Recent content by brianinbwangju

I am sorry that I don't have much knowledge, so I may be making a convoluted mess of this but... I don't see the rope torque cancelling the friction torque. In the attachment I labelled what I think are the forces acting on the wheel, and what direction the forces are causing the wheel to...

Oh, right, here is the attachment

Homework Statement A 14kg bicycle is on a 12 degree incline. A tension scale is connected by a rope to both the rear wheel circumference and a metal bar that is extending the bicycle frame rearward. The line of the rope is at a 30 degree angle to the metal bar. What will be the reading on...

T = rF F2 and F3 are both acting on the cylinder, so F2 = F3 F1 = F4 F3 = 1/2 F4

I'll go a little further in my interpretation to see if it sounds right. In the bicycle there is a force, F1 = m*g*sin12, that is pulling the bicycle down the incline. The same force is pulling down the cylinder. On the pulley there are 2 opposing forces. F2, the friction force of the...

By the way, this seems to show that a approaches g as m1 approaches 0

There is a tension scale on the bike part also. It is connected to the rear tire on one side and seat post on the other side. The scale is supposed to come off as a tangent to the rear tire.

Homework Statement What will the scale read in these 2 examples? The incline is 12 degrees. The bicycle weight is 14kg. The cylinder weight is 14kg. Homework Equations Force of 2nd pulley 'arm' is 1/2 weight Force on incline is force X sin(angle) The Attempt at a...

Hi. Thank you so much for posting this thread which I am so interested in! I this the right way to find the answer: earth's gravity*rider's mass*18m*(1/12)m = horizontal force*30m*(1/6)m (9.8ms2)(70kg)(18m) / 12m = horizontal force (30m) / 6m 205N = horizontal force