# Resultant Force's point of application in 2D

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#1
If you have an object with its weight and another force acting on it, how can you determine the point of application of its resultant? One way i thought of was to find the direction of the resultant by creating a force triangle or resolving. Then once we have the direction of the resultant, resolve both the weight and the force in that direction at their respective points of application. Then we can determine the point of application by choosing a point then pretending you are finding the centre of mass in 2d, so x coordinate of center of mass=(force * x cord+ weight * x cord )/(resultant force)
then do similar for the y coordinate. Is this the easiest way to find the point of application?
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#2
an example of the use of this would be to find the resultant direction and point of application then find where this intersects with the base of the box. This gives the position of the normal contact force. correct?
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6 years ago
#3
(Original post by MEPS1996)
If you have an object with its weight and another force acting on it, how can you determine the point of application of its resultant? One way i thought of was to find the direction of the resultant by creating a force triangle or resolving. Then once we have the direction of the resultant, resolve both the weight and the force in that direction at their respective points of application. Then we can determine the point of application by choosing a point then pretending you are finding the centre of mass in 2d, so x coordinate of center of mass=(force * x cord+ weight * x cord )/(resultant force)
then do similar for the y coordinate. Is this the easiest way to find the point of application?
The point of application of the resultant of two forces will be the point of intersection of those forces or the point on the line of action nearest to that point of intersection.
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#4
(Original post by brianeverit)
The point of application of the resultant of two forces will be the point of intersection of those forces or the point on the line of action nearest to that point of intersection.
how did you work this out? sorry for late reply :/
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6 years ago
#5
(Original post by MEPS1996)
how did you work this out? sorry for late reply :/
Because otherwise there would be a turning moment.produced.
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#6
(Original post by brianeverit)
Because otherwise there would be a turning moment.produced.
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#7
(Original post by brianeverit)
Because otherwise there would be a turning moment.produced.
how about finding the point of application of the resultant of 3 forces though?
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6 years ago
#8
(Original post by MEPS1996)
how about finding the point of application of the resultant of 3 forces though?
The same applies if they are concurrent otherwise there is a resultant plus a couple. Note that if 3 forces acting on an object hold it in equilibrium then they must be concurrent. This is often useful in problem solving by enabling you to solve a problem by drawing a triangle of forces rather than having to resolve forces..
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#9
(Original post by brianeverit)
The same applies if they are concurrent otherwise there is a resultant plus a couple. Note that if 3 forces acting on an object hold it in equilibrium then they must be concurrent. This is often useful in problem solving by enabling you to solve a problem by drawing a triangle of forces rather than having to resolve forces..
so if they are not concurrent then to find the resultant we use my method at the top of this thread?
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