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anonymous

  • one year ago

Two forces with magnitudes of 25 and 30 pounds act on an object at angles of 10° and 100° respectively. Find the direction and magnitude of the resultant force. Round to two decimal places in all intermediate steps and in your final answer.

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  1. anonymous
    • one year ago
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    Resolve the force vectors into horizontal and vertical components first |dw:1439676785822:dw|

  2. anonymous
    • one year ago
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    \[F_x=F \cos \theta\] \[F_y=F \sin \theta\]

  3. anonymous
    • one year ago
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    Then add the components in each direction to get the resultant force in that direction. The resultant force is (from the Pythagorean theorem) \[|F|=\sqrt{F_{x}^2+F_{y}^2}\] and the angle is \[\theta = \tan^{-1} \frac{ F_y }{ F_x }\]

  4. anonymous
    • one year ago
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    so confused

  5. anonymous
    • one year ago
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    Each of the forces makes a right triangle. The given force magnitudes are the hypotenuses of the triangles. You need to use the angles (10° and 100°) and the magnitudes to find the horizontal and vertical sides. |dw:1439677744565:dw|

  6. anonymous
    • one year ago
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    For example the 25 lb force acts at 10°. The horizontal component is \[F_x=25 \cos 10°\] and the vertical is \[F_y=25 \sin 10°\] Do the same thing to find the components of the 30 lb force

  7. anonymous
    • one year ago
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    fx= 30cos100deg fy=30sin100deg

  8. anonymous
    • one year ago
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    Now add the 2 Fx's and the 2 Fy's to get the components of the resultant force.

  9. anonymous
    • one year ago
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    Fx = 25 cos 10° + 30 cos 100° Do Fy

  10. anonymous
    • one year ago
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    fy=25sin10+30sin100

  11. anonymous
    • one year ago
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    Use this to find the magnitude \[|F|=\sqrt{F_{x}^2+F_{y}^2}\] and this to find the direction \[\theta = \tan^{-1} \frac{ F_y }{ F_x }\]

  12. anonymous
    • one year ago
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    ohh okay thank you!

  13. anonymous
    • one year ago
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    you're welcome

  14. anonymous
    • one year ago
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    Why am I not getting any of my answer choices? What am I doing wrong?

  15. anonymous
    • one year ago
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    @peachpi

  16. anonymous
    • one year ago
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    what did you get?

  17. anonymous
    • one year ago
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    IFI=410.66 and theta=16.47

  18. anonymous
    • one year ago
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    I got Fx = 19.41 N and Fy = 33.89 N \[|F|=\sqrt{19.41^2+33.89^2}=39.05~N\] \[\theta=\arctan \left( \frac{ 33.89 }{ 19.41 } \right)=60.2°\]

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