# Resultant force

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)

## The Attempt at a Solution

#### Attachments

• IMG_20150922_074842.jpg
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billy_joule
It's best to type out the problem statement in full. We have no idea what you are trying to do.

• goldfish9776
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

billy_joule
for Fy , I have 90-170(15/17) = -60N( in upwards direction )

How many forces have a Y component?

How many forces have a Y component?
90, 170, 20

billy_joule
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

billy_joule
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.

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.
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.

billy_joule
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?

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...

billy_joule
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:

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.

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....

billy_joule
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?

Fx looks correct.

Which direction (up or down) does the Y component of the 20N vector point?
Downwards

billy_joule