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5.61

$$ \sum F_{y} = 0 $$ $$ \sum F_{y} = T_{1y} + T_{2y} - mg $$ $$ mg = (4400N)\sin{60} + (4400N)\sin{40} $$ $$ mg = 6640N $$ Why is this wrong?


solution

$T_{1x}$ has to be equal to $T_{2x}$ $$T_{1x} = 4400N cos(60)$$
$$T_{2x} = a cos(40)$$
$$4400N cos(60) = a cos(40)$$
$$a = \frac{4400N cos(60)}{cos(40)}$$ $$mg = 4400N sin(60) + a cos(40)$$

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closed as off-topic by JMac, Aaron Stevens, Kyle Kanos, user191954, John Rennie Apr 2 at 14:07

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – JMac, Aaron Stevens, Kyle Kanos, Community, John Rennie
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Please note that homework and exercise questions should focus on a conceptual question, and show some effort. See the meta discussion here. What factors have you thought about when determining if it's right or wrong? Have you tried to check these values in a different context to make sure they make sense? $\endgroup$ – JMac Apr 2 at 12:49
  • $\begingroup$ @JMac is there anywhere I can ask questions like this? With book name and question number $\endgroup$ – user644361 Apr 2 at 12:50
  • $\begingroup$ If you have access to the stack exchange chat (I'm not sure if you do or not yet), I can give you some advice in the problem solving strategies chat. $\endgroup$ – JMac Apr 2 at 12:53
  • $\begingroup$ @JMac Talk in chat requires 20 rep, though you might be able to invite OP into a chat room, which I believe gives them write access to that room. $\endgroup$ – a CVn Apr 2 at 12:54
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Because the force on each wire isn't the same. You can't just plug in $4400N$ as the force on each. Try balancing the vertical and horizontal components separately.

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