Forces and angles question

[Jonnybmac]

I'm baffled by forces and angles having not been too familiar with them. I'm going to explain some of what I've worked out and would like someone to explain if it's correct, not understandable or somewhere on the right lines.

I've been doing research on a lintel, one of which called a catnic, and how altering the backplate from vertical a number of degrees reduces the load capability, and I'm provided with examples and an explanation which i couldn't understand much, being not familiar with cosine. I made a little diagram to understand how coside works(according to fig (b) in the next pic), and how if Ø = 0° then cosine = 1 or 100% force going in Fh; and 90 degrees = 0 or 0% force going in Fh direction.

cos(60) = 0.5

So i'm given an example to explain

Answer to activity 1:10 is Fv = Facos(90-Ø)

I then decide to take what I've been shown and try to put it into something that I could understand clearer, but I don't actually have any means of answering if I got it right or not.. So I've drawn a diagram and would like someone to explain if I'm correct?

I've applied a downward force Fp of 200N, which travels to Fy then finishes at Fv.

Fp = starting force(vertical)
Fa = applied force
Fy = force down angle
Fv = vertical force
Fh = horizontal force

But this is where i get stuck. are these figures right and what are they telling me about how strong this is compared to a straight piece?
If I'm correct, there are forces being sent 86.6N to the right Fh and 150N down Fv. But how does that say about the load capability? its strength has reduced to "173.2/200 = 87%"?

I hope this makes sense to someone

 

[Illgresi]

What angle is Fp at? Is it meant to be vertical?

 

[Jonnybmac]

Yeah, sorry. Fp and Fh are vertical and horizontal

 

[Illgresi]

Ok.

Fy = Fa / cosΦ = 200 / cos 30 = 230.94 N

Therefore, the angled part is experiencing a greater load than the vertical part. Assuming it has the same physical capabilities (i.e. it's made of the same stuff as the vertical part), it is comparably less able to withstand force than the vertical components.

 

[Jonnybmac]

So everything I did was backwards? Ive tried to figure out why you used divide rather than multiply?

The top angle, if the force Fa is 200 pulling downward, then Fa = Fy/cos(30)

We use 173.2 as Fy

173.2 / cos(30) = 200

And to rearrange the formula to solve Fy its Fa * cos(30)

200 * cos(30) = 173.2

Where have I gone wrong?

 

[Illgresi]

So, according to the SOHCAHTOA rule, the angle Φ = arccos (Adjacent/Hypotenuse).

If you notice from the diagram above, Fa, Fy and Fh form a right angle triangle where Fa is the adjacent, and Fy is the hypotenuse, therefore:

Φ = arccos (Fa/Fy)

Apply cos function to both sides:

cosΦ = Fa/Fy

Rearrange so Fy (the unknown) is the subject:

Fy = Fa / cosΦ

Edit:

In case you weren't aware arccos = cos^-1

 

[Jonnybmac]

So going by your rule, it should look like this? And to solve for fx direction I use cos again to solve for the adjacent side knowing the hypotenuse and cos

 

[Illgresi]

Yep, that's correct.

 

[Jonnybmac]

Thanks for that I appreciate it!

 

[Illgresi]

No problem!