Recent content by rlay023

Thank you very much. The rotation is opposite with the bottom rope being wrapped and the top rope exiting the drive. Does that change the location of the pulley? Also, could I mount a smaller idler pulley on a pole as show that would feed the bottom rope around the pulley at a higher point.

The birds nest in the fishing reel isn't the problem as it is only looped around the drum once; however the rope always finds away to cross over itself within a few revolutions. I didnt have the entire rope under tension at the moment but here are a few pics that may help illustrate:

I have the bottom idler tensioned with a ratchet strap. Are you thinking that isn't tight enough? I will take some pictures this evening and post.

Hello - We are building a backyard tow rope for my course. All of the electrical components are working as designed; however, the pulley can't consitently grip the rope when only placed around the rim. When looped around 1.5 times the rope continually shifts underneath itself causing the motor...

In doing further research on the forum, I stumbled upon a torque related post relevant to the same project - https://www.physicsforums.com/threads/help-with-motor-for-pulling-uphill.957569/ Following that as guide, I come up with the following solution: F = G * sin (theta) + R , where: G =...

Awesome. Thanks you so much. I had sized a 5HP to be on the safe side.

Thanks again for all your help. Depending on the how this is done provides very different answers, which I didn't expect to happen. Approach 1: If I add .43310kw to the original .82147kw, I get 1.25457kW or 1.68 bHP. Approach 2: If I take the normal force of 2619.4258N and add the...

Good catch. I used .1 in the actual equation, as the upper limit of the static friction coeficient - just transcribed it wrong. Sorry for the basic question here but (assuming the work/power calcs are done correctly on the friction part), how do I factor in the .433k1w from friction. Do I...

Something along the line of: F = uN = .01 * (2619.4258) = 261.94N where N = Cos(18.9)*272.2*9.8 Power = (F*D)/t = (261.94N)*(75.2m)/45.5sec = 433.10W

That is where I am stuck. The examples I have found are only on flat surface with friction or slope surface with no friction.

Thanks for your replies. Let's see if I can restate this in all the same units. What is the required HP to pull 272.2kg up a 14.0M incline with a slope of 18.9% in 45.5 seconds? Using the previous example and the response above, I arrive at 1.10. Work = mgh = 272.2(9.8)(14.0) Time = 45.5...

Sure thing. Here you go: https://www.physicsforums.com/threads/building-a-ski-rope-tow-need-help-with-motor-sizing.655897/

Hi Everyone - I found a previous post here that has helped me size an engine for a ski tow rope but wanted to take that a step further an add friction into the equation. If I followed that problem corrected, the tow rope requires 821W calculated as follows: power = work/time...