Finding the Magnitude and Direction of Force one with the Resultant Given

  • Thread starter Nacha
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  • #1
Nacha
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Homework Statement



Determine the magnitude and direction theata of F1 so that the resultant force is directed vertically upward and has a magnitude of 800N.

Homework Equations


Question 2-37.gif



The Attempt at a Solution



I have difficulty of where to start the problem? do i sum up the forces of the x-axis with unknowns in it and forces of the y-axis with the unknown forces...where should i start....and how do i start solving the problem????

Please someone help me with this....

Thanks,

Nacha
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

I have difficulty of where to start the problem? do i sum up the forces of the x-axis with unknowns in it and forces of the y-axis with the unknown forces...where should i start....and how do i start solving the problem????

Hi Nacha ! Welcome to PF! :smile:

Yes … split the force F1 into Fx and Fy

then do x and y separately …

finally combine Fx and Fy to give you F1 and theta. :smile:
 
  • #3
Nacha
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Hi,

Thanks for helping me, but one more question, my Fx and Fy for F1 is unkown as shown in the diagram that i have posted earlier, so how do i go about solving for the magnitude of F1 and the direction of F1. Sorry for the trouble....

thanks
Nacha
 
  • #4
tiny-tim
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Hi,

Thanks for helping me, but one more question, my Fx and Fy for F1 is unkown as shown in the diagram that i have posted earlier, so how do i go about solving for the magnitude of F1 and the direction of F1. Sorry for the trouble....

thanks
Nacha

Hi,

Have you got Fx and Fy yet?

You can get them from the diagram.

Then use Pythagoras, and tangents. :smile:
 
  • #5
sandra_abe
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can anyone explain how to solve this problem in complete details? i don't understand the explanations. i have the same problem too. :confused:
 
  • #6
tiny-tim
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welcome to pf!

hi sandra! welcome to pf! :wink:

you need to get all the y-components adding to 800, and all the x-components adding to zero

that will give you two equations involving two unknowns (F1 and θ), which should enable you to find the unknowns

what do you get? :smile:
 
  • #7
sandra_abe
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owhh.. how about to solve this question:

determine the magnitude and direction of P so that the resultant of P and the 900N force is a vertical force of 2700N directed downward.

is it using the same method also? it's almost the same right?
 
  • #8
tiny-tim
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right! :smile:
 
  • #9
sandra_abe
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thanks so much tiny-tim! I'm doing my engineering mechanics tutorial problems right now.. if i have any problem again, i'll ask your help later.. ;)
 
  • #10
sandra_abe
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err... do you online often? then, i'll know when can i ask for your help..
 
  • #11
tiny-tim
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sandra, just post a new thread any time :smile:

somewhere in the world, there'll be someone awake and logged in ready to help, whatever time it is :wink:
 
  • #12
sandra_abe
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tiny-tim

okie! will do.. :approve: :smile: :wink:
 
  • #13
Zlash
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i got a similar problem but i don't have a resultant force :\ not sure on how to solve I've found theata using a parallelogram
theata is the one between the X axis and F1
 

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  • #14
tiny-tim
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Welcome to PF!

Hi Zlash! Welcome to PF! :smile:

(have a theta: θ :wink:)

Sorry, I don't understand the question …

you have three forces (100N 350N and R), and also a line called X at 30º …

if X is equivalent to F1 in Nacha's question, then changing the magnitude of X will give different directions for the resultant R (but a unique magnitude for each direction) …

if neither the magnitude nor the direction of R is given, I don't see how you can answer the question :confused:
 
  • #15
Zlash
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Hi Zlash! Welcome to PF! :smile:

(have a theta: θ :wink:)

Sorry, I don't understand the question …

you have three forces (100N 350N and R), and also a line called X at 30º …

if X is equivalent to F1 in Nacha's question, then changing the magnitude of X will give different directions for the resultant R (but a unique magnitude for each direction) …

if neither the magnitude nor the direction of R is given, I don't see how you can answer the question :confused:

nah X' is a vector that cuts through.
in the quiz I am doing it doesn't say anything about a resultant but I am assuming we have to use the one given in the textbook. so the resultant measured from the positive X' -the line- is 600N
i found θ as 60. θ being between F1 and the X axis( not x' )
im not sure how i can work through forces based on different axis kinda messes me up =/
im trying it now
 
Last edited:
  • #16
haaaaa87
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0
hi all
∑F x= 400cos30 – 600 cos36.87 + F1 sinØ =0
∑F y= 400sin30 – 600sin36.87 + F1 cosØ =0
R=√(〖fx〗^2+〖fy〗^2 )
800=√((〖298.35+f1sinØ)〗^2+〖(560+f1 cosØ)〗^2 )

As we see ; we have two equation and two unknown so we can get the magnitude and direction of f1
 

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