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Two bars $A D$ and $D G$ are connected by a pin at $D$ and by a spring $A G$ Knowing that the spring is 12 in. long when unstretched and that the constant of the spring is 125 lb/in., determine the value of $x$ corresponding to equilibrium when a 900 -lb load is applied at $E$ as shown.

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Physics 101 Mechanics

Chapter 10

Method of Virtual Work

Work

Potential Energy

Hope College

University of Winnipeg

McMaster University

Lectures

02:08

In physics, work is the transfer of energy by a force acting through a distance. The "work" of a force F on an object that it pushes is defined as the product of the force and the distance through which it moves the object. For example, if a force of 10 newtons (N) acts through a distance of 2 meters (m), then doing 10 joules (J) of work on that object requires exerting a force of 10 N for 2 m. Work is a scalar quantity, meaning that it can be described by a single number-for example, if a force of 3 newtons acts through a distance of 2 meters, then the work done is 6 joules. Work is due to a force acting on a point that is stationary-that is, a point where the force is applied does not move. By Newton's third law, the force of the reaction is equal and opposite to the force of the action, so the point where the force is applied does work on the person applying the force. In the example above, the force of the person pushing the block is 3 N. The force of the block on the person is also 3 N. The difference between the two forces is the work done on the block by the person, which can be calculated as the force of the block times the distance through which it moves, or 3 N × 2 m = 6 J.

03:23

In physics, mechanical energy is the sum of the kinetic and potential energies of a system.

11:04

Two bars $A D$ and $D G$ a…

05:11

Two $5-\mathrm{kg}$ bars $…

All right, I'm gonna try to solve this one without, um, doing it on paper first. So, um, we have a diagram, so I'm going to draw the diagram. Um, it's got a spring force here and here. I wish I would have changed the color of that. Hey, b c d be See, de, they're all supposed to be the same, and they're all supposed to be. I'm gonna call them l, which is Ah, 200 millimeters. And, um, I'm actually going to call it 0.200 meters because that's the same as 200 millimeters. All right, then we've got to in here. Let's label G f e de A B c. All right. Um, these are all l's in here. Also, um, I'm gonna set up my axis so that this is X, and this is why. And then we've got a force that wish I would have drawn my forces in red, so I'm gonna draw them in red right now. We got a force acting downward at E, and I'm just gonna call that P, which is 900 Newtons. Determine the value of X. Ah ha X is here. Okay, I was wondering what the question was. Okay, so I'm going to need the y value at point. Okay, so that appears to be Wait a minute. We're not given a fatal here. So I am going to say, Fada, is this which would be the same as this. All right, Why equals Oh, signed data. All right. So that would be why at a Why at G is going to be 1234 There are five of the Els sign fade, and it's negative. Why have a is positive. Y g is negative. I'm also gonna need why e. Which is 123 negative three l's signed data. Let's go ahead and do the derivatives Delta. Why? At a derivative of Sinus co signs was just Elko. Sign data Delta. Why? At G ups Delta Theta is negative five l co sign beta Delta data. Delgado. Why? E is negative. Three l saying fada Delta data. So, um, Delta U is zero. Um, the force at a is negative. So negative f sub s Elko ST Data Delta theta um, plus f sub s. And this is it G now negative five l co sign Fada Delta theta. Um, and then e is downward. So minus he times negative three l Same fada. Zelda Fada. All right. Um l l l That cancels out Hussein data Dahllof data could send data dolla theta Hussein data Delta theta. That cancels out Also. So now it says negative f sub s minus five f sub s. That would be negative. Six f sub s. Um, so I'm going to write six f sub s, and that's gonna equal three p. So p is to f sub s yet. OK, so piers to offs of s all rights. Um, what else do we know here? The spring is 300 millimeters long when it's unstrapped hte UK. Well, let's figure out what why sub a minus why sub g would be because that is the stretch length. So why sub a is l sign data? Um, why sub G is negative. Five l signed data. So why said a minus? Weiss of G is going to be, um, six. El signed Fada. So this is, um, ex, actually. So the spring force is que Times X. Wait a minute. No, it's not case at times X. It's this spring force is K Times Delta X. So it's going to be X minus 303 100 millimeters. Um, right. The spring Force is K times, Delta X, the stretch and Delta X is going to be X minus that. Okay, so now we could just put all this together. P equals two f of S f so vast, but SMS is K times X minus 0.300 meters. We know that p is 900 Newtons and we know that K is 5000 newtons per meter. And so now we just need to solve that for X. So 900 equals two times 5000 which would be 10 1000 times x minus 0.3 Some zeros here, nine hundreds that would be 0.9 equals X minus 0.3. So X is going to be 0.3 plus 0.90 pulling 39 meters which is, of course, the same as 390 millimeters

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