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...
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 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...
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...
Sure. My simpler method was to consider the system as a whole and take moments about A and then about C, these lead straight to R2 = 3W/2 and R1 = 5W/2. Then, looking at the conditions for no slipping at A and then C leads to F1 <= 5W/3 and F2 <= W and so slipping will occur at C as frictional...
Could I please ask for help with the following:
Here's my diagram:
The forces at the hinge (green) are internal forces.
For the whole system resolving vertically gives:
R1 + R2 = 4W
and horizontally gives:
F1 = F2
For the rod of weight 3W only, taking moments about B gives:
F1 * L *...
Thanks very much indeed. I had swapped S and F1. I had written S <= u * F1 whereas, as you say, I should have written F1 <= u S. And indeed this leads to the book's answer. Thank you very much.
Thank you very much kuruman for your reply.
So, notwithstanding the signs, would you (or others) think my analysis of taking moments for CD only about D looks correct? I ask because here I differ in the actual form of the answer. The given answer is 2u but I arrive at 2/u. If you\others agree...
Could I please ask for help with the following question:
The last part follows easily from the first part.
Answer from back of book for first part is:
2/(3u') <= tan(Ɵ) <= 2u
What I have done is the following:
Here's my diagram (I have separated the components to show the internal forces...
Thanks very much. I was so close with my method but didn't know it. Will look for the simpler route you hint at also and then I'll move move on to the next part, that of having to actually find the magnitudes of S and F1.
Thanks again,
Mitch.
Please could I ask for help with the following:
Here is my diagram, I show the rod displaced from the sphere so as to label the internal forces acting on each of the rod and the sphere:
In the diagram below I have added the line through DE at angle ꞷ the the horizontal, and a few other...
I see it!
I would never have gotten it without your help. In particular, setting the equations in terms of W/ω, I don't see how I would ever of thought of that.
Thank you very much indeed.
Mitch.
Source https://www.physicsforums.com/threads/to-locate-the-centre-of-gravity-of-a-rod.1010047/
Thank you very much for your reply. I would never have thought of that approach. So here is my previous diagram, with 'a' added:
I can combine my previously derived equations with your suggested terminology to give:
(am using u for the symbol 'mu')
a = d(1 + 2u)
and
a = L - d(1 + 3u)
and I...
Could I please ask for a help on how to attack this question?
A heavy rod AB of length L can be made to balance across a small smooth peg C when a weight of 2W is suspended from A. Alternatively, it can be made to balance across the peg with a weight of 3W suspended from B. If the distance AC...
Please could I ask for help with the following?
Two points, A and B on a horizontal ceiling are at a distance of 2a apart. A uniform rod CD of length a and weight W is suspended from A and B by two light strings AC, BD. A particle of weight (2/5)W is attached to the rod at D, and the system...
Thanks for the reply. From my diagram if I take moments of rod AD about D I get:
T1 * l/sqrt(2) - T * l/sqrt(2) - w * (l/2)*(1/sqrt(2)) = 0
Can divide both sides by l/sqrt(2) to give:
T1 - T - w/2 = 0
Resolving vertically for the whole system I have that T1 = 4w and so the equation above...
Could I please ask for help with the following question?
Four uniform rods of equal length l and weight w are freely jointed to form a framework ABCD. The joints A and C are connected by a light elastic string of natural length a. The framework is freely suspended from A and takes up the shape...
Well, I'm a little closer to the answer:
Firstly, yes, as Steve4Physics says, A is also a point of contact. And by resolving horizontally on the system as a whole I also get that X = 8w/45. So that completes the first part of the question.
I still believe that my analysis of the cylinder...
Could I please ask for help with the following:
Here's a diagram:
(in what follows, for clarity, I write L to represent the lower case L of the diagram).
My diagram wrongly shows u (the coefficient of friction) to be the same at both points of contact. Since this image was taken I have...
Can I please ask for help regarding the following:
A uniform rod AB of length 3L is freely hinged to level ground at A. The rod rests inclined at and angle of 30 degrees to the ground resting against a uniform solid cube of edge L. Contact between the rod and the cube is smooth and contact...
Could I please ask for a pointer in the right direction?
Here's a less cluttered diagram:
So it's a purely geometrical argument we need. Which relationship between a and h will yield VA horizontal.
In going from A to G that's a distance of (using Pythagoras) of sqrt( a^2 + (5h/16)^2 ).
So...
I see now that indeed all of the equations I had derived are true regardless of the orientation of the object and so always true, hence 1 = 1,
So somehow I need to use the extra condition\constraint that VA is horizontal (where V is the vertex of the cone). Well, that tells me that the angle...
Could I please ask for help with the following:
Given: The centre of gravity of a uniform solid right circular cone of vertical height h and base radius a is at a distance 3h/4 from the vertex of the cone.
Such a cone is joined to a uniform solid right circular cylinder of the same material...
There is a second and final part to this question, and I am confused as to the final diagram, here is the additional question:
A particle of weight W is attached to a point P of AB where AP = (2/3)a and the body now settles in equilibrium with the midpoint of BC vertically above A. Prove that...
Thanks very much everyone for your help. I see it now. My incorrect diagram sent me off on a much more complicated path. This is a very simple problem which was made into a tough one by my malformed diagram.
Lesson learned (hopefully!) thanks to all the great help,
Mitch.
Thanks for the help so far. Firstly, sorry, my typo, it is cot not cos.
Here's a sketch with BC vertical.
Is the following the best way to go (sorry, preview not working for me so I am not using LaTeX in case I make errors):
A(0,0), B(2a cos(α), -2a sin(α)), C(2a cos(α), 2a sin(α))
x1 =...
Thanks for the help so far. Firstly, sorry, my typo, it is cot not cos.
Here's a sketch with BC vertical.
Is the following the best way to go (sorry, preview not working for me so I am not using LaTeX in case I make errors):
Orange dot is midpoint of AB.
A(0,0), B(2a cos(α), -2a sin(α))...
Could I please ask for help with the following question:
A uniform lamina of weight W is in the shape of a triangle ABC with AB = AC = 2a and the angle BAC equal to 2ᾳ. The side AB is fixed along a diameter of a uniform solid hemisphere of radius a. The plane of the lamina being perpendicular...
I followed haruspex up to his last equation. From there we know that BP = P - B and BD = D - B and so as x = BP / BD then x = (P - B) / (D - B) so we substitute this into haruspex's last equation and the (B - D) terms cancel, as do the P terms. We are left with WM = (W/3)*(Q + D + B) = (W/3)Q +...
Thanks very much for your help. I am interested in this method but I can't follow it fully. Indeed I can see that there must be a point (say E) on CD that forms a triangle with BC which is 1/3 of the total area of the trapezium. I don't see though how this implies that the centroid of that...