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Show that equilibrium is neutral in Prob. 10.7.

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Show that equilibrium is neutral in Prob. 10.1.

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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

here we are revisiting an earlier problem using our new techniques of using the expressions for potential energy and the equilibrium condition to more efficiently find this force P and then also showed that the equilibrium is neutral. Uh, here we have free rods A, B, C, a, D and D E F G, coupled by two pins at A and D, and we also have Point B is attached by a pin to a fixed pivot point in the same is true of point F. Then we have a Siris of forces that weaken effectively. Trudeau's weights hanging out of our system. 40 Newton's at point g 20 Nunes E 68 d a. D a. Unknown force p at sea. So what we want to do here is find expressions for all of our potential energies and get them in terms of Fada so that we can then use the equilibrium condition where, in the first derivative of potential energy with respect to Fada is equal to zero. So first thing we want to do is to find the data. So we're gonna go ahead and define that as between the horizontal going through B and up here If Rod you know BC were to rotate up like that, we're going to find this as Fada, and that's again we would expect to see. To go up here is we have forces down on A and D and e G is less than those much less torque. We're expecting this thing to twist this way. So we're going to find Fada as that. And given that we expect this to be fairly static, we can actually go ahead and get all of these in terms of the same fada. So let's go ahead and get after that. And we're going to go ahead and use the expressions for potential energies for these forces, pretty much the same as the potential energy due to gravity, because for treating them pretty much just wait. So the of F N is equal to just f sub n times Weiss event, which is the vertical displacement of each of the points A B, C, D E f G. And it's worth noting here that be an f are fixed points, so there will be just no displacement, i e They can't do any virtual work, so they're not gonna have any effect on our system. So we're gonna go ahead and skip them. But we'll go ahead and start with defining our potential energy of point C So the subsea equals length of this arm. Zero length RBC that is 0.6 m times sine fada to get our just all right here. Why succeed? Uh and just times are force of P. Then you can see geometrically that point A is gonna have this little angle here, As this is half the length here, A B is half the length of B C. This is gonna be We can see that y c equals two times y A. Which leads us to be able to say that our sorry negative, because it's going down over here. Visa A equals negative 0.3 m times sine fada times its force which is 80 Newtons and in like fashion we can proceed a Z A and D arc connected by this bar aiding We can presume that, uh y a equals y de, which leads us to quickly conclude that V D is also equal to negative 0.3 m. Sign Fada times It's force of 60 meetings and then you can see from Point F. The fixed pivot point of bar D E F G pointy is twice as far out as Point E, so we can conclude that y d equals twice the displacement of y e. Which leads us to believe that the potential energy for E could be expressed as negative 0.15 m again. Time sign of the same fada and times its force at that point of 20 Newtons. And then finally, as egos down and deficit six point g must go up so we can conclude that why sub G equals negative y c e leading us to conclude that the potential energy for point G due to the force there is equal to 0 15 m times. Sign Fada Times 20 New or its force, which is not 20 but 40 Newton's. Sorry, I started copying the lion Buff Uh, and now we can go ahead and write the expression for our total potential energy, which is just all of those terms combined. And I'm gonna go ahead and pull out the sine theta because it's in literally every term. So we've got negative 0.3 m times 80 Newtons plus 06 m times P minus 0.3 m times 60 Newton's minus 0. 15 meters times 20 Mutants Who US 0.15 m Times 40 Newton's and quantity. I ran out of space but want to close the preface e there. And now we're going to go ahead and simplify it. Calculate all out those constants and this simplifies quite easily to just V equals 0.6 m times p minus 39 Newton meters times sine fada. So we can go ahead and take the derivative D V. With respect to Di Fada is equal to again just this constant of 0.6 m times p minus 39 Newtons per meter. And the derivative of sign is obviously cosign Fada and we can use the equilibrium condition to go ahead and set this equal to zero. And since we know also know that we want the current equilibrium of liquid of linkage to remain ie in this ve remains. The system doesn't distort it all. E fada equals zero, which tells us that coastline fada equals one which tells us that we need this clause or this expression to equal zero So we know that 0.6 m times peeing in close 39 Newtons per meter, leading us to discover that P equals 65 Newtons for simple division on. Now we know P we want Thio. Go ahead and show that this equilibrium is in fact neutral and again to check you plug this 65 back into here or even intervene. You get just zero time to co sign Fada or zero time Sign Fada, which is definitely zero Um so it's an equilibrium. So by definition, for an equally room to be neutral, the V must be constant and therefore the first derivative with respect to theta, the second derivative with respect to Fada and so on. All derivatives with respect to all equal zero And what do we have here? So with our Nuveen, we're gonna plug in R 0.6 times 65 minus 39 which surprised zero tons sign Fada which is equal zero, which is a constant and all the derivatives air constant. So therefore we can go hey and say that the equilibrium is neutral and that's it. Very much pretty efficient just using identifying which forces could potentially do virtual work expressing their potential energies in terms of Fatah and utilizing the equilibrium condition. And then again

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

Vector Mechanics for Engineers: Statics and Dynamics

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Problem

Show that equilibrium is neutral in Prob. 10.7.

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Vector Mechanics for Engineers: Statics and Dynamics

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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

Marshall Styczinski

Zachary Mitchell

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

Marshall Styczinski

Zachary Mitchell

Meghan Miholics

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

Christina Krawiec

Marshall Styczinski

Zachary Mitchell

Meghan Miholics

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

Vector Mechanics for Engineers: Statics and Dynamics