Recent content by riddle

Ok. I'm getting the same thing. I understand how this works now. But I'm still having trouble believing that the publishers messed up so bad. Can you just have a look at what they did. I'm really tired right now, and I won't be able to use the book for a week or two so I won't be able to look...

Oh. I forgot to mention that I'd got the first part. Yes. The root is in between 2 and 3. I just substituted "x" with 3 and 2 and saw that f(3) / f(2) = -s. i.e., the sign changed, so there had to have had been a number which had f(x) = 0, i.e., the root of the equation. My first approximation...

Homework Statement Show that a root of the equation x3 - 3x - 5 =0 lies in the interval [2,3], and then find the root using linear interpolation correct to one decimal place. Homework Equations n/a The Attempt at a Solution This is my first ever time using interpolants ( well at...

OH! It's a right triangle. I thought I saw that, but then I remembered you telling me that the triangle in the drawing was different from the force triangle. And it is different. But! It's still a right triangle. But one question. How do you draw the triangle you told me to draw. With the...

lol, I've never been good with to-scale diagrams. meh. I don't see any of the angles inside the triangle. All I know is that the two angles opposite the one in the "center" are equal, meaning the "center" angle is 180 - twice one of the angles. (I can't wait till all this seems like kid's stuff)

Why do I have a feeling that this (thumbnail) isn't right. And if it is right. I don't see how it helps. *sigh*

Ah, I understand. But, I still don't see how you work out the angles. Can you please draw it for me? I would really appreciate it.

But isn't the vector of the resultant force parallel to the horizontal?

Ok, now I'm just uber confused. At first I thought that the angles in the diagram (see thumbnail) were the angles that the tension made, but then I realized that the triangle is isosceles, so the angles would have to be the same, but that's not the case. What are the angles created by the...

I don't understand. Oh and this is what it says on the answer page thing: force on pulley = 22.7 N force acts at an angle 45° to each plane i.e. 15° to vertical

I made a triangle, used the angles to and saw that the triangle was a right triangle, using pytahgagoras i found the magnitude of the force exerted on the pulley. I don't know how to find the direction.

Yes. The tension is approximately 16.064 N. Making the net force exerted on the pulley around 22.719 N. But I still don't see how I find the angle.

Homework Statement Figure 3 shows a particle X of mass 3 kg on a smooth plane inclined at an angle 30° to the horizontal, and a particle Y of mass 2 kg on a smooth plane inclined at an angle 60° to the horizontal. The two particles are connected by a light, inextensible string of length...

Oooh. So I guess that these are "free vectors". EDIT: From Gerbens reply I'll take it as a yes, since they can be moved around. Thanks guys.

What's a "triangle of forces" (triforce, lol) I've attached a snip from my textbook. I don't get how they make that triangle, and how they get those angles. Can someone please help.