Recent content by jepe

Try this: http://www.aps.anl.gov/asd/me/Calculators/ElasticBeam2.html It does not show any equations, but gives you quick result

try this: http://www.aps.anl.gov/asd/me/Calculators/ElasticBeam2.html

Try this :

Yes, you can use the horozontal forces to solve the problem. The clue is that, since the bar is in equilibrium, the total horizontal force = zero. Hence, the horoizontal components of the ropes are equal and in opposite directions. Work it out from here

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set up the momentum balance about the zero mark of the meterstick: 90*9.8*82+120*9.8*50-210*9.8*x=0 210 (=90+120)*9.8 being the reaction force at the pivotpoint x = the distance of the pivot point from the zero mark work out x and you have the answer for question 1.

You are absolutely right, a metacenter height of just 0.29 cm is certainly not the most stable condition. I understand that you have a box with the dimensions you specified and you actually put it in the water. If everything is absolutely correct, the positive metacenter should result in zero...

Hi, here is your answer again. Scribanti's formula, as you want to use it, only works if you have a negative stability, hence if the metacentre is below the centre of gravity. If you add weight to the exact point of the centre of gravity until you have a depth of 13.7 cm, the metacentre is...

Referring to the previous pdf file:http://www.ocp.tudelft.nl/mt/journee/Files/Lectures/OffshoreHydromechanics_Intro.pdf The stability chapter is referring to ships and barges which are designed to always have a positive stability. A positive stability means that the metacenter M is always...

I assume that w in your thread means the circular frequency ω (in rad/s)? Assume the vertical location of the weight from its equilibrium position is z (z pointing upward is positive) The mass-spring system can be described as a harmonic motion: z = za . sin (ω t), where za is the...

sorry, figure 5.2 should be figure 2.5

The basics of the Scibanti Formula can be found in: http://www.ocp.tudelft.nl/mt/journee/Files/Lectures/OffshoreHydromechanics_Intro.pdf Note that in this pdf file: B = breadth which corresponds to W = width in your thread B is also centre of buoyancy (bit confusing, but should be clear in...

OK, let's do it with the new numbers: For floating objects you need to know another point to check if the object has a positive stability (floats without tilt). this is the Metacentric height. Here is the solution: W = width = 15.4 cm L = length = 27.4 cm H = height = 27.0 cm T = draft of...

The system is an undamped mass-spring system. the natural period of the system is: T = 2*pi*sqrt(m/k) = 2*pi*sqrt(128/64) = 2*sqrt(2)*pi This may help you to calculate the position at any time using the equation: x(t) = 6*cos(2*pi*t/T) (inch) for t=0 : x(0) = 6 inch (start position) for...

For floating objects you need to know another point to check if the object has a positive stability (floats without list). this is the Metacentric height. Here is the solution in metric units: W = width = 39.1 cm L = length = 69.6 cm H = height = 68.6 cm T = depth of the box in water =...