are you sure?
as the degrees get smaller, can't you say that the width becomes the height and the height become the width?
did i misunderstood the problem because that is the way i visualized it(eg: @ angle 0 the width will be 16" and the heigth 1")
the way that you worked your problem will...
*******I changed the heigth from 14" to 16" by accident*******
HallsofIvy, i uderstand that he is asking about the width. I just found it easy to find it using the area which doesn't change and divide it by the heigth at a particular angle.
I also realized that i made a big mistake.
I read...
16* 90/90 = 16 (height)
16*60/90 = 10.666... (height)
you had a rectangle of area 16inches squared
so now: 16inches squared/10.6666...inches = 1.5 inches (width)
zoner7, do not be ashamed to post anything online. this site is made to get help/advises from others. Looking at your errors can really help you understand the subject more. So, next time, to be ashamed, just post them.
I also want to give you my up most respect for teaching yourself about...
o0o ok. now i am going to solve this using what alphysicist said about another person pulling Jane up a weigthless vine.
I am now going to assume that Tarzan, Jane and cheeta are the maximum weight on the vine that the other "person" can pull up. ****I am doing this like a pulley problem****...
ok so, the vine is like a rope, it is not moving at all. Jane does all the work. She does not hold on to a certain spot on the vine and it pulls her up. is that correct?
dudes, i am so sorry. I thought Jane did not climb but hung on onto the vine which acted like a spring :(. lol Now i see why you folks said that energy was not conserved. I was on a totaly different topic.
Ok so now let me get this right. the question asks to find the time that Jane will climb...
i am 100% sure that i am correct.
to find the velocity, I used Ke= 1/2*m*v^2 and soleved for v
2x*9.81m/s^2*60m=1/2*2x*v^2.
and yes, you can use the potential energy as the kenetic energy. Think of droping Jane with the same acceleation(replaces gravity) that she went up. At the top she will...
The 1st part looks ok.
i don't know too much about rotational inertia. but i saw 2 things that torubled me: the UNITS and the Direction. For your answer for B, you really dint give much of a direction and the units were off. When you did :
" 1/3(60kg)(1.4m)^2*(omega)= 39.2 kg m*3.95/sec= 154...