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anonymous

  • one year ago

A 100-N sack of potatoes is suspended by a rope. A person pushes sideways on the sack with a force of 40 N. What is the tension in the rope?

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  1. anonymous
    • one year ago
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    This is a force question so my first advice is draw a force diagram, NEVER try and do a force question without first drawing a diagram. In my experience, if you can't draw a diagram; then you don't properly understand the problem. Also, with a diagram you can more quickly hone in on just where you are weak in these types of problems. Now this strikes me as an equilibrium problem, so I would suggest you treat this situation as your horizontal push is balanced by the horizontal component of the tension in the opposite direction. In other words, once you have your force diagram, write the force balance equations in the horizontal (x) direction and the vertical (y) direction. Why do I think this? Well, think... if your horizontal push didn't balance the ball eventually then your push would no longer be horizontal! (The ball is travelling in a circular arc... eventually it will go up to far and your push will slip off) I will be online for the next hour or two, so try this and see if you can figure out the rest :). If you get stuck post here again and I will give you another push.

  2. anonymous
    • one year ago
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    Just checking in how are you doing... Let me know if you are stuck and if so where.

  3. anonymous
    • one year ago
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    Alright well I have to go now.... In practice its best to try and reason through these things and get help only when needed otherwise you will never actually learn how to do it because you will never stumble and learn what you need to do/not do.... That being said I will post the solution below and I STRONGLY suggest you follow my hints above to try and reason it out FIRST and only look below to check your answer.

  4. anonymous
    • one year ago
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    Drawing the force diagram: You will find you have tension (in the rope) directed at an angle theta from the vertical (the side doesnt matter)

  5. anonymous
    • one year ago
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    The forces on the mass are its weight mg directed downwards, the tension (mentioned above and the pushing force directed horizontally to the side

  6. anonymous
    • one year ago
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    Writing the force balance equations should give you that the pushing force (assuming it is directed to the right) of 40N will balance the horizontal component of the tension T directed to the left: or 40N=Tsin(theta)

  7. anonymous
    • one year ago
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    The weight of 100N is directed downwards is balanced by the vertical component of the tension directed upwards... 100N=Tcos(theta)

  8. anonymous
    • one year ago
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    Dividing these equations gives: \[\frac{ 40N }{ 100N } = \frac{ \sin \theta }{ \cos \theta } = \tan \theta \ \ \ \ \ \rightarrow \ \ \ \ \ \theta =\tan^{-1} (\frac{ 2 }{ 5 }) \approx 21.8^{o}\]

  9. anonymous
    • one year ago
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    Now that you have the angle, selected either one of the equations (both give the same answer) and plug it in and solve for T: \[T \sin(21.8^{o}) = 40N \ \ \ \ \ \rightarrow \ \ \ \ \ T=\frac{ 40N }{ \sin(21.8^{o}) }\approx107.7N \approx107N\] Where the final simplification is required since the largest number of sig figs in the problem is 3.... therefore the answer should reflect that

  10. anonymous
    • one year ago
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    The most easiest way i tgink is what you have to do is just find hypotaneuse of 100 and 40 you will get the answer..

  11. anonymous
    • one year ago
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    True Shiraz... if you reconstruct my logic that is exactly what I did (ultimately) to get to the answer... but I was assuming that this question was a result of being unfamiliar with the problem solving process of force questions which is why I went into the detail I did

  12. anonymous
    • one year ago
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    @Shiraz-Khokhar additionally.... this is a really easy question to practice constructing a force diagram and balancing the forces and going through the process.... so when you get to a problem you cant just solve for the hypotenuse, these skills will allow you to quickly and effectively solve the problem

  13. anonymous
    • one year ago
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    and a note to the poster... my method definitely seems longer but I assure you it took less than a minute to do it properly which (if you also have to explain why the solution is the hypotenuse of the right triangle which most physics teachers will require you to do) is as quick if not quicker and more illustrative

  14. anonymous
    • one year ago
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    Right @plasmaFuser ..both the methods are right...but you gave the detail which was pretty cool..and also enhanced my concepts...bravo!!

  15. anonymous
    • one year ago
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    Thank you :D if you like please feel free to give me a medal!

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