Bicycle and Pulley, part 2

[brianinbwangju]

1. The problem statement, all variables and given/known data
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 the tension scale?


2. Relevant equations
T = rF
Fslope= m*g*sin(angle)
x component of F = cos(angle) * mag F



3. The attempt at a solution
I am basing this solution on my solution for Bicycle and Pulley, part 1,
http://www.physicsforums.com/showthread.php?t=148577
but I didn't get any feedback to know whether part 1 was correct. Also I am a little more hazy on part 2, but I want to know if I am right so I'll proceed.

F1 is the force pulling the bicycle down the incline
F1 = m*g*sin(12) = 14kg * sin(12) = 2.96kg


There are 2 forces opposing F1 :
F2, the friction force of the wheel on the incline
F3, the force from the rope
Both forces act on the rear wheel

Considering the direction of the forces acting on the bicycle,
F2 + F3 = F1
F2 = cos(30)* F3 = .866*F3
(Should that be the magnitudes of the forces?)

F1 = 2.96kg
F2 = 1.3
F3= 1.6

So, the tension scale should read 1.6kg

Note: The tension scale on the bicycle of part 1 read 2.96. This means, in part 2, there will be less tension on the rope than in part 1

[brianinbwangju]

Oh, right, here is the attachment

[OlderDan]

This should be no different from the bicycle part of the first problem. Only friction keeps the bike from moving. The rope torque must cancel the friction torque.

[brianinbwangju]

This should be no different from the bicycle part of the first problem. Only friction keeps the bike from moving. The rope torque must cancel the friction torque.


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 rotate.

In Part 1,
F1 is the component of gravity pulling the bicycle down the ramp
F2 is the force of friction of the ramp
F3 is the force of the rope pulling on the bike frame
F4 is the resulting force of the rope pulling on the bike frame
Here I think F4 cancels F3

In Part 2,
F1 is the component of gravity pulling the bike down the ramp
F2 is the force of friction of the ramp
F3 is the force of the rope pulling on the bike frame
F4 is the resulting force of the rope pulling on the bike frame
Here I think F1 is causing the wheel to rotate one way, while F2, F3, and F4 are causing the wheel to rotate the other way

[OlderDan]

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 rotate.

In Part 1,
F1 is the component of gravity pulling the bicycle down the ramp
F2 is the force of friction of the ramp
F3 is the force of the rope pulling on the bike frame
F4 is the resulting force of the rope pulling on the bike frame
Here I think F4 cancels F3

In Part 2,
F1 is the component of gravity pulling the bike down the ramp
F2 is the force of friction of the ramp
F3 is the force of the rope pulling on the bike frame
F4 is the resulting force of the rope pulling on the bike frame
Here I think F1 is causing the wheel to rotate one way, while F2, F3, and F4 are causing the wheel to rotate the other way

The key point here is that the force of the rope acting on the bike is NOT contributing to the force needed to keep the bike from moving down the plane. There are equal and opposite forces acting on the bike at the ends of the rope between the wheel and the frame. This is no different from applying the brakes to keep the wheel from moving, and it is no different from having the rope attached anywhere else on the bike. The rope must porovide a torque that opposes the torgue due to friction.