Understanding Vector Components of a Downward Facing Force

[ThomasHW]

If anyone could explain to me how to do the problem below that would be great. I don't understand how the vector \vec{F} can be split into components when it's facing straight down?

http://tunerspec.ca/school/question6.jpg

 

[learningphysics]

It's the parallelogram rule for adding vectors... from the tail of F draw a line parallel to BC... extend AB so that it intersects this line...

Do you see the triangle?

 

[ThomasHW]

No I don't... I don't see how F, AB, and BC can fit together to make a triangle...

 

[learningphysics]

I uploaded a picture drawing the components of F parallel to AB and BC. Let me know if it makes sense:

http://www.photoleech.com/action.php?do=show&imgid=image/21149

click on the picture to make it bigger.

 

[ThomasHW]

Ah, thanks.

So to solve it, I should use the lengths given to find an angle, and then solve for F from there?

 

[learningphysics]

Ah, thanks.

So to solve it, I should use the lengths given to find an angle, and then solve for F from there?

exactly. that should do it...

another method:
an indirect way to approach this... suppose the AB member is exerting a 5.19kN force (upward and to the right at joint B)... if you do the sum of forces at the joint B = 0... and solve the equations... you can get the magnitude of F that way also...

 

[ThomasHW]

I got an answer of 6.64kN for F. If you did the question, did you get that or does that sound right?

If you'd like to see my work, let me know and I'll scan it.

 

[learningphysics]

I got an answer of 6.64kN for F. If you did the question, did you get that or does that sound right?

If you'd like to see my work, let me know and I'll scan it.

Yeah, I'm getting something different...

one thing... the two components of F aren't supposed to be perpendicular... it looks kind of perpendicular in my drawing, but they're not supposed to be perpendicular...

 

[learningphysics]

Ah, thanks.

So to solve it, I should use the lengths given to find an angle, and then solve for F from there?

Yeah, you'll need two angles, because the two components aren't perpendicular.

 

[ThomasHW]

http://www.tunerspec.ca/school/triangle.jpg

Is this the triangle I should be using? Or am I making it wrong?

 

[ThomasHW]

Bump!

 

[learningphysics]

Bump!

that diagram doesn't look right to me... the force diagram should look like the one I drew... only thing is that it's not a right triangle...

Try to see which angles in the force diagram are equal to the angles in the structure... try to find equal angles...

 

[ThomasHW]

What lines did you use in the force diagram that you drew? I don't get how you came up with that diagram...

 

[learningphysics]

What lines did you use in the force diagram that you drew? I don't get how you came up with that diagram...

The way to get that diagram... draw a line through the tail of F (the end without the arrow) parallel to BC.

draw a line extending AB upwards and to the right...

The two lines intersect.

 

[learningphysics]

Does this diagram make sense:

http://www.photoleech.com/action.php?do=show&imgid=image/21154

 

[ThomasHW]

Oh, I thought you were using the tip-to-tail method to figure out F by moving CD and AB into a triangle.

I don't remember how this method works, any chance you could explain it? Or send me to a page that will?

 

[learningphysics]

Oh, I thought you were using the tip-to-tail method to figure out F by moving CD and AB into a triangle.

I don't remember how this method works, any chance you could explain it? Or send me to a page that will?

You can resolve a force into any two directions as long as they're in the same plane and they aren't parallel... the components don't necessarily have to be perpendicular. For example you can resolve a force into horizontal and vertical components... but this isn't the only way. In this case you can resolve F into a component parallel to AB and a component parallel to BC... the two components add to F.

I've marked down two pairs of angles (one pair purple, one pair green... do you see why they are equal?

http://www.photoleech.com/action.php?do=show&imgid=image/21156

I've marked the components in red, and F in blue...

 

[ThomasHW]

Ahh, yes I do see that they're equal.

I don't remember ever doing this method though, so I'm not sure what to do next...

 

[learningphysics]

Ahh, yes I do see that they're equal.

I don't remember ever doing this method though, so I'm not sure what to do next...

Find the angles in green and purple... once you do that, you can use the sine rule or cosine rule to get the magnitude of F... ie the blue side... you know the red side opposite the green angle is 5.19kN.

 

[ThomasHW]

Ok, here is what I did:

http://www.tunerspec.ca/school/triangle1.jpg

x = 69.44^{\circ}

theta = 33.69^{\circ}

fishy = 76.87^{\circ}

\vec{F} = 5.4kN

Is that what you got?

 

[learningphysics]

Ok, here is what I did:

http://www.tunerspec.ca/school/triangle1.jpg

x = 69.44^{\circ}

theta = 33.69^{\circ}

fishy = 76.87^{\circ}

\vec{F} = 5.4kN

Is that what you got?

No, but your angles are right. Did you use the sine rule? I did F/sin(76.87) = 5.19/sin(33.69) and solved for F.

 

[ThomasHW]

I used F/sin(76.87) = 5.19/sin(69.44).

That was just a dumb mistake. I'm not sure how I ended up using that angle, oh well.

\vec{F} = 9.11kN

Right?

 

[learningphysics]

I used F/sin(76.87) = 5.19/sin(69.44).

That was just a dumb mistake. I'm not sure how I ended up using that angle, oh well.

\vec{F} = 9.11kN

Right?

yup. that's right.

 

[ThomasHW]

Awesome, thanks a ton for the help. :)