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Okay, so we have this light suspended from a ceiling by two ropes and you can see the angles that they're making with the ceiling up here. And there's a tension and one there's attention to grow up to. And so if we think about this in terms of components, well, the light is going to be a applying a force down. It's gonna wanna fall. This is really due to gravity. So there's a force of Â£50 down. T two is gonna be pulling the light this way and then also up holding it up, T one is going to be pulling the light this way and then also pulling it up. So the light is what's in, uh is in what's called static equilibrium. And so all that means is that all of these arrows air balancing out So the horizontal arrows have to equal each other, and the vertical arrows have to equal each other. And so what we wanna do is use a little bit of trigonometry here, notice that these right triangles, so the horizontal and vertical components of T one and t two we confined with just a little bit of trick and So if we add up, uh, t one t two and then the weight of the light, we should get zero so we can represent that as Okay, So if we start with t one, the horizontal component of T one is going to be negative because it's pushing to the left 15. It it's the angle, whatever t one is and then cosign of 15. So remember, we're actually looking for t one. But whatever it is, uh, to get the horizontal component, we just multiply by co sign of 15 and then the horizontal component is t one signed 15 and it's positive because it's pointing up and then we can happy to so same thing this time the horizontal is gonna be positive, but it will be t two times co signed 25 and then the vertical will be teaching. Signed 25 and then well, Adam, the weight So the weight is just going straight down. So there's zero force in the horizontal and there's negative Â£50 in the miracle because it's going down okay, and then this should equal 00 Well, why should equal 00? Because we want there to be zero net force. So this is just adding up all the forces as vectors and the net should be easier zero. And so what we get is we get a system of two equations and two unknown, so we have. So if we just kind of could add all of these horizontal components, they have equal zero. So negative T one co sign 16 plus t to a sign of five plus zero has to equal zero. And then also t one signed 15 62 signed 25 minus 50 has equal zero. In this again is just a system of two equations. Two unknowns, actually. Just put this in my calculator you could. Another way to do is just to use substitution of some offers 82 then substitute inverted to Or you could just set up a matrix. However you wanna do it, it's just Cem algebra review. And so what you get is you get the T one is about 34 point or 7 to 8. Um, in t two is about 36 7405 That's

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