Return to Answer

Since the connecting rod is not rotating we can conclude the net torque on the rod is zero. This means:
$$N_F l - N_R l = 0 $$ $$N_F = N_R $$

This sounds counterintuitive to me since I can't see the difference between this system with an arbitrarily small rod mass, and 2 wheels rolling down separately without a rod between them, since all the rod does (in my opinion) is giving us the constraint that the distance between the centres of the wheels stay constant, a constraint already satisfied in the 2 separate wheels case naturally, assuming same initial speeds.

This is an excellent argument that $N_F = N_R $ in agreement with the conclusion above. I am not sure why you are convinced that $N_F \ne N_R$? Is that what your teacher said the answer is?

A massless rod changes nothing, UNLESS there is friction between the rod and the wheel.

Clearly $N_F l > N_R l$ in the above image, but this is only because the brakes are being applied. If there is no braking force, then $N_F = N_R.$

$N_F + N_R = m_{total} g \cos(\theta)$ From these equations we calculate different NF and NR, something I'm having trouble understanding.

If $N_F + N_R = m_{total} \ g \cos(\theta) $ then $N_F = N_R $
is a perfectly valid solution.

The torque ($m \ g \ \sin(\theta) R $) that acts on the wheels, that causes their angular momentum to increase as they roll down the slope, does not act on the connecting rod or alter the normal forces on the wheels, if there is no friction between the wheels and the connecting rod.

Clearly $N_F l > N_R l$ in the above image, but this is only because the brakes are being applied. If there is no braking force, then $N_F = N_R.$

$N_F + N_R = m_{total} g \cos(\theta)$ From these equations we calculate different NF and NR, something I'm having trouble understanding.

If $N_F + N_R = m_{total} \ g \cos(\theta) $
then $N_F = N_R $
is a perfectly valid solution.

Since the connecting rod is not rotating we can conclude the net torque on the rod is zero. This means:
$$N_F l - N_R l = 0 $$ $$N_F = N_R $$

Your own argument in the OP that a massless rod connecting the wheels does not change the normal forces, is absolutely correct and is a perfect argument to support the conclusion that $N_F = N_R $. I am not sure why you are convinced that $N_F \ne N_R$? Is that what your teacher said the answer is?

The torque ($m \ g \ \sin(\theta) R $) that acts on the wheels, that causes their angular momentum to increase as they roll down the slope, does not act on the connecting rod or make the normal forces on the wheels unequal, if there is no friction between the wheels and the connecting rod.

Since the connecting rod is not rotating we can conclude the net torque on the rod is zero. This means:
$$N_F l - N_R l = 0 $$ $$N_F = N_R $$

This sounds counterintuitive to me since I can't see the difference between this system with an arbitrarily small rod mass, and 2 wheels rolling down separately without a rod between them, since all the rod does (in my opinion) is giving us the constraint that the distance between the centres of the wheels stay constant, a constraint already satisfied in the 2 separate wheels case naturally, assuming same initial speeds.

This is an excellent argument that $N_F = N_R $ in agreement with the conclusion above. I am not sure why you are convinced that $N_F \ne N_R$? Is that what your teacher said the answer is?

A massless rod changes nothing, UNLESS there is friction between the rod and the wheel.

Clearly $N_F l > N_R l$ in the above image, but this is only because the brakes are being applied. If there is no braking force, then $N_F = N_R.$

$N_F + N_R = m_{total} g \cos(\theta)$ From these equations we calculate different NF and NR, something I'm having trouble understanding.

If $N_F + N_R = m_{total} \ g \cos(\theta) $ then $N_F = N_R $
is a perfectly valid solution.

Since the connecting rod is not rotating we can conclude the net torque on the rod is zero. This means:
$$N_F l - N_R l = 0 $$ $$N_F = N_R $$

This sounds counterintuitive to me since I can't see the difference between this system with an arbitrarily small rod mass, and 2 wheels rolling down separately without a rod between them, since all the rod does (in my opinion) is giving us the constraint that the distance between the centres of the wheels stay constant, a constraint already satisfied in the 2 separate wheels case naturally, assuming same initial speeds.

This is an excellent argument that $N_F = N_R $ in agreement with the conclusion above. I am not sure why you are convinced that $N_F \ne N_R$? Is that what your teacher said the answer is?

A massless rod changes nothing, UNLESS there is friction between the rod and the wheel.

Clearly $N_F l > N_R l$ in the above image, but this is only because the brakes are being applied. If there is no braking force, then $N_F = N_R.$

$N_F + N_R = m_{total} g \cos(\theta)$ From these equations we calculate different NF and NR, something I'm having trouble understanding.

If $N_F + N_R = m_{total} \ g \cos(\theta) $ then $N_F = N_R $
is a perfectly valid solution.

The torque ($m \ g \ \sin(\theta) R $) that acts on the wheels, that causes their angular momentum to increase as they roll down the slope, does not act on the connecting rod or alter the normal forces on the wheels, if there is no friction between the wheels and the connecting rod.

Since the connecting rod is not rotating we can conclude the net torque on the rod is zero. This means:
$$N_F l - N_R l = 0 $$ $$N_F = N_R $$

This sounds counterintuitive to me since I can't see the difference between this system with an arbitrarily small rod mass, and 2 wheels rolling down separately without a rod between them, since all the rod does (in my opinion) is giving us the constraint that the distance between the centres of the wheels stay constant, a constraint already satisfied in the 2 separate wheels case naturally, assuming same initial speeds.

This is an excellent argument that $N_F = N_R $ in agreement with the conclusion above. I am not sure why you are convinced that $N_F \ne N_R$? Is that what your teacher said the answer is?

A massless rod changes nothing, UNLESS there is friction between the rod and the wheel.

Clearly $N_F l > N_R l$ in the above image, but this is only because the brakes are being applied. If there is no braking force, then $N_F = N_R.$

$N_F + N_R = m_{total} g \cos(\theta)$ From these equations we calculate different NF and NR, something I'm having trouble understanding.

If $m_{total} g \cos(\theta) = 100N$ then $N_F = N_R = 50N$
and $50N + 50N = 100N$ is a perfectly valid solution. Why do you exclude this possibility?

Since the connecting rod is not rotating we can conclude the net torque on the rod is zero. This means:
$$N_F l - N_R l = 0 $$ $$N_F = N_R $$

This sounds counterintuitive to me since I can't see the difference between this system with an arbitrarily small rod mass, and 2 wheels rolling down separately without a rod between them, since all the rod does (in my opinion) is giving us the constraint that the distance between the centres of the wheels stay constant, a constraint already satisfied in the 2 separate wheels case naturally, assuming same initial speeds.

This is an excellent argument that $N_F = N_R $ in agreement with the conclusion above. I am not sure why you are convinced that $N_F \ne N_R$? Is that what your teacher said the answer is?

A massless rod changes nothing, UNLESS there is friction between the rod and the wheel.

Clearly $N_F l > N_R l$ in the above image, but this is only because the brakes are being applied. If there is no braking force, then $N_F = N_R.$

$N_F + N_R = m_{total} g \cos(\theta)$ From these equations we calculate different NF and NR, something I'm having trouble understanding.

If $N_F + N_R = m_{total} \ g \cos(\theta) $ then $N_F = N_R $
is a perfectly valid solution.