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# Problem of Acceleration of a point on a moving wedge ... watch

1. Well folks I am not sure how to get to the answer. I'll post the question then my working out. The question is trying to find the acceleration of a point on a piece of string as the triangular wedge underneath moves with constant acceleration.
Attachment 694434694436

My attempted working out on the next thread.

EDIT: It wants you to find out the acceleration at the point B. The mass is not significant (numerically or to the question hence -> ), but think of it more as a placeholder.
Attached Images

2. Yeah it's not great. I understand how the wedge moves with respect to time, and I understand the ratio of how it will travel. As the wedge moves across, the particle will presumable want to move in the opposite direction at the same rate, and with components as I discussed.

Anyway I'm clearly getting something wrong and I'd really appreciate some input / advice.

Tagging some of the most helpful guys I know:

EDIT: I know that I can get a*t^2 * sec (x) /2 as my displacement, and then double differentiate that but that's not the answer in the back of the book.
3. (Original post by DrSebWilkes)

Yeah it's not great. I understand how the wedge moves with respect to time, and I understand the ratio of how it will travel. As the wedge moves across, the particle will presumable want to move in the opposite direction at the same rate, and with components as I discussed.

Anyway I'm clearly getting something wrong and I'd really appreciate some input / advice.

Tagging some of the most helpful guys I know:

EDIT: I know that I can get a*t^2 * sec (x) /2 as my displacement, and then double differentiate that but that's not the answer in the back of the book.
Hi.

I don't have time to work through this now since I am going away today in ~30 mins, but I would note that the string is inextensible, so at any time, the displacement of the particle up along the slope (ie the diagonal component) will be equal to the horizontal displacement of the wedge from the starting position (assuming the string is horizontal in the first portion).

You can rationalise this by noting that as the triangle moves, extra string will be overhanging horizontally, and as the string is inextensible then this is how much less string is then going down the slope.
4. (Original post by DrSebWilkes)
...
Load moves up the slope with same acc as wedge, as they cover equal distance at time t. Find load's resultant acc wrt ground (a*):

Horiz a* = a - [email protected] = a(1 - [email protected])
Vert a* = [email protected]

(a*)^2 = a^2[(1 - [email protected])^2 + sin^[email protected]]
= 2a^2(1 - [email protected]) = 4a^2sin^2(@/2)

=> a* = 2asin(@/2) ; 0° < @ < 90°
5. (Original post by K-Man_PhysCheM)
the displacement of the particle up along the slope (ie the diagonal component) will be equal to the horizontal displacement of the wedge from the starting position (assuming the string is horizontal in the first portion).
Yeah I noticed the whole string thing and that it obviously wouldn't just go straight up. Maybe I overcomplicated it? Hmm anyway yeah thanks for chipping in. Got one further question, if it's not answered, if that's alright?

(Original post by Physics Enemy)
Load moves up the slope with same acc as wedge, as they cover equal distance at time t. Find load's resultant acc wrt ground (a*):

Horiz a* = a - [email protected] = a(1 - [email protected])
Vert a* = [email protected]

(a*)^2 = a^2[(1 - [email protected])^2 + sin^[email protected]]
= 2a^2(1 - [email protected]) = 4a^2sin^2(@/2)

=> a* = 2asin(@/2) ; 0° < @ < 90°
Whoooo nice thanks! I can't believe you did that in so few lines of working hahaha me trying to integrate ... eughh

Anyway what I want to know now, given I can see the proper - and easy way - to do it, if you were given it in terms of displacement, like how I found it, how would it be done? OR perhaps really what I'm trying to ask is why when I split the displacement into how the particle would move horizontally and vertically did it not work?

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