I understand now! I took what you told me, about my signs and horizontal forces, and I got my answer of 44.5 revolutions per minute, and then for part c) I and got 29.9 rev/ min. I appreciate all the help you gave me, man. Thanks, Ty.
To find centripetal acceleration is a = v^2/r, but I do not have v. I do know that v and a are at perpendicular to one another though. Right?
I know centripetal force (F= (mv^2)/R) is a horizontal force acting on the mass. By the way, what about centrifugal "force"; does it play a role...
Thank you again.
< http://people.hws.edu/faux/Phys1/practice2.pdf >
Question number 5:
5. The 4.00 kg block in the figure to the right (see link) is attached to a vertical rod by means of two strings. When the system rotates about the axis of the rod, the strings are
extended as shown...
Thank you, Doc Al. Weirdly enough it accepted my answer for d, even though I used 1.0 as the acceleration.
Would you mind if I asked you another question?
That's why I like the good old fashioned "write-it out" technique.
If I was going to find d then, can I use:
vf^2 = vi^2 +2ad
or would it be wrong to assume I started from rest?
Bugger of a question, ain't it?
Is there anyway I can do c) without using acceleration? If I do, I am going to assume that a = 1.0 since I, myself, also can't see anything wrong with it. But you'd know better. What equation can I then use.
Thanks, Ty.
- Thank you. sincos was an error. It was meant to be just cos.
The questions asks me to express myself to two significant digits, and when I do whether I input 19 degrees or 18.77803322 degrees, I get the same, "1.0". Do you know what it could be?
Thanks, Ty.
Homework Statement
A box of textbooks of mass 25.0 kg rests on a loading ramp that makes an angle "a" with the horizontal. The coefficient of kinetic friction (muk) is 0.230 and the coefficient of static friction (mus) is 0.340.
a) As the angle "a" is increased, find the minimum angle at...