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Could I get some input on how to fix this problem please? Thanks in advance! The exercise for which I need the model. The simulink model together + the error that needs to be fixed + Matlab code used: Note: Gain 1 = 1/Ti

Hello, so we have two potitions right, if we take ##\theta = 90## as the first position (i.e. both rods are flat) and then the second position at ##\theta = 0##. I totally understand the exercise, not difficult. The only issue I am having is the torsional spring... it says that it is uncoiled...

or.... do we consider all of the metalhydroxides (that do not hydrolise) to be strong bases because they will form OH- ions which is basically considered a strong base?

So I know that I can get whether an acid is strong or not from a table where if ##pK_a## < 0 than it is a strong acid, but there is no such table for bases, so how do I determine whether a base is strong or weak? Also.. for metalhydroxides MOH, which will be strong and which are weak bases...

yes ? so what do you mean by that, I already said that I am taking into account the moment of ineratia of the whole plate right? because if I were to look at it in 2D it would just look as if it's a rod which it is not no?

Will the moment of intertia be the same then? and yeah but I can only use principle of work and energy here, haven't seen conserv of energy for rigid bodies yet.

hello guys, I wanted to ask whether I can just consider/think about this as being rotation around a fixed axis in a plane representing it as if it was 'just' a rod. This is mainly so that for the kinetic energy in the second position is where if we think about it in just a plane. Is this...

This is what i have done

hey, thanks a lot! I guess I'll just always be checking with ##\psi##. but.. I have another question about this. Because in HIbbelers book engeneering dynamics he uses ##\psi = arctan(r/dr/d\theta)## so that is what I use, but.. the problem here is that I get -72.23 degrees which is different...

so this is what the FBD is.... but to be fair, to me this one looks as if the normal force in the direction of the radial line, yet it isn't???? here in the solution, it's not along the radial line, whys that???

I get - 60 degrees which indeed makes it 30 degrees now..... thanks a lot!! Need to watch out on those two mistakes I just made here.

oh dear......

how come is it not?

##r = cos(\theta)## and for the given, r = 0.866 and for rdot --> ##rdot = sin(\theta) * \theta(dot)## = -1

oh, that is what they mean.., okay thanks a lot. I thought that it was downwards or something.

nope... I am getting -40.89

btw @erobz, this is in feet, but look at the FBD, haven't they forgotten to include the gravity force...

actually you might be right.. I got something in the 40's, but I had the wrong equation for r=f(##\theta##) one sec let me try again.

well, so if you look at this example the angle psi, the formula I mentioned above is used for the normal force angle.

yes yes of coursre that I know, and that is why I am asking why I can't use the formula to determine the angle between the radial line r shown and the normal force. Instead they use 30 degrees here, but how do you determine that?

what do you mean? the path's center is not in the pin of the arm guide right? and the force F is the force exerted by the arm guide on the can/ball or whatever.

for more clarity: here, 30 degrees is used, but why is that?

so I was wondering. there is this normal force on the can from the path. And there's this formula to find the angle between the radial line and the tangent or also between the normal force and either the radial or theta axis. the formula is ##\psi = r/dr/d\theta##. The thing is that here they...

thanks a lot! but I think you were totally correct in your previous comment. :) Knowing that in the extreme cases it's not horizontal anymore gives us an indication that there actually is an angular acceleration for the middle link.

Hello that is indeed what I did. I had 9 unkowns, 5 forces with 5 eqn's of motion from the 3 bodies and the rest from the kinematics. but.. the thing is. I took AB's angular acceleration as an unknown and I indeed got a nonzero value. And it says in the picture that only the angular velocity is...

well let's put it like that, so on two of the links there's no angular acceleration given. The following is where I had my doubts: So for link AB for the moment equation, the doubt was whether the angular acceleration will also be 0 at that point then or whether I should take it as an unkown...

it'll 'rotate' about A clockwise, and yes my apologies. oh.. wait so I'm confused. If I would consider the extreme case of A and B horizontal to the right, I'd say that it rotates about A. But when I look at it by considering that B will drop faster due to the link being smaller than I see a...

OH I see, thank guys I get it now! .. Had a bit of difficulty 'actually' visualizing it. And I have a related question to this one actually, one that problem from a couple days ago, which I still haven't figured out. If we look at this structure, will the link AB also undergo translation due to...

Change of angular velocity over dt, but what i meant was, if I where to actually show that the angular velocities are the same, can I say that a_A=0.9 x alpha_C due to rolling without slip, or is that only for the acceleration at the COG

how do I know that both angular accelerations are the same for both wheels here? should I apply relative motion analysis for the acceleration at A(with ##a_x,A and a_y,A##) and B(with ##a_{x,B} and a_{y,B}##) here, or is just a_A=r*alpha_C and a_B = r*alpha_D enough from which a_A=a_B and thus...

hey @DrClaude (or anybode else :) )I didn't want to open another thread for this small question but I wanted to ask the following: it's about question (a): I was wondering what the reason 'd be not to write ##Na_2Cl_2## instead of 2NaCl? my possible explanation, because the ionic compounds...

What I assumed as well, thank you.

Thanks guys, appreciate it! also just found a comfirmation here, just wasn't sure. https://chemequations.com/en/?s=Na2CO3+%2B+HCl+%3D+2NaCl+%2B+CO2+%2B+H2O&ref=input

Hello, Can I get some clarification on why it is NaCl and not 2NaCl as a product please? Thanks in advance! I had ##Na_2CO_3 + 2HCl## --> ##2NaCl + CO_2 + H_2O## instead. Why is this not correct? I have this as shouldn't the number of chlorine and sodium atoms be the same in reactant and...

Hello there, can I get some help with (b) please? first of all I wanted to ask.. can is it permitted to use different systems in one exercises? like f.e. for conservation of energy to find the velocity of ##m_3## I used as system only ##m_3## but for the collision I used the 3 masses as the...

well the question that arises, whether it is correct that this would be the minimum speed I guess then? forgot to add minimum in the previous post. like.. how do I know from my solution that it would not be the minimum solution?

@Orodruin dudeeeeee... look at this please.. at what he did at 1:40 ... youtube vid still don't get how this makes my solutions wrong:rolleyes: because.. isn't mine the initial velocity as well?

Thank you, that is interesting. I don't understand how you think in such a way by taking ##-x_0## and stuff, I would never ever do such a thing xD And thanks for the conclusion, you are right.. good attitude to have to my own answers as well tbh. But yeah, the thing is that this exercise has...

appreciate it! Will def use those next time.

just as I thought.. but the answers convey a different story unfortunately.. ##v_0 = 9.76m/s## and ##\theta = 58.3## degrees

oh @Orodruin because I know the final answers, that is why :)

euhh becuase that is what I got as an answer? pulling the intial velocity and angle AND the time as unkowns throught he equations

Hello there, I don't understand what I'm doing wrong I don't get the correct answer, but have done the same analysis 3x already and still get the same... Some input would be appreciated thanks in advance. Note: y-axis is upwards and x-axis is to the right. 3 unkowns i.e. 3eqs. ##x = x_0 +...

I am sorry.. but have you read what I have said.. I have NEVER studied geometry (yet). so I do NOT know those first principle derivations. I know that it is important in physics, but I'll come that later on as I have referred to in post ##22 So inevitably as I haven't studied the real deal yet...

well @nasu to be honest, my problem was that I didn't have math for approx. 1.5 years for the last 3 years(including geometry) and next to that I didn't have any science and also my Math was only 2 hrs a week. So it's basically all self-teaching, but haven't reached geometry yet. Skipped it due...

hahahaha

thank you that is indeed what I use (but intuitively) to determine the angle for the mg-force on a slope.

so from that theorem we get ##\phi## == ##\theta##, correct?