How Do You Calculate the Resultant Force in This Vector Problem?

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Homework Help Overview

The discussion revolves around calculating the resultant force in a vector problem involving multiple forces acting at different angles. The forces include two acting in the X quadrant and one acting straight down on the Y axis.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct components of the forces in the X and Y directions, questioning the inclusion of certain forces in the summation. There are attempts to clarify the calculations for Rx and Ry, with some participants expressing confusion about the application of trigonometric functions.

Discussion Status

There is ongoing clarification regarding the calculations and the correct application of trigonometric functions. Some participants have provided guidance on the components of the forces, while others are seeking assurance on their understanding and next steps.

Contextual Notes

Participants express uncertainty about the correct signs for the components and the implications of their calculations. There is mention of language barriers affecting comprehension of the advice given.

Dellis
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Hi, I need some guidance regarding this exercise, I attached a visual representation of the exercise

Homework Statement



Find the Resultant Force for the following system of forces shown


FORCES SHOWN on a Vector:


F1= 5 kips on the X quadrant in a 60 degree angle

F2= 5 kips straight down on the Y axis

F3= 8 kips on the X quadrant in a 30 degree angle



Homework Equations



These right?

Rx= [tex]\Sigma[/tex]Fx= F1 + F2 + F3
Ry=[tex]\Sigma[/tex]Fy= F1 + F2+ F3

(theta)Qx= Tan- Ry/Rx


The Attempt at a Solution




I want to know if I am going in the right direction, if not please guide me, this is what I tried.


Rx= [tex]\Sigma[/tex]Fx= F1 + F2 + F3

=+5(sin 60 )- 5 + 8(cos 30 )

= -6.16 kips


Ry=[tex]\Sigma[/tex]Fy= F1 + F2+ F3

= -5( sin 60)-5 + 8 (cos 30)

= -8.571 kips
 

Attachments

Last edited:
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In ΣFx, there is no F2 and F1x = F1*cosθ.
In ΣFy, F1x is positive, and F3x = F3sinθ and it is negative.
 
rl.bhat said:
In ΣFx, there is no F2 and F1x = F1*cosθ.
In ΣFy, F1x is positive, and F3x = F3sinθ and it is negative.

So like this right?


Rx= [tex]\Sigma[/tex]Fx= F1 + F3

=+5(cos 60 ) x 8(cos 30 )

= answer hereRy=[tex]\Sigma[/tex]Fy= F1 + F2+ F3

= 5( sin 60)-5 + 8 (sin 30)

= answer here
 
Note down whether the components are on the +ve or -ve axis.
 
This is the next steps right?R= [tex]\sqrt[]{}[/tex] Rx+ Ry

= answer herethen...(theta)Qx= TAN-Ry/Rx

= answer here
 
Last edited:
rl.bhat said:
Note down whether the components are on the +ve or -ve axis.

What do you mean?, my english understanding isn't that great can you say that in another way.

I just want assurance that I am on the right path, thanks for the help btw I appreciate it.
 
Last edited:
Dellis said:
So like this right?


Rx= [tex]\Sigma[/tex]Fx= F1 + F3

=+5(cos 60 ) x 8(cos 30 )

= answer here


Ry=[tex]\Sigma[/tex]Fy= F1 + F2+ F3

= 5( sin 60)-5 + 8 (sin 30)

= answer here

=+5(cos 60 ) x 8(cos 30 ) This step should be

=-5(cos 60 ) + 8(cos 30 )
5( sin 60)-5 + 8 (sin 30) This step should be
5( sin 60)-5 - 8 (sin 30)
 
rl.bhat said:
=+5(cos 60 ) x 8(cos 30 ) This step should be

=-5(cos 60 ) + 8(cos 30 )
5( sin 60)-5 + 8 (sin 30) This step should be
5( sin 60)-5 - 8 (sin 30)

I had it like that then got confused by your advise that said " *cos " and I took that as you saying multiply then I added an X, sorry lol.

Thanks for clearing that up, so the next steps I showed you in post#5 above is the right path correct?, that would pretty much wrap this exercise up, I will proceed once you tell me.
 
Last edited:
Carry on with post5#
 
  • #10
Alright thanks, I appreciate the help and guidance :)
 

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