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Force Vector Addition

  • Thread starter Luongo
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1. If the resultant force acting on the bracket is to be 790N directed along the positive x axis, determine the magnitude of F



2. Basically I got the components of the two vectors above and below F summed them up. I got THE RESULTANT X-component: 600cos45+325(5/13)+790cos(theta)



3. Same with the y component of the resultant HOW DO I GET THETA? I cant get the magnitude of F which is sqrt(Frx^2+Fry^2) without knowing that angle between x-axis and F. Does anyone know how to get that angle and what i do? Thanks!
 

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tiny-tim

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Hi Luongo! :smile:

(have a theta: θ and try using the X2 tag just above the Reply box :wink:)
… I got THE RESULTANT X-component: 600cos45+325(5/13)+790cos(theta)[/b]

3. Same with the y component of the resultant HOW DO I GET THETA? I cant get the magnitude of F which is sqrt(Frx^2+Fry^2) without knowing that angle between x-axis and F. Does anyone know how to get that angle and what i do? Thanks!
(i assume that should be an = instead of that second + in the equation?)

I'm confused :confused: … you have a value for 790cosθ, don't you? ok, divide by 790 and use arccos.

(If you didn't have 790, and had only Fcosθ and Fsinθ, you would get tanθ by dividing, and F2 by squaring and adding :wink:)
 

PhanthomJay

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1. If the resultant force acting on the bracket is to be 790N directed along the positive x axis, determine the magnitude of F



2. Basically I got the components of the two vectors above and below F summed them up. I got THE RESULTANT X-component: 600cos45+325(5/13)+790cos(theta)
This equation is incorrect. The resultant force is 790 N in the x direction. You should have written 600cos45 + 325(5/13) +Fcostheta =790.
3. Same with the y component of the resultant
set up a similar equation with the y component of the known and unknown forces, all of which must sum to ______?
HOW DO I GET THETA?
solve for the 2 unknowns F and theta using the solution of the 2 equations with 2 unknowns (substitution method works best).
 
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This equation is incorrect. The resultant force is 790 N in the x direction. You should have written 600cos45 + 325(5/13) +Fcostheta =790. set up a similar equation with the y component of the known and unknown forces, all of which must sum to ______? solve for the 2 unknowns F and theta using the solution of the 2 equations with 2 unknowns (substitution method works best).


well whats the answer? i dont understand i got
sqrt(Frx^2+Fry^2) = 790 shouldnt it but i still got 2 unknowns this makes no **** sense
where Frx = 325(5/13)+600cos45+Fcos(theta)=790 and Fry= 325(12/13)+Fsin(theta)-600sin45=0
treating as a system i get theta = 27.3 degrees??
 
Last edited:
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Think about the resultant force and it's direction. This should allow to you solve for Fry. (No calculation required)

Then you'll have two equations and two unknowns which can be solved by a number of methods.
 
120
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i got it! thanks guys!! couldnt have done it without your help
 
1,096
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Glad to hear you figured it out.

Try this question again later and see if you can generate the same equations as PhantomJay provided for the x and y components of the resultant vector, otherwise you're simply going to make the same mistake across all questions in which addition of vectors is required.
 

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