a, but when I use that and find for the Tension of the string or the angle it is hanging from verticle, it says I'm wrong. Any hints on how to find the Tx would be greatly appreciated.

EDIT: Figgured this one out for sure, I was using 4.5 m/s^2 as a straight leg along the horizontal axis, and gravity as the other leg instead of the hypotenuse as it was meant to be used (the answers were 27.334 degrees and 6.094N tension on the string, incase you were wondering). I think I have the other one now as well, I haven't worked it yet though.


2. A simple accelerometer is constructed by suspending a mass m from a string of length L that is tied to the top of a cart. As the cart is accelerated the string-mass system makes an angle of theta with the vertical.

Assuming that the string mass is negligible compared to m, derive an expression for the cart's acceleration in terms of , and show that it is independent of the mass m and the length L. (Answer using g for the acceleration due to gravity, theta for , and L, and m, as needed.)

I was completely lost on this one for a little while, but I came up with Ty=Tcos(theta) = g*m and Tx = Tsin(theta) which, from what I understood from my teacher was m*a, so I just went through the problem again and realized I completely lost again, yay...

anyway, thanks in advance for any help.

If you can give me specific help, thats great, but, they're really quite similar problems, all I'm having trouble with is finding the tension of the rope in the x direction while it is at an angle from verticle, if you can point me in the direction of how to do this, (or tell me that it is infact m*a, and the computer is wrong) that would be all I need.

 

[stan394]

why don't you just do your own homework?

 

[Zeeky Boogy Doog]

Originally posted by: stan394
why don't you just do your own homework?

lol, I'm trying, i got the other 24 points... these two questions have me completely stumped

 

[stan394]

i just re-read your questions, i may be able to help you 7 years ago while i was still in college. but now.. sorry

 

[Zeeky Boogy Doog]

Originally posted by: stan394
i just re-read your questions, i may be able to help you 7 years ago while i was still in college. but now.. sorry

:laugh:
thanks anyway

I really think I can get both of these questions, it just boils down to finding the Tension in the X direction, on both of them, and like I said before, from what I understood from my teacher, Tx is m*a, but using that tells me I'm wrong, so...

 

[gsethi]

I took my last physics class couple of years ago..so kinda rusty but I think you have to use this:

Tx = F * Cos (theta)
Ty = F * Sin (theta)

remember....F = m * a

hope that helps

 

[Gibsons]

I think you have the wrong tension for the vertical component on the string? If the van were to accelerate downhill at 9.8m/s2, then the vertical tension in the string would be zero, right?

 

[Zeeky Boogy Doog]

Originally posted by: gsethi
I took my last physics class couple of years ago..so kinda rusty but I think you have to use this:

Tx = F * Cos (theta)
Ty = F * Sin (theta)

remember....F = m * a

hope that helps

I think thats what I was trying, perhaps I need to use a uniform acceleration though, I was using gravity for the y component and the acceleration of the car/cart as the x component? ok, I'm giving that a try on the car problem, that might be it...

 

[gsethi]

for acceleration, you have to treat them as Ax and Ay also since it it also a vector (I think Acceleration is a vector).

so:
Ax = A Cos (theta)
Ay = A Sin (theta)

vertical acceleration will be gravity + Ay component of the van's acceleration.
horizontal acceleration will be Ax component of the van's acceleration.

do the same thing with force...Fx and Fy.

TIP: always treat 2 dimensions separately. If you use x and y any place, keep using them separately till the end and combine them only at the end. Dont go back and forth.