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Question: A box of of height 2m, width 6m and length 6m is half filled with water. Then it is given a horizontal acceleration of $\frac{g}{2}$ and vertical accleration of $\frac{g}{2}$. Find the angle made by the free surface with the horizontal and the length of the exposed portion of the top of the box? Also find the pressure at the centre of the box.


My attempt: $$tan\theta=\frac{a_x}{g_{eff}}$$ $$tan\theta=\frac{a_x}{g-a_y}$$ $$tan\theta=\frac{\frac{g}{2}}{\frac{g}{2}}$$ $$tan\theta=1$$ $$\theta=\frac{\pi}{4}$$ For the second part of the question, we could equate the volume of the water before and after the acceleration. Before the acceleration, the water is contained in the shape of a cuboid of 6*6*1 dimensions and after the accleration, it is in the shape of a trapezoidal prism since a 3d right angled triangle of the required dimensions cannot fit in the box. Let a be the exposed portion and b be the base length. Let h be the new height of the 3d trapezium. $$height*width*length=width*(\frac{1}{2}*(a+b)*h)$$ $$6*6*1=6(\frac{1}{2}*2*(a+b))$$ $$a+b=6$$ I think the figure would look like this:-

enter image description here

Taking the slope into account, $$a=2$$ $$b=6-a=4$$ So the exposed portion should be 2m. But my textbook says it is 4m. Can anyone explain my how that could be or is it a printing mistake.

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closed as off-topic by John Rennie, Kyle Kanos, JamalS, Sebastian Riese, Gert Jan 23 '16 at 1:33

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  • $\begingroup$ I think the b in your figure should a + b, yes? $\endgroup$ – Mark Hill Jan 21 '16 at 17:55
  • $\begingroup$ no, a+b is the total length of the container, b is the length consumed by the water on the base and a is exposed length on the top. $\endgroup$ – Abhishek Mhatre Jan 21 '16 at 18:06
  • $\begingroup$ well I think I have a simple answer for you. $\endgroup$ – Mark Hill Jan 21 '16 at 18:06
  • $\begingroup$ can you edit your picture somehow to clearly show the top bottom and exposed parts of the container and water please? $\endgroup$ – Mark Hill Jan 21 '16 at 18:13
  • $\begingroup$ To be honest, I think your entire initial set up may be wrong. Because you're looking at this from a 2D perspective when it is a 3D result. $\endgroup$ – Mark Hill Jan 21 '16 at 18:37