Recent content by psyclone

Thank-you for your post. Obviously, max deflection can not occur at BM = 0. In this case the fixed moments are not equal, therefore it won't occur at centre of beam. Given, I've used 'three moment equation(s)' to arrive at the given moments at b and c, and reaction at b (reaction b is under...

Homework Statement Using continuous beam theory, constructing BM diagram from points b to c, to calculate the max deflection. I only found a have a single solution, though the BM digram show two points of zero bending. I can provide the solution. [edit: Rb = 685 N] Homework Equations...

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B2 needs to be reversed I agree. And we are discussing E and B1, but I was referring to all the fields that needed to be changed, mentioned in the question.

So what your saying, If the cross product of v & B is equivalent to a force. Then the electric field must be reversed, when a negative ion is introduced, to have zero net forces acting on the ion. please find attached.

Homework Statement Please refer to attachment for solution. Can you help with the following (please refer to attached). If I understand the problem, both magnetic fields B1 & B2 need to change direction, due to the change in current (I) by replacing the positive with a negative ion. Your...

Solved it! \frac{P}{p} = \frac{1}{2}(a + d(p-1)) (eq 1) Use sum of Q, to let a be the subject. a = 2\frac{Q}{q} - d(q-1) (eq 2) Sub, eq 1 into eq 2 d = 2 \frac{Q}{(p-q)} ( \frac{P}{p} - \frac{Q}{q} ) Do the same for Sum Q & R, solve for d. d = 2 \frac{Q}{(p-r)} (...

Homework Statement Hi All, I found this problem, The sum of p, q, r terms of an Arithmetic Progression, are P, Q, R respectively: show that \frac{P (q - r )}{p} + \frac{Q (r - p )}{q} + \frac{R (p - q)}{r} = 0 Homework Equations 3. The Attempt at a Solution [/B] My thoughts on how to...

p { margin-bottom: 0.25cm; line-height: 120%; } I'm trying to simplify the problem, otherwise I have to take into account separate vertical & horizontal forces.So what your saying is, simply let the buoyancy force act through centre gravity of the volume which is occupied by the portion wood...

I'd like to note.

Thanks for your post. Yes, buoyancy force is equal to weight of displaced fluid, thanks to Archimedes. But, taking a second to think about introducing a Fwa term. This term is necessary because the weight of the water (ignoring air pressure) above the beam will add to the moment acting on the...

Hi, This problem has been driving me mad! Can anyone simplify the physics of this problem? Because I can't solve due to to many unknowns. Question ref: Engineering Materials, Benham, Crawford & Armstrong. Please refer to attachments.

Thx for your post. but.. W(-1/(2e^(-3/2))) should be W(-1/(2e^(-5/2))), but your working is correct!

Hi, I had a little problem which I'd like to have answered by anyone who has a-little time to spend! How to solve for x, given the following; 2Ln(x+4)+1=x ? the solutions are x= -3.91.. & 5.50.. Wolfram-alpha solves the problem using 'Lamberts W-function', which makes sense because using...

sorry for not getting back sooner. Thanks for responses, To both of the above comments. Correct. The intersection is equal to 0. If we consider the problem as a tree diagram, there are 6 branchs of X=6 given the first 4 shots are successfull. and given; Pr(x=6) = 0.2936 (exactly 6...