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catsrule332

  • 2 years ago

The mass of the ball is 0.050kg, and at the instant shown the value of the tension is 4N. Find the net force on the ball and find the acceleration vector (give the x and y components). *T is the tension in the string and Fg is the gravitational force.

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  1. imalittletoyrobotfromthefuture
    • 2 years ago
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    hi abby

  2. eashmore
    • 2 years ago
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    We need to better understand the orientation of the system. Is the object being pull across the ground, it the object hanging, is the object rotation? If it is rotating, what plane is the rotation in? Horizontal or vertical?

  3. catsrule332
    • 2 years ago
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    The object is swinging in a vertical circle.

  4. eashmore
    • 2 years ago
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    Ah. Perfect.

  5. eashmore
    • 2 years ago
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    Now, we need to know where the ball is located at the instance shown. Can you give me an angle, say 30 degrees counter-clockwise from the horizontal?

  6. eashmore
    • 2 years ago
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    |dw:1332714065917:dw|Something like this would help.

  7. catsrule332
    • 2 years ago
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    Here I will give you the a picture:|dw:1332714090745:dw|

  8. srinidhijha
    • 2 years ago
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    u should hv stated earlier that ball is swinging in circle

  9. eashmore
    • 2 years ago
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    Now we can work it. Notice that tension and gravity act orthogonally to each other. Therefore, the tension force is equal to the centripetal force\[F_c = m{v^2 \over r} = m r \omega^2 = T\]

  10. eashmore
    • 2 years ago
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    Understand?

  11. eashmore
    • 2 years ago
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    The net acceleration can be expressed as\[a_n = \sqrt{~a_c^2 + g^2 } = \sqrt{~ \left(v^2 \over r \right)^2 + g^2 }\] Notice that \(a_c\) acts in the x-direction and g acts in the y-direction.

  12. srinidhijha
    • 2 years ago
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    net forc is 4.03 N by Pythagoras theorem root ( 4^2 + .0.5^2) = 4.03 N

  13. eashmore
    • 2 years ago
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    Answers aren't helpful. Please try to explain your methodology. Refer to the new CoC.

  14. eashmore
    • 2 years ago
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    http://openstudy.com/code-of-conduct

  15. srinidhijha
    • 2 years ago
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    please correct me I want an answer too Its a humble request please

  16. catsrule332
    • 2 years ago
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    srinidhijas method using the T-mg=ma seems to make sense, and seems like it would be correct. Eashmore I don't understand your method completely, if you could try to explain it a bit more that might help.

  17. eashmore
    • 2 years ago
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    The Force vector and the acceleration vector will have the same direction. Therefore, \[F = a_n \cdot m = \sqrt{T^2 + F_g^2}\]

  18. srinidhijha
    • 2 years ago
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    |dw:1332714774000:dw|

  19. eashmore
    • 2 years ago
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    T - mg = ma is not valid here.

  20. srinidhijha
    • 2 years ago
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    yes u did not said earlier tht it was swingin in a circle atsrule

  21. catsrule332
    • 2 years ago
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    alright, sorry for not making that clear. eashmore the last equation you posted makes sense.

  22. eashmore
    • 2 years ago
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    In the x-direction, we have a tension force and only a tension force. T = 4N In the y-direction, we have a gravitational force and only a gravitational force. Fg = 0.005 * 9.81. The magnitude of these two forces is found from Pythagorus theorem\[F = \sqrt{~ T^2 + F_g^2} = \sqrt{~ 4^2 + (0.05 \cdot 9.81)^2}\] To find the acceleration in each direction, realize that the tension force is caused by the centripetal force. \[T = m \cdot a_x\]and that the gravitational force causes an acceleration in the y-direction\[F_g = m \cdot a_y = m \cdot g\] We can use these two equations to find the acceleration in the x and y-directions. Then, using the same methodology as we did to find the net force, we can find the net acceleration as\[a = \sqrt{~ a_x^2 + g^2}\]

  23. catsrule332
    • 2 years ago
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    Thank you both! I understand it now.

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