- #36
CAF123
Gold Member
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JollyJed said:0kN?
Yes. Now what are you left with?
JollyJed said:0kN?
The picture. You are right in that 16kN is all that is left.JollyJed said:No to the picture or no to the 16kN left over?
If I understand you correctly, that is simply not true. If the 5kN forces were pointing in the same direction, their resultant would not be zero. Hence direction matters.what i was told at college was the vectors could go in any direction, and would always give the same result..
CAF123 said:The picture. You are right in that 16kN is all that is left.
If I understand you correctly, that is simply not true. If the 5kN forces were pointing in the same direction, their resultant would not be zero. Hence direction matters.
No, because you added them head to head.JollyJed said:Ahh yes i understand what you mean now mate! so if one of the 5kN forces was facing the opposite way to the other i would be right?
The resultant is simply a vertical line pointing in the -ve y direction with magnitude 16kN.so when drawing it graphically would i draw it the same? and obviously 'measure' the 16kn line?
cheers for all the help mate!
CAF123 said:Why can't you use what here?
JollyJed said:Move the vectors like a did in the previous pictures? I am pretty sure the arrows are just pointing to where the origin is, which they were in the example but i could still move them..
Yes, 16kN is the Resultant vector magnitude.JollyJed said:16kN is the resultant?