- #1

AzimD

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- 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##?