Recent content by fgb

Plotting the graph of speed x time and thinking about what is the meaning of the area in this graph might make the problem easier. The speed x time graph usually comes in handy when you're dealing with mixed accelerations. :)

Adding to what Doc Al said, the weight of blocks B and C is what drives the system to accelerate - when you write T1 = mA*a, for instance, the influence of blocks B and C is embedded in the value of the acceleration (different blocks B and C would result in a different acceleration).

sankalpmittal, could you post the correct answer? I know it is kind of cheating, but it might be useful to find out where we are getting it wrong :P

At first I had thought that the correct answer was D. Since as the block moves it feels a force due to the rope (which has a horizontal component after the rope tilts a little) there are external forces and no conservation of linear momentum. Tricky question since you are told to consider a...

v = \sqrt{\dfrac{2mgh}{M+\frac{M^2}{m\cos^2{\theta}}}} , maybe? Not sure my answer is correct, but here are a few hints: Use conservation of linear momentum (horizontal) to relate V (velocity of big block) to the HORIZONTAL component of v (velocity of small block). Use also conservation of...

I have not solved the problem, but shouldn't the answer be a function of θ? For θ=90°, for instance, we know the answer should be zero, since m would be in freefall and would not "push" the triangular block.

UPDATE: I had misread the problem, nevermind what I said before :P UPDATE2: I read it right, I erased my comment for nothing ¬¬ Will re-write. Sorry. Something like that: In these "simple" problems, forces essentially come from gravity or contact. An external force will come from contact to...

Well, you see, if you had x(t) = Asin(2t), then x''(t) = -4[Asin(2t)] = -4x(t) Notice that the function in brackets is x(t)! Therefore, with the A in front, x(t) = Asin(2t) also satisfies our condition for x(t), ie, x''(t) = -4x(t) even with the A in front - it is a solution. And it is just a...

So, you have the following differential equations \dfrac{d^2x}{dt^2}=-4x \\ \dfrac{d^2y}{dt^2}=-4y Notice that they are very similar, so I will describe my solution to x(t), but that can be extended to y(t). The formal solution requires a little background on differential equations. If...

The ones I called x(t) and y(t)? :)

I would take a slightly different approach. You have therefore that a_x = -4x \\ a_y = -4y where the x subscript is for the x component, and so on. Therefore \dfrac{d^2x}{dt^2}=-4x \\ \dfrac{d^2y}{dt^2}=-4y So for x(t), ie, the x position of the particle with respect to time, you...

For clarity, I would like to point out that, in the formula P2 = P1 + ρgh, h in this equation it is not really the height of the point whose pressure you want to know - e.g., at sea level it is not zero, h here is the height of the column of air above that point, which is smaller for higher...

Oh, I had not paid attention to that. UPDATE: Oh, the formula for F had something missing, too.

Weird, I agree with your equations but they don't give an answer for Question 19 - unless I'm making the same mistake you are. It seems to me that it is a simple question, you want x = 0i + 0j, which gives -2t² + 3t = 0 -2t² + 4t = 0 and there is no time that satisfies both equations. If you...

It seems to me that you just forgot a term in the chain rule to differentiate your expressior for the speed. Realize that as r is changing, so is \theta, so \theta is also a function of time. You have v = -\dfrac{b\omega}{\sin{\theta}} a=\dfrac{dv}{dt} =...