Recent content by gnits

Thanks, that's a relief. The book has rarely been wrong and so I was all too ready to doubt myself. Much appreciated.

A very simple (I thought!) question: I'm just looking at the first part, finding the reaction at the hinge A. Here is my annotated diagram, with the reaction and A resolved into it's X and Y components, the force at E labelled as Fe and the length of ED labelled as L. Considering the body...

No, nothing else is given. I'll assume that 60⁰ was meant to be given and go from there. Thanks.

That's very helpful to hear. You see, in the book the diagram is presented without any angle being given. I was thinking that I must be missing something. I don't think that it is solvable without the angle being given. In this case Ѳ = 60⁰.

Could I please ask for help in how to do this question. Is it in fact well formed, can it be solved as it is or do I need more information? Q. Find the external forces and the force in each rod in the following framework of light rods which is supported and A and C: So, I need to find Fa...

Yes, of course, thank you. The reaction force R must be perpendicular to CB, and so using my already derived formula sqrt(3)*Rx + Ry = W and replacing Rx with R cos(30) and Ry with R sin(30) I find that R = w/2 as required. Thank you very much indeed.

Could I ask for a hint as to where to go next with this question please? I've done this first part, to find the reaction on the wall. Here's my diagram: I've labelled the internal forces at B in red. In green I've shown the reaction at the ring. So I need to find sqrt(Rx^2 + Ry^2) = R. So...

Thanks all for the replies. They have helped me to solve to problem.

Thanks very much for that. Indeed, if I include the torque produced by BC's weight everything works out. This torque is given by W * a sin(Ө)I guess I am having trouble justifying to myself, why, when I am considering forces on the ring only, I need to include W due to the weight of BC in this...

Could I please ask where I have gone wrong with my reasoning in the following question: The answers in given in the book are: (1/2)W tan(Ө) W vertically (1/2)W tan(Ө) horizontally Here is my diagram: Considering the system as a whole: (In the text below "Ya" and "Xa" are the forces at...

OK, found it. Was indeed an arithmetical slip. Thanks so much for all of the help. My main lesson here has been to be mindful of which parts of the system are rigid. Thanks again, Mitch.

I see, yes, all joints are hinged. So I must consider a rigid part. ABC is not such a part. AB is. If I take moments for AB about B I get: 2a*F*cos(Ө) = 2a * F1 * cos(Ө) + 2a*F2*sin(Ө) + wa * sin(Ө) which simplifies to: 2F = 2 * F1 + 2* F2 * tan(Ө) + w * tan(Ө) Now I substitute in for F and...

Unfortunately, this approach hasn't worked. So, if I consider part ABC and take moments about B (so as to exclude the reaction force on the pivot at B) I get: F4 * 4a * sin(Ө) + F * 2a * cos(Ө) = F1 * 2a * cos(Ө) + F2 * 2a * sin(Ө) + w*a*sin(Ө) + w1 * 2a * sin(Ө) which simplifies to...

Thanks for your reply, It is not the couple forces (F) as I am resolving vertically, and they both act horizontally. Ahh, would it be the force at the pivot B? If so then I don't want to introduce this unknown into my equations. So I could consider ACB and take moments about B. This would...

Please could I ask with help with the following question: I have done part (a) and agree with the answer given in the book of 2a (2w + w1) sin(Ө) It is part (b) where I am stuck. Here is my diagram: (In green are show the forces of the couple - but they are not needed in my following...