Determine the point of application of a resultant force

In summary: F_y - r_y F_x ##.One equation in two unknowns (do you understand why there is one degree of freedom left over ?).
  • #1
SumDood_
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6
Homework Statement
Given two Forces and their position vectors, find the resultant force and the point of application of the resultant Force.
Relevant Equations
moment = Force * distance
I am struggling with part b of the question attached in the screenshot. For part a, I simply add the components of the given forces.
I tried calculating the moments using vector cross multiplication, but I don't know what to do after that or even if that step is useful.
 

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  • #2
SumDood_ said:
but I don't know what to do
a. Did you make a sketch ?

SumDood_ said:
I tried calculating the moments
b. Please post your working -- and why you think this is useful

c. Post a typed problem statement, not a screenshot. See rules and guidelines

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  • #3
By the way, I agree with your ##{\bf F} = \left (2{\bf \hat\imath}+ 2{\bf \hat\jmath}\right )\ ## N.
 
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  • #4
a. Did you make a sketch ?
b. Please post your working -- and why you think this is useful
I did make a sketch now and attached my working. I calculated the moments because that is the material I am taking at the moment in class. I realized that the question does not mention that the system is in equilibrium, which is what I have used in previous questions.
 

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  • #5
Okay! No beauty prizes, but that's not the intention anyway. So you found (I hope) the ##\bf F## total and you found ##\vec\tau_1 = -1\; {\bf \hat k} \ ## Nm and ##\vec\tau_2 = -33 \;{\bf \hat k} \ ## Nm. Useful, because the resultant force and torque must both be the same.

I assume you found ##{\bf F} = \left (2{\bf \hat\imath}+ 2{\bf \hat\jmath}\right )\ ##

So how did you (?) find that
1673353934207.png
?

SumDood_ said:
the question does not mention that the system is in equilibrium
There is no system in this case. It's just the addition of two forces: a mathematical exercise.

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  • #6
Ah, I see. You didn't find: the screen shot is from the solutions manual. :confused:
No wonder I agreed... :smile:

So at this point you have a total ##\vec F## and a total ##\vec \tau## and I think your question boils down to "how do I find an x- and a y-coordinate for "the point where the resultant acts".
Am I right ?

##\ ##
 
  • #7
BvU said:
Okay! No beauty prizes, but that's not the intention anyway. So you found (I hope) the ##\bf F## total and you found ##\vec\tau_1 = -1\; {\bf \hat k} \ ## Nm and ##\vec\tau_2 = -33 \;{\bf \hat k} \ ## Nm. Useful, because the resultant force and torque must both be the same.

I assume you found ##{\bf F} = \left (2{\bf \hat\imath}+ 2{\bf \hat\jmath}\right )\ ##

So how did you (?) find that
View attachment 320129 ?There is no system in this case. It's just the addition of two forces: a mathematical exercise.

##\ ##
Thank you for the explanation! These are practice questions and I have the solutions for them. Why are your moments negative? From my working, they should be positive. Can you please explain that? From what I have read online, the cross multiplication of two dimensions generates an output that is on the third dimension which you have specified in your working, but why is it negative?

I'm also struggling with this:
BvU said:
Useful, because the resultant force and torque must both be the same.
Why must they both be the same?
 
  • #8
BvU said:
I think your question boils down to "how do I find an x- and a y-coordinate for "the point where the resultant acts".
Am I right ?

##\ ##
Yes!
 
  • #9
SumDood_ said:
Why are your moments negative?
comes out as a result of
SumDood_ said:
vector cross multiplication
I learned ##\vec \tau = \vec r\times\vec F## and I learned $$\vec a\times \vec b=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\
a_x&a_y&a_z\\b_x&b_y&b_z\end{vmatrix}$$

And (for me) the sketch is a confirmation of the negative sign of the results: both forces turn clockwise, i.e. negative sense.

SumDood_ said:
Why must they both be the same?
So that you have a single resulting force that does the same as the two forces combined

BvU said:
how do I find an x- and a y-coordinate for "the point where the resultant acts"
Vector sum of forces is ##{\bf F} = \left (2{\bf \hat\imath}+ 2{\bf \hat\jmath}\right )\ ##
Sum of torques is ##\vec\bf \tau = -34 \;{\bf \hat k} \ ##Nm.

From ##\vec \tau = \vec r\times\vec F## we get for a point ##\vec r## of action:##\tau_z = r_x F_y - r_y F_x ##.
One equation in two unknowns (do you understand why there is one degree of freedom left over ?).

so we can finish the exercise by picking any two terms from the boxes that satisfy this condition...

##\ ##
 
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  • #10
BvU said:
comes out as a result of

I learned ##\vec \tau = \vec r\times\vec F## and I learned $$\vec a\times \vec b=\begin{vmatrix}\hat\imath&\hat\jmath&\hat k\\
a_x&a_y&a_z\\b_x&b_y&b_z\end{vmatrix}$$

And (for me) the sketch is a confirmation of the negative sign of the results: both forces turn clockwise, i.e. negative sense.So that you have a single resulting force that does the same as the two forces combinedVector sum of forces is ##{\bf F} = \left (2{\bf \hat\imath}+ 2{\bf \hat\jmath}\right )\ ##
Sum of torques is ##\vec\bf \tau = -34 \;{\bf \hat k} \ ##Nm.

From ##\vec \tau = \vec r\times\vec F## we get for a point ##\vec r## of action:##\tau_z = r_x F_y - r_y F_x ##.
One equation in two unknowns (do you understand why there is one degree of freedom left over ?).

so we can finish the exercise by picking any two terms from the boxes that satisfy this condition...

##\ ##
Understood, thank you for your help!
 
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  • #11
In summary: a force does not have a point of action, but a line of action.

##\ ##
 
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Related to Determine the point of application of a resultant force

What is the point of application of a resultant force?

The point of application of a resultant force is the specific location on an object where the single resultant force can be considered to act, producing the same mechanical effect as all the individual forces acting on the object combined.

How do you determine the point of application of a resultant force in a two-dimensional system?

To determine the point of application of a resultant force in a two-dimensional system, you typically use the principles of moments (torque). You calculate the moments about a reference point for all the individual forces and then use these to find the coordinates of the point where the resultant force acts, ensuring that the total moment about the reference point is the same as the sum of the moments of the individual forces.

What is the formula for finding the point of application of a resultant force?

The formula for finding the point of application (x, y) of a resultant force in a two-dimensional plane involves calculating the moments of the individual forces about a reference point. For forces F1, F2, ..., Fn acting at points (x1, y1), (x2, y2), ..., (xn, yn), the coordinates (x, y) of the point of application are given by:\[ x = \frac{\sum (F_i \cdot x_i)}{\sum F_i} \]\[ y = \frac{\sum (F_i \cdot y_i)}{\sum F_i} \]where \(\sum F_i\) is the sum of all forces and \(\sum (F_i \cdot x_i)\), \(\sum (F_i \cdot y_i)\) are the sums of the moments of the forces about the y and x axes, respectively.

Why is it important to determine the point of application of a resultant force?

Determining the point of application of a resultant force is crucial for understanding the mechanical behavior of structures and objects. It allows engineers and scientists to predict how an object will react under the influence of multiple forces, ensuring stability and preventing failure in structural designs. It is also essential in the analysis of equilibrium and dynamics in physics and engineering.

Can the point of application of a resultant force be outside the object?

Yes, the point of application of a resultant force can be outside the object, especially in cases where the object is subjected to a couple or a system of forces that create a turning effect. In such scenarios, the resultant force and its point of application can be located in a position that is not physically on the object, representing the equivalent effect of the force system on the object's motion or equilibrium.

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