Recent content by Yalanhar

##x[n] = (\frac{1}{2})^{-2} u[n-4]## ##h[n] = 4^{n} u[2-n]## So I plotted x[k] and h[n-k] in picture but x[n] is 0 for n < 4, therefore ##y[n]## only has value for n >= 4. Therefore my sum is like that: ##y[n]=\sum_{k=4}^{\infty} 4^{n-k} (-\frac{1}{2})^k## ##y[n]=-4^{n}...

Tnks

Well, yes. I don't know if can change dr to v(t)dt

Yes Calculate the work done by that force after time t

Is my solution correct? $$ F(t) = at + b $$$$W=\int_t F(t)dr, ~~~~~~~~~ ~~dr = v(t)dt$$$$W=\int_t F(t)\cdot v(t) dt$$ $$ f = \frac{dp}{dt}$$ therefore $$v(t) = \frac{1}{m}(at^2/2+bt)$$then $$W = \int_t \frac{at+b}{m}\cdot\left(\frac{at^2}{2}+bt\right)dt$$ $$W = \frac{1}{m}\int_t...

For the block ##m(A-2Acos\beta cos\alpha) = Tcos\alpha - fsen\alpha)## (1) For the wedge ##MA = 2Tcos\beta +fsen\alpha -Tcos\alpha## (2) (2) + (1) ##MA + m(A-2Acos\beta cos\alpha) = 2Tcos\beta## (3) Using non inertial frame, and the force balance of the block on tangente ##2mAcos\beta =...

I don't understand how to balance the forces of block and wedge while they have diferent horizontal accelerations. m is ##A - 2Acos\beta cos\alpha## M is ##A##

@Chestermiller is pointing out that there is an error in your equation (2) where you treated both masses as having the same horizontal acceleration ##A##. If I want to stick to the non inertial frame then I can't use block+wedge system, I'll try on Earth's reference

But that is on a inertial reference frame right? If yes, then my equation 1 is wrong. I'm trying to solve on the wedge reference of frame, that's why A is perpendicular to g, as show on the upper right corner. But now I'm stuck again

Wedge + block X axis ##0=(m+M)A - 2Tcos\beta## Therefore ##T=\frac{(m+M)A}{2cos\beta}## (2) yeah true

Thanks! I got the answer! Here my full resolution: Block on X axis ##2mAcos\beta = mgsen\alpha +mAcos\alpha -T (1)## Wedge + block X axis ##0=(m+M)A - 2Tcos\beta## Therefore ##T=\frac{(m+M)A}{2cos\beta}## (2) (2) in (1) ##2mAcos\beta = mgsen\alpha +mAcos\alpha - \frac{mA+MA}{2cos\beta}##...

I've worked this problem out to completion, and get the answer given by your book. If, in the equation ##a = 2Acos\beta##, you mean that A represents the horizontal acceleration of the wedge and a represents the acceleration of the block relative to the wedge (in the direction tangent to the...

I don't know the name in english. Geometric link? As the wires are inextensible, using the "geometric link" we can relate the accelerations. In this image's case when A travels D, then B travels D/2. Because both wire travels D/2 + D/2 = D. Inextensible. In the wedge case, when the box travels...

When the box travels a ## X## distance, the wedge travels ## \frac{X}{2}##. So ##a = 2A## Using the wedge as a non inertial frame: I didn't use (4). Using (2) on (3) and then on (1) I got: ##2mA=mgsin\alpha +mAcos\alpha + \frac{-mgcos\alpha sin\alpha +mAsin^2\alpha +MA}{2cos\beta -...

[QUOperpendicularmr post: 6260489, member: 639870"] I can see that the parallel axis theorem would be useful if the axis of rotation is in the same plane as the hexagon. Is that the case, or is the axis perpendicular to the hexagon? The problem as you have stated it in the original post leaves...