Recent content by cordines

No, on wheel A there is only one horizontal force (which is taken to the left). This force acts at the midspan of the member AB as denoted by the parameter r/2. The arrow on the force F indicates that it is a vector. The arrow below the force F indicates how it acts and its direction. The...

@pongo38. Note that the moment and force in red are both given. What exactly do you mean by "the unknown load F (along line of action lF)" are parallel but separated"? @nvn. Yes, I threw out one equation and solved 15 equations with 15 unknowns and then used the other equation as a check to see...

Homework Statement Please refer to the attached diagram. I am trying to find the unknown load F (along line of action lF) that will keep the system in equilibrium as well as compute all reaction forces developed at the joints. Homework Equations Summation of forces in the horizontal and...

Thanks milesyoung. I will try to implement that in Simulink, don't know if I will succeed since I'm new to this. I will get back to you. Thanks again!

The problem is, how can I do that by Simulink?

Yes, since the frequency is directly proportional to the rotational speed.

No, I'm not that familiar with electric generators. However we did do the equivalent circuits of the generators/motors in my first year in mechanical engineering albeit it being a tiny bit. With regards to the others, I'm not that familiar.

Generator is AC.

It's a permanent magnet electrical generator, having a power output of 1.8 kW.

Hello, I am trying to build a SIMULINK model in Matlab in order to determine the RPM of a wind turbine. Can someone please tell me how I can obtain the RPM of a wind turbine from the voltage fluctuations of a generator? Any good sites in the subject are also welcome. Thanks in advance.

To prove: a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } = ( a x c ) x ( b x d) Proof: a x { b x ( c x a ) } = {a . (c x d) }b - (a.b)(c x d) b x { c x ( d x a ) } = b x { (c.a)d - (c.d)a } = (c.a)(b x d) -...

b x ( c x d ) = (b . d)c - (b . c)d a x { b x ( c x d ) } = (a x c)(b . d) - (a x d)(b .c) ...[1] or should [1] stay as: a x { b x ( c x d ) } = { a . (c x d) }b - (a.b)(c x d) c x ( d x a ) = (c . a)d - (c . d)a b x { c x ( d x a ) } = (b x d)(c . a) - (b x a)(c . d) ...[2] d x (a...

As in... To prove: a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x ( a x b) } + d x { a x ( b x c ) } = ( a x c ) x ( b x d) Proof: a x ( b x c ) = ( a . c )b - (a . b)c a x { b x ( c x d ) } = { a . (c x d) }b - (a.b)(c x d) ... [1] b x ( c x a ) = (b . a)c - ( b...

Actually at first that's what I did by substituting a x ( b x c) = - b x ( c x a) - c x (a x b ) into d x { a x ( b x c ) } and using the triple vector definition for the other three, however it was all but in vain. Any other suggestions on how I can substitute i) in ii) ?

i) Show that: a x ( b x c) + b x ( c x a) + c x (a x b ) =0 I managed to this, by expanding each term using the definition of the triple vector product i.e. a x ( b x c) = (a.c)b-(a.b)c and adding the results. ii) and deduce that a x { b x ( c x d ) } + b x { c x ( d x a ) } + c x { d x (...