Tension in Rope Wrapped Around a Rod

[AzimD]

Homework Statement

Two unequal blocks of masses m1 and m2, m1 > m2, are suspended by a rope over a
fixed rod. The axis of the rod is perpendicular to the figure (only its cross section is
shown). The coefficient of static friction between the rope and the rod is µs, and the
coefficient of sliding friction is µk, µk < µs. The mass of the rope can be ignored.
For this problem , you can use the result from Chapter 8, Example 8.11 in the course
textbook, titled ”The Capstan”, where it is shown that when the rope is about to slide
the tension at point B in the rope, TB, is related to the tension at point A in the rope,
TA, by:
TB = TAe
−µsθ
where θ is the angle subtended by the portion of the rope in contact with the rod. In
this problem, the angle θ is θ = π. (Note: points A and B are the points where the
rope loses contact with the surface of the rod and we assume the cross section of the
rod to be a perfect circle).
(a) Let T1 be the magnitude of the force of tension exerted by the rope on block 1.
Is TA greater, less than, or qual to T1?
(b) What is the value of m1 for which the rope starts sliding? Express your answer
in terms of µs and m2.
(c) Now assume that m1 is large enough so that the rope starts to slip and the masses
start to move. What is a, the magnitude of the acceleration of the masses after
sliding has begun?
(Hint: Just when the masses start moving, the relationship between TA and TB
becomes TB = TAe
−µkθ
, where µs is replaced by µk. You can show this by following
similar logic used in solving example 8.11 in the textbook.)
Express your answer in terms of some or all of the following: µk, m1, m2, and g.

Relevant Equations

$T_B = T_Ae^{−\mu_sθ}$

On question B, I've attempted a solution that I have posted. However, I don't think it's correct. Am I allowed to treat this rope that wraps around the rod as a "negative" force and simply attach it to the other side of the equation such that $m_1g=m_2g+\mu_sN$?

 

[erobz]

On question B, I've attempted a solution that I have posted. However, I don't think it's correct. Am I allowed to treat this rope that wraps around the rod as a "negative" force and simply attach it to the other side of the equation such that $m_1g=m_2g+\mu_sN$?

For part b) you want to satisfy equilibrium for each mass using the given relationship for the tensions at point A and B.

 

[haruspex]

The "normal" force between the rope and the rod is distributed around the arc and points in different directions at different elements. You cannot just write $f_{s_{umax}}=(m_1+m_2)g\mu_s$.
You are given the equation for the maximum difference between the two tensions.

In part a) you were asked how those tensions relate to the forces on the masses. What was your answer? Use that in part b).

 

[AzimD]

The "normal" force between the rope and the rod is distributed around the arc and points in different directions at different elements. You cannot just write $f_{s_{umax}}=(m_1+m_2)g\mu_s$.
You are given the equation for the maximum difference between the two tensions.

In part a) you were asked how those tensions relate to the forces on the masses. What was your answer? Use that in part b).

Okay, I understand that now. So to attempt this again I know that:
$T_1=T_A$ and $T_2=T_B$. I also know that $T_B=T_Ae^{-\mu_s\theta}$

$m_1$ at the point where it begins to slide would mean that $T_1$ and $T_2$ are equal correct?

So would that mean:
$m_1g=m_2ge^{-\mu_s\pi}$
and so:
$m_1=m_2e^{-\mu_s\pi}$

 

[kuruman]

$m_1=m_2e^{-\mu_s\pi}$

You are told that $m_1 > m_2.$ Is your equation consistent with that?

 

[haruspex]

$m_1$ at the point where it begins to slide would mean that $T_1$ and $T_2$ are equal correct?

No… why would it?
You already have the condition for being about to slide: $T_B=T_Ae^{-\mu_s\theta}$. If $T_B$ is less than that then it cannot restrain $m_1$.

 

[AzimD]

@haruspex @kuruman I realized I had a couple of mathematical errors as well as an error in conceptualizing the problem. Here's my reattempt at it:
$T_1=T_A=m_1g$ and $T_2=T_B=m_2g$ and $T_B=T_Ae^{-\mu_s\theta}$ so $T_A=T_Be^{\mu_s\theta}$
Thus:
$m_1g=m_2ge^{\mu_s\pi}$
$m_1=m_2e^{\mu_s\pi}$
But since we're discussing the point at which it will begin to slip it should be:
$m_1>m_2e^{\mu_s\pi}$
Is this now correct?

 

[AzimD]

I also attempted to solve part C. From the FBD's of each block I got:
$T_B-m_2g=m_2a$ and $m_1g-T_A=m_1a$
using the relationships provided, I was able to obtain the answer:
$a=g\frac{m_1-m_2e^{\mu_k\pi}} {m_1+m_2e^{\mu_k\pi}}$

 

[haruspex]

@haruspex @kuruman I realized I had a couple of mathematical errors as well as an error in conceptualizing the problem. Here's my reattempt at it:
$T_1=T_A=m_1g$ and $T_2=T_B=m_2g$ and $T_B=T_Ae^{-\mu_s\theta}$ so $T_A=T_Be^{\mu_s\theta}$
Thus:
$m_1g=m_2ge^{\mu_s\pi}$
$m_1=m_2e^{\mu_s\pi}$
But since we're discussing the point at which it will begin to slip it should be:
$m_1>m_2e^{\mu_s\pi}$
Is this now correct?

I also attempted to solve part C. From the FBD's of each block I got:
$T_B-m_2g=m_2a$ and $m_1g-T_A=m_1a$
using the relationships provided, I was able to obtain the answer:
$a=g\frac{m_1-m_2e^{\mu_k\pi}} {m_1+m_2e^{\mu_k\pi}}$

Yes, all correct.

 

[AzimD]

Thank you all! @erobz @kuruman @haruspex