The take home problems require a good understanding of...

The take-home problems require a good understanding of homework due on 10/24/22 and the problems solved in the classes. 1. A stone beam is in structurally static equilibrium due to a hanging stone pillar that is attached to the beam but does not touch the ground, and the entire structure rests on a pillar that is on an extensive- curved stone. Beam mass $m_b$= 15000 Kg; Each of the pillars' mass $m_p$=5000 Kg (same mass pillars); Pillar 1 is at a 2 m distance from the center of gravity (c.o.g) point of the beam as shown in the below figure. Stone Beam Hanging pillar Pillar on the ground 1.1. Draw accurate freebody diagrams separately for the forces acting on the beam and pillars. (Points 30). 1.2. How far away is the hanging pillar attached from the center of gravity (c.o.g) point of the beam? (Points 30). 1.3. Calculate all the forces acting on the beam and both pillars. (Points 30). 2. A solid stone ball of mass 2000 kg of radius 0.75 meters is placed one meter away to the right side of the c.o.g point on the beam. (Beam mass and the mass of the pillar touching the ground are the same as in problem 1. The hanging pillar is placed at the distance you get from the previous calculations; answer to 1.2. The hanging pillar mass is unknown at this time.) Hanging pillar Pillar on the ground Stone Beam 2.1. Draw accurate freebody diagrams for the forces acting on the beam, two pillars, and the ball. (Points 40). 2.2. What is the mass of the hanging pillar that keeps the structure in the equilibrium state? (Points 40). 2.3. Calculate all the forces acting on the beam, pillars, and ball. (Points 40).

Transcript

00:01 In this problem, you have a ball on a horizontal smooth surface.

00:04 It means no friction.

00:06 You have two weights, q and p, and we'll use those to indicate the load, the weight.

00:12 The ball's weight is w.

00:15 First part of this problem is to find what the angle theta is for equilibrium, and then define the reaction of the horizontal surface, frictionless horizontal surface, it has on the what the reaction force on the ball from the horizontal surfaces.

00:33 So let's look at the free bi diagram for the ball for w.

00:41 So we have the ball, we have the weight straight down, tensions always pull, so i'll call it t sub r for right, and then we have t sub l, and let's remember this angle of theta here.

00:59 Now, smooth surface means there's no reaction, action, no, nothing has to arise in any attempt to move it horizontally, but it's not the case.

01:12 These two will cancel out on the horizontal.

01:16 So all we get here is r straight up.

01:20 It's a vertical force.

01:22 It's just basically the normal force.

01:24 Normal force on the ball from the horizontal table because there's nothing else.

01:32 So nothing else has to arise.

01:37 So make my axis x, y, traditional and let's start looking at the equations.

01:46 So f net x minus tl sine theta because the i here is the x component here's the y.

01:55 X is opposite in this case and in the negative x direction so it's minus tl sine theta plus tr the r and w forces they don't have any x components so there's zero x components for both so it's equal to zero because we're we're told to be in equilibrium.

02:16 So, sine theta is tr over tl.

02:23 And i'm going to call this, i'm going to call this equation one.

02:31 Now, you might say, isn't tr, what do we know about tl, tr and tl? well, that's, you probably already know what the answer to that is.

02:39 Let me, let me draw the free value diagrams for them, but they're trivial for q.

02:49 I don't need to circle so you have the weight q down, that's the load, tl plus y direction up.

03:03 So this is from f -net y, tl minus q is equal to zero.

03:10 So as you expect, tl is equal to q.

03:13 It's obvious in that case, even though no, if you're ever in doubt, you just do the free buy diagram.

03:19 And you can do the same thing for the free by diagram for p.

03:24 And it's exactly it looks the same except for just changing symbols.

03:29 And i should mention the l and the r, if you didn't realize it, left, right, so it stood for.

03:36 And this is p here plus y direction up.

03:41 So f net, why, like i said, no real difference.

03:46 Tr minus p is equal to zero.

03:49 So, tr is equal to p.

03:53 So one becomes then, now this is obviously an equal.

03:58 Equilibrium.

03:59 If you had acceleration and so on, then the nets aren't zero.

04:04 But that's not worry about that here.

04:07 So one becomes sine theta, because i give it to you in terms of p and q and all that and w.

04:14 We'll do the same thing.

04:15 P over q.

04:17 That's just r -t -r -p, t -l -q.

04:23 Take the inverse of that.

04:25 So if you knew p and q, take that ratio, take the inverse, get your theta.

04:32 So theta inverse sine, p over q, and that's, that will give us our angle for equilibrium.

04:42 That basically is an indication that the two tensions cancel out.

04:49 There is no, there is no reaction force necessary because they cancel.

04:53 If they didn't cancel, that's something else.

04:57 I mean, then you'd have to have a horizontal force, reaction force, for this to be equilibrium but we're told it's smooth so they have to cancel that's that's that the equilibrium okay so that's getting the theta now it wants also what is r the value of r the magnitude of the reaction force technically it's the by component of the reaction force is the only let's look at f net y r is in positive y the weight is in negative y then we have t l cosine theta is to 0...

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