Recent content by janelle1905

Okay... I'm just wondering if the Force of tension for C-B and A-B are equal? And if not, how do you figure out what each is (with them being different?) Thanks :)))

Okay...thanks for your help...and sorry for my delay in responding, but I still haven't had much success with this question. I have the torque equation as: T=FN(10)sin65-22(2.8)x1/2cos65=0 as well as: T=FNr=(95)(9.8)r But I don't know how to incorporate the mass of the man into the normal...

Hi - sorry for the delay in my response as I have been away for a few days. Anyways - just to make sure I understand what you're saying: Fnet = ma = FT-Fccos53.1 - Fcsin53.1 so: ma = mv2/r - mv2/r(cos53.1) - mv2/rsin53.1 and then solve for v? Thanks for your help :)

Homework Statement A ski jumper travels down in a slope and leaves the ski track, moving horizontally with a speed of 25m/s. The landing incline below her falls off with a slope of 33 degrees. a. How long is the ski jumper airborne? b. Where does the ski jumper land on the incline...

Okay, thanks for all your help!

First off, yes you are correct about A being at the bottom, C at the top of the string etc... Regarding the forces on the bead, this is what I think: For the x-axis there are two forces, a positive force (FT) and negative force acting in the opposite direction (Fc). Therefore, FT=FC = mv2/r

Regarding part 1, I did draw a FBD, and I know there is a force, Fc=mv2/r acting in the opposite direction as the Force of tension 2. I calculated the dimensions of the triangle, and r=30. But I am just wondering what do I set mv2/r equal to ? Thanks for your help :)

Oh okay - thanks I understand what you're saying now. For the eq'n: 1/12 ML2+ M(0.2)2 Will the L be equal to the full distance of 1.0m, or will it equal 0.30 (the distance from the end to the axis)?

The course I am in is a very basic course, and as such, calculus is outside the scope of this course. The only formulas we have been given are for moment of inertia (i.e. I=1/12ML2, etc.) Can the problem be solved using those formulas? Thanks very much for your help though :)

Homework Statement A 100g bead is free to slide along an 80cm piece of string ABC. The ends of the string are attached to a vertical pole at A and C, which are 40cm apart. When the pole is rotated about its axis, AB becomes horizontal. a. Find the tension in the string b. Find the speed of...

Okay - so does the torque equation have to be re-written to include the mass of the man? And is that then the unknown?

I'm not sure what the parallel axis theorem is, as we haven't covered it in any of the lessons for this course...Is there a way to solve the problem without that theorem? Thanks :)

Ok thanks. However, wouldn't the formula for both terms on the right side of the equation be 1/12ML2 (I.e., LA and LB, respectively) Also, if the L is determined by the distance from the centre of gravity, does that mean LA=0.5? or does is it still equal to 1.0? Sorry for so many...

Okay - I understand it now. I plugged in and got the same answer for both part a and b this time. Thanks very much for your help!

Okay I see what you were saying - but you get the same answer either way, correct?