Calculate Theta for 0 Vertical Component of P

[Ry122]

http://www.users.on.net/~rohanlal/untitl.jpg
In the above diagram how do I calculate a value for theta that makes P's (P is the resultant) vertical component 0?

 

[Astronuc]

Is the 120 N vector vertical? Is the magnitude of 80N or the 20° fixed?

Can one rotate the vectors (triangle) to get P horizontal?

 

[Ry122]

http://users.on.net/~rohanlal/q2-9.jpg
Here is the diagram and question in the textbook
Two forces act on the bracket.
determine the angle theta that will make the vertical component of the resultant of these two forces zero.

The bottom force is 175lb and the top one is 145lb

 

[tiny-tim]

(Goodness, what a horrible diagram - your 20 degrees looks more like 70 degrees - and is that line supposed to be vertical?)

If P's vertical component is 0, then P is horizontal.

So doesn't that make theta 90º?

 

[tiny-tim]

one known angle and two known sides

Ah! Just seen your new diagram (you beat me by a few seconds!)

ok - now draw a force triangle - it will look nothing like the one in your first post.

Then you'll have one known angle and two known sides, so you use ordinary trigonometry to find the other angle(s).

 

[Astronuc]

The bottom force is 175lb and the top one is 145lb

First determine the vertical component of each force.

Then for the resultant vector to have a zero vertical component, the two vertical force components must be equal in magnitude, but opposite in direction.

So is the 175 lb force at an angle of 50° and the angle \theta is the angle between the force vectors? If so, then the angle of the 145 lb force with the horizontal is just \theta-50°.

 

[Ry122]

I understood Astronuc's method.
175sin50=134.057
145sinX=-134.057
X=67.54 degrees (this is the answer in the back of the txt book)

Using tim's method the answer was very close (67.6 degrees) but i don't understand how it found the correct answer. I don't understand how using a trig ratio (cosine) automatically found a resultant and angle that had no vertical component.

 

[tiny-tim]

Draw a parallelogram!

Using tim's method the answer was very close (67.6 degrees) but i don't understand how it found the correct answer. I don't understand how using a trig ratio (cosine) automatically found a resultant and angle that had no vertical component.

Hi Ry!

Odd isn't it - two apparently different methods giving the same result?

To see why they do, draw a parallelogram made of two of the triangles (one a reflection of the other).

That's two copies of my method!

Now draw in the perpendiculars to the diagonal from the top and bottom points - you now have two different triangles on the bottom, and a reflection of those same two triangles on the top, with the left and right triangle swapped over - we'll call that Astronuc's parallelogram.

In my parallelogram, the diagonal represents the third force: you want that force to be horizontal, so you just adjust theta until that diagonal is horizontal.

In Astronuc's parallelogram, these new perpendicular lines represent the vertical components: you want those components to be equal and opposite, which, after you've made my adjustment, they are.

Astronuc only uses the two left triangles - but his top left triangle is the same as his bottom right triangle, and his two bottom triangles together are the same as my triangle!

That's why they give the same result!

Isn't geometry wonderful! ​

(Technically, Astronuc's method is better, because you can use it with more than three forces, but my method has the advantage - especially for me - that it's more foolproof!)