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Resolving and reducing forces.

  • Thread starter mm391
  • Start date
  • #1
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Homework Statement


I am trying to find the resultant forces in the attached picture.


Homework Equations





The Attempt at a Solution



The resultant of a couple of forces = 0
F in x direction = 5cos(45)
F in y direction = -5sin(45) - 2

Therefore:
R = ((5√2)/2)i + ((-(5√2)/2)-2)j
R = √(3.53^2+(-5.53)^2) = 6.56kN

is this correct does the distance of the 2kN force come in to play somewhere. If so where?
 

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Answers and Replies

  • #2
tiny-tim
Science Advisor
Homework Helper
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hi mm391! :smile:

looks ok so far …

now you need to find the line along which it acts (of all the possible parallel lines) :wink:
 
  • #3
66
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Thanks Tiny Tim but I am still al ittle confused.

What do you mean "find the line along which it acts" ? What line. All I can see it that the 2kN force acts 6 meters from point P but I am not sure how to incorporate that into the question.

Thanks
 
  • #4
tiny-tim
Science Advisor
Homework Helper
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a force that is (say) 10 N north-east can be applied anywhere along a rigid body

if applied on the left side, it will turn it clockwise, if applied on the right side, it will turn it anti-clockwise

the point of application (or the line of application, through that point) matters, and when specifying the resultant of forces, you must also specify the line of application :wink:
 

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