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lizki

  • 3 years ago

Can someone send me a link,where finding torque is explained simply ?

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  1. amistre64
    • 3 years ago
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    Force times distance ..... right?

  2. lizki
    • 3 years ago
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    yes ,but i find them hard when we also have cos37 and sin37!

  3. amistre64
    • 3 years ago
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    thats part of vectors then, and not really torque ... im assuming is there a more specific problem that we can relate these to?

  4. lizki
    • 3 years ago
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    This example is solved in my book,but i cant understand the procedure! -The force applied to a 0.2m long spanner is 15N.Find the torque produced by this force.(take cos37=0.8 and sin37=0.6)

  5. amistre64
    • 3 years ago
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    then tag me in your post so I can have a look see later please

  6. lizki
    • 3 years ago
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    |dw:1361799280435:dw|

  7. amistre64
    • 3 years ago
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    we want to find the part of the force that is perpendicular to the spanner

  8. amistre64
    • 3 years ago
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    |dw:1361799368979:dw|

  9. amistre64
    • 3 years ago
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    the amount of force applied at a right angle to the spanner is therefore: 15 sin(37) and the distance is already given as : .2m you said?

  10. lizki
    • 3 years ago
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    yes 2 m

  11. amistre64
    • 3 years ago
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    so your question is not about torque ... but rather how to find the required adjustment of the force.

  12. amistre64
    • 3 years ago
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    which is just your right triangle material :)

  13. lizki
    • 3 years ago
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    But the topic in the book is 'TORQUE' but anw,thenkyou :)

  14. amistre64
    • 3 years ago
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    youre welcome; and yes, torque is the new material that they are adding to the old material from previous sections :) good luck

  15. Iamgmg90
    • 3 years ago
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    \[Torque = \overrightarrow{F} \times \overrightarrow{r} \times Sin \Theta \]

  16. lizki
    • 3 years ago
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    thankyou alot

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