Resultant Force - Find Fx,Fy & Ans of 52.8N

AI Thread Summary
The discussion revolves around calculating the resultant force and its components, Fx and Fy, with a given resultant of 52.8N. Participants analyze the calculations for Fx and Fy, noting discrepancies in the values and signs used. A key point of contention is the resolution of the 20N vector, with clarification needed on its Y component and the assumptions regarding the orientation of the triangles involved. The conversation emphasizes the importance of correctly interpreting the vector components to arrive at the accurate resultant force. Ultimately, resolving these components correctly is essential for achieving the correct answer.
goldfish9776
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1. The problem statement, all variables and given/known dat
the ans given is 52.8N .
I tried to let Fx and F y =0.
for Fx , I have 120-20(12/13) -170(8/17)=21.5N (to the right)
for Fy , I have 90-170(15/17) = -60N( in upwards direction )
but my ans is incorrect = 63.7 N ( by using resultant force formula)

Homework Equations

The Attempt at a Solution

 

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It's best to type out the problem statement in full. We have no idea what you are trying to do.
 
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billy_joule said:
It's best to type out the problem statement in full. We have no idea what you are trying to do.
Sorry, I have forgotten about the problem. The problem is find the resultant force
 
goldfish9776 said:
for Fy , I have 90-170(15/17) = -60N( in upwards direction )

How many forces have a Y component?
 
billy_joule said:
How many forces have a Y component?
90, 170, 20
 
goldfish9776 said:
90, 170, 20

...And do they all appear in your ∑Fy calculation?
 
So my new fy =90-170(15/17)-20(5/13)(15/17)= -66.7n
I made 20(5/13)(15/17) becoz I resolve 20n to become force which is parallel with 170, then I resolve it to y axis
 
goldfish9776 said:
So my new fy =90-170(15/17)-20(5/13)(15/17)= -66.7n
I made 20(5/13)(15/17) becoz I resolve 20n to become force which is parallel with 170, then I resolve it to y axis
That's not right. You can find the Y component of the 20N force directly from the info shown on the vector itself, no need for any info from the 170N vector.
Also, you have an incorrect sign in that expression.
 
billy_joule said:
That's not right. You can find the Y component of the 20N force directly from the info shown on the vector itself, no need for any info from the 170N vector.
Also, you have an incorrect sign in that expression.
You mean 20(5/13) ? But , when I resolve the force in that way, it's not exactly the y-axis , right? It's slanted.
 
  • #10
goldfish9776 said:
You mean 20(5/13) ? But , when I resolve the force in that way, it's not exactly the y-axis , right? It's slanted.

I think it's safe to assume the the sides of the three triangles given are referenced to the x and y axis, otherwise, what use would they be?
(that is, the lines with length 12, 12 and 8 are horizontal, lines 15, 5 & 5 are vertical)
Your previous working implies you've already made that assumption. So I'm not quite sure why you didn't make it for the 20N vector?
 
  • #11
billy_joule said:
I think it's safe to assume the the sides of the three triangles given are referenced to the x and y axis, otherwise, what use would they be?
(that is, the lines with length 12, 12 and 8 are horizontal, lines 15, 5 & 5 are vertical)
Your previous working implies you've already made that assumption. So I'm not quite sure why you didn't make it for the 20N vector?
What do u mean by12, 12 and 8 are horizontal, lines 15, 5 & 5 are vertical)
I m confused now...
 
  • #12
goldfish9776 said:
What do u mean by12, 12 and 8 are horizontal, lines 15, 5 & 5 are vertical)
I m confused now...

There are 3 small right triangles in your image, One on the incline, one on the 20N vector and one on the 170N vector.
the length of two of each triangles sides are labelled. These labelled sides are either horizontal or vertical.
The lines labelled 12, 12 and 8 are horizontal, lines 15, 5 & 5 are vertical. If this were not true, then you could not solve the question at all! It is the only information you have to find the relative direction of the force vectors.This post implies you do not believe the side of a triangle with length 5 is vertical:

goldfish9776 said:
You mean 20(5/13) ? But , when I resolve the force in that way, it's not exactly the y-axis , right? It's slanted.

I am saying ; both lines labelled with length 5 are vertical. So 20(5/13) is the 20N vector resolved on the Y axis.
 
  • #13
for Fx , I have 120-20(12/13) -170(8/17)=21.5N (tothe right)
So, which of the sign is incorrect? I have no clue at all...
 
  • #14
goldfish9776 said:
for Fx , I have 120-20(12/13) -170(8/17)=21.5N (tothe right)
So, which of the sign is incorrect? I have no clue at all...

Fx looks correct.

Which direction (up or down) does the Y component of the 20N vector point?
 
  • #15
billy_joule said:
Fx looks correct.

Which direction (up or down) does the Y component of the 20N vector point?
Downwards
 
  • #16
Correct...post #7 suggests otherwise..
 
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