Tension in rope wrapped around a rod

[ThEmptyTree]

Homework Statement

Two unequal blocks of masses $m_1$ and $m_2$, $m_1 > m_2$, 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 $\mu_s$, and the
coefficient of sliding friction is $\mu_k$, $\mu_k < \mu_s$. The mass of the rope can be ignored.
(a) Let $T_1$ be the magnitude of the force of tension exerted by the rope on block 1.
Is $T_A$ greater, less than, or equal to $T_1$?
(b) What is the value of m1 for which the rope starts sliding? Express your answer
in terms of $\mu_s$ and $m_2$.
(c) Now assume that $m_1$ 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?
Express your answer in terms of some or all of the following: $\mu_k, m_1, m_2$ and $g$.

Relevant Equations

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, $T_B$, is related to the tension at point $A$ in the rope,
$T_A$, by:
$$T_B = T_Ae^{-\mu_s\theta}$$
where $\theta$ is the angle subtended by the portion of the rope in contact with the rod. In
this problem, the angle $\theta$ is $\theta = \pi$. (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).

Just when the masses start moving, the relationship between $T_A$ and $T_B$
becomes
$$T_B = T_Ae^{-\mu_k\theta}$$
, where $\mu_s$ is replaced by $\mu_k$. You can show this by following similar logic used in solving example 8.11 in the textbook.

 

[ThEmptyTree]

This is the first problem where I couldn't find a solution on the internet whatsoever. Am I doing gud?

 

[haruspex]

Homework Statement:: Two unequal blocks of masses $m_1$ and $m_2$, $m_1 > m_2$, 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 $\mu_s$, and the
coefficient of sliding friction is $\mu_k$, $\mu_k < \mu_s$. The mass of the rope can be ignored.
(a) Let $T_1$ be the magnitude of the force of tension exerted by the rope on block 1.
Is $T_A$ greater, less than, or equal to $T_1$?
(b) What is the value of m1 for which the rope starts sliding? Express your answer
in terms of $\mu_s$ and $m_2$.
(c) Now assume that $m_1$ 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?
Express your answer in terms of some or all of the following: $\mu_k, m_1, m_2$ and $g$.
Relevant Equations:: 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, $T_B$, is related to the tension at point #A# in the rope,
$T_A$, by:
$$T_B = T_Ae^{-\mu_s\theta}$$
where $\theta$ is the angle subtended by the portion of the rope in contact with the rod. In
this problem, the angle $\theta$ is $\theta = \pi$. (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).

Just when the masses start moving, the relationship between $T_A$ and $T_B$
becomes
$$T_B = T_Ae^{-\mu_k\theta}$$
, where $\mu_s$ is replaced by $\mu_k$. You can show this by following similar logic used in solving example 8.11 in the textbook.

View attachment 288149

View attachment 288160View attachment 288161

Very good.

 

[ThEmptyTree]

Very good.

Thanks. !