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$\bullet$$\bullet$ Find the tension in each cord in Figeure 5.46 if the weight of the suspended objectis 250 $\mathrm{N}$ .

a. The tensions in each cord are, $[183 \mathrm{N}],[224 \mathrm{N}]$ and $[250 \mathrm{N}]$b. The tensions in each cord are, $[682 \mathrm{N}], \quad 836 \mathrm{N} |$ and $[250 \mathrm{N}]$

Physics 101 Mechanics

Chapter 5

Applications of Newton's Law

Motion Along a Straight Line

Motion in 2d or 3d

Newton's Laws of Motion

Applying Newton's Laws

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all right, so it's always nice to start off with. You're free body tohave. You're free body diagrams handy when you start up. So here are the situations for for an B and so tensions are the only forces in both situations here. So the 1st 1 you have attention from wire A and test for more people attacking upwards and both inclined at different angles from the from the horizontal intentions from C acting directly downwards, uh, and and B, you have tension from B also inclined at 45 degrees to the horizontal axis upwards with attention from a cycle downwards at an angle of 60 degrees from the vertical and then t of sea is all sack and downwards tension for sea. Uh, that's, I think, directed downwards. So the best way to approach this problem would be obviously to resolve forces on DH. This would be done in both X and y directions. So So let's look at that. Basically just knew from the second law we want to find the net force. And so So actually, let's start with situation there, um, net force in the X and in the extra hand x and y are actually see a Is this an equilibrium? So zero net force in the X direction. So this implies that the ex component of sir tea, eh? Co sign there to x x forces here X compartment ta, which is tear co signed 30 on the ex composed of team B, which is TV Coast 45 would be sad equal. Thank you. And so we have that TV is equal to 1.225 cans. Ta be a relation between those great Ah. Then we're looking at resolving in the white direction again. Net force is zero in that direction. So this means that the why component of T n t. V s a tear signed 30 plus TB signed 45 would be equal to T c. Okay. And TC, we do know it to be 200 50 2 50 once. Okay, s o we find here. And TV. We know the relationship, Tio tio. So we plugged that in. There's a t A. So we take ta common and you've signed 30 from the first term. The second term, you have 1.225 side 45 because we're just replacing plugging in the relationship between TV and Tere There on that is equal to 2 50 Okay. And so you have the TA is 182 new ones. And they're far TB, will you? 224 News. Excellent. And some for part b means a loo in here, uh, sames something Result forces. Ah, And so let's let's start off with the ex direction. The net force zero there. And this means that TV coast 45 same horizontal part of T B tears acting opposite to that. So that's recalled to t heir. This time it will be sine 60. Um, this is also the same co signed 30. The reason for that is that this angle sixties is with respect to the vertical. So you want you want you want this component here, right? Well, the horizontal component. And for that if you if you rotate your head 90 degrees, you'll see that you'll need a TA signed six. He will give you that composed so could have also used just this angle, which is 30 and ta co signed 30 in the usual way would give us the same answer CO signed 30 inside. Six. He are the same, by the way, since 30. So okay, so that's on. So we get a relationship between TV or TA, which happens to be the same as partner 1.225 times Tears. What TV It's so now we resolve in the wife direction again. No net force. This gives us that t be signed. 45 plus t heir Sorry, T ai ko Sign 60 which is the same by the eyes. Signed 30 is equal to T C, which is 2 50 again. So we take ta Common said you have tear times 1.225 sign 45 us co sign 30 is he called 2 50 and so ta comes out to be in this case 680 mutants therefore T b d 830 70 And that's a

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