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A vertical force $\mathbf{P}$ of magnitude $150 \…

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Problem 34 Easy Difficulty

Rod $A B C$ is attached to blocks $A$ and $B$ that can move freely in the guides shown. The constant of the spring attached at $A$ is $k=3 \mathrm{kN} / \mathrm{m}$ and the spring is unstretched when the rod is vertical. For the loading shown, determine the value of u corresponding to equilibrium.

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Physics 101 Mechanics

Vector Mechanics for Engineers: Statics and Dynamics

Chapter 10

Method of Virtual Work

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Video Transcript

considers slender rod a B of a weight w drawn here from the center of Mass, which is attached to two blocks A and B, which can move freely in the guides as shown, uh, A and B are also attached by a spring has shown. And we know that the sprint constant of the spring is K and that the spring is unstrapped when a B is horizontal, as in, when this angle Fada that they've defined is equal to zero. So we're called that fit and not for the equilibrium fitted not is equal to zero, which also tells us that the equilibrium length of the spring since a B will be horizontal means that the spring is also the length of the baby which is given to us is l. So I'm gonna call that X not is also equal toe l now neglecting the weight of the blocks. We want to derive an equation in terms of our givens. Fada w l okay, that will be satisfied with Rod is in equilibrium on. This is a fairly straightforward thing to dio. What we want to do is identify which forces could do any virtual work on our system express their potential energies in terms of a single variable. In this case, we'll go with this Fada they've defined for us. Such that weaken therefore easily use the equilibrium condition that the first derivative of potential energy what respects that single variable in this case, beta is equal to zero, and we can use that known equilibrium condition to then solve for the fate a corresponding to that equilibrium. And first things first. It's pretty simple. What forces are at work here? It's just going to be the weight of the Rod, A B and this spring force from this spring that's being stretched out by Rod A. Be moving away from horizontal. Uh, and we know the expressions for both of us. And again, those were the only forces that could do any virtual work on our system. Because while the guides A and B are in provide, like normal force supports, they're not doing any displacement on our system. So therefore, no work and we're going to just look at the spring and the weight. And so our total potential energy is the elastic potential energy of the spring. Both the potential energy due to gravity i e the potential energy of the weight and we know the expressions for both of those three elastic is one half k x squared where k is the spring constant and X is the amount of the spring has been distressed either stressed or compressed from its equilibrium Length on then Plus are just the force are w times d y position of the center of mass. We're gonna call that wise a B and again we want to get these all in terms of Fada so we can go ahead and say that our X for this spring which is gonna be basically ah segment B C plus segment a see the two sides of our triangle is going to be equal to l times l being the hype oddness of our triangle here the length of a B So l times cosign fada e are adjacent plus l times sine fada e r opposite minus one Because we're minus and l the equilibrium length and R y A b is similarly just going to be negative. L over to sign Fada. Why is it negative? Because I'm just gonna We're gonna define the, uh okay below are equilibrium origin point of when the springs equilibrium point of fit equal, not as our origin. It doesn't really matter as long as the two potential energies are opposing each other because there clearly not additive. In this scenario, the spring is not for helping the weight fervor Stretch it out. It's opposing it. Um, so then we can sub these in on get our general expression for the potential energy, so v equals one half k times to quantity. L Times CO sign data plus sign Fada minus one and quantity and bigger quantity squared minus again just w times l over to sign Fada. And again, this is just minus because we need them to be opposing. And that's how we've defined our coordinate system is this is the negative one. So we can go ahead and take our first derivative with respect, to fade out, to go ahead and get after using the equilibrium condition to find the equilibrium positions. And this is just gonna pull out a k l squared. The half cancels from the power rule coming down. These are all pretty easy derivatives because they're just basic trig and chain rule. So we've got, uh, quantity cosign Fada plus sign fada minus one times It's changed the quantity CO sign Fada minus sign Fada and that quantities. And then again, this is just minus w l over to cosign. Faded is super easy trick derivatives. And we can clean this up a bit to get it to be thief First derivative, where prospective data is equal to K L Squared Times Co sign to Fada minus cosign Fada plus sign Fada and then just again minus W L over to co sign Fada. And now we want to go ahead and use that equilibrium condition and said this equal to zero and rearranged a little bit to get our like terms on different sides of the equation. But we get a final equation in terms of K l W and fada of K L Times Equality Sign, Fada Plus co sign to Fada equals the constants in front of Okay l plus W over two times Kassian Fada. And that is our general equation for our system that is satisfied when it is in equilibrium

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Ferdinand Beer, Jr., E. Russell Johnston, David Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

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Problem

A vertical force $\mathbf{P}$ of magnitude $150 \…

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Related Courses

Vector Mechanics for Engineers: Statics and Dynamics

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Top Physics 101 Mechanics Educators

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Watch More Solved Questions in Chapter 10

Video Transcript

Ferdinand Beer, Jr., E. Russell Johnston, David Mazurek

Vector Mechanics for Engineers: Statics and Dynamics

Christina Krawiec

Zachary Mitchell

Jared Enns

Meghan Miholics

This textbook answer is only visible when subscribed! Please subscribe to view the answer

Vector Mechanics for Engineers: Statics and Dynamics

Christina Krawiec

Zachary Mitchell

Jared Enns

Meghan Miholics

Ferdinand Beer, Jr., E. Russell Johnston, David Mazurek Vector Mechanics for Engineers: Statics and Dynamics

Christina Krawiec

Zachary Mitchell

Jared Enns

Meghan Miholics

Ferdinand Beer, Jr., E. Russell Johnston, David Mazurek

Vector Mechanics for Engineers: Statics and Dynamics