Tension in a Rope from Lowering Down and Friction Braking Torque at Stop

[Herbid]

Homework Statement

Tension of Rope from Lowering Down and Friction Braking Torque at Stopping (With Full Power Braking)

Relevant Equations

F = m.a

(The weight scale = 10 kg).
After initial force, now the mass is lowering down at constant speed of 1 m/s.

The motor with pulley's radius of 1 meter is exerted 100 Newton m to keep constant V.
The motor speed is about 9.55 rpm.

* All other weights, inertia and frictions of the system is neglected

The dynamic friction / torque braking of Electric Mechanical disc brake at motor's shaft is 150 Nm.
If the motor sudden / abruptly stopping (full braking) that moving mass_by Emergency Stop Push Button,
what is the (max) Tension of the rope?

Is the rope's Tension the same if the dynamic friction / torque braking 200 Nm or 300 Nm?

 

[Alex Petrosyan]

Don’t mean to be that guy, but usually you need to submit an attempted solution.

 

[Herbid]

Don’t mean to be that guy, but usually you need to submit an attempted solution.

Well, I think the force tension is 150 N.
F = Fbrk - Fmass
F = 150 N - 100 N
F = 50 N

Then F total is,
Ftot = F + Fmass
Ftot = 50 + 100
Ftot = 150 N.
========

But I wonder if the speed much slower, 0.1 m/s and if the speed higher 10 m/s,
by the equation above the tension of rope at braking is still the same = 150 N.

 

[haruspex]

Well, I think the force tension is 150 N.
F = Fbrk - Fmass
F = 150 N - 100 N
F = 50 N

Then F total is,
Ftot = F + Fmass
Ftot = 50 + 100
Ftot = 150 N.
========

But I wonder if the speed much slower, 0.1 m/s and if the speed higher 10 m/s,
by the equation above the tension of rope at braking is still the same = 150 N.

I don't follow your calculation (what is F?), but 150N looks right.
The answer is counterintuitive because you have to assume both the rope and the spring inside the scale are inextensible, which cannot be true in practice. That is why it feels like the answer should depend on the speed.