What is the impact of changing the reference angle in a static friction problem?

[Jzhang27143]

Homework Statement

A disk of mass M and radius R is help up by a massless string. Let there now be friction between the disk and the string, with coefficient u. What is the smallest possible tension in the string at its lowest point?

The problem is problem 8 from here: http://www.personal.kent.edu/~fwilliam/Chapter 1 Statics.pdf

Homework Equations

Fnet = 0 at equilibrium. fs <= uN

The Attempt at a Solution

I understand how to do the problem when the angle theta is defined to be 0 at the lowest point and pi/2 at the right end. In this case, the result is T(0) >= Mg/2 e^(-u*pi/2). However, I wanted to see what would happen when theta is 0 at the right end and -pi/2 at the lowest point since it shouldn't matter. As in the solutions, the equation for tension as a function of theta should still be T(theta) <= T(0) e^(u*theta). In this case, T(0) = Mg/2 so T(theta) <= Mg/2 e^(u*theta). So if you plug in theta = -pi/2, T(-pi/2) <= Mg/2 e^(-u*pi/2). But here, the direction of the inequality is different. Was I supposed to switch the sign somewhere?

 

[CWatters]

The problem is problem 8 from here: http://www.personal.kent.edu/~fwilliam/Chapter 1 Statics.pdf

The wording of problem 8 appears to be different. It says..

8. Tetherball *
A ball is held up by a string, as shown in Fig. 1.12, with the string tangent to the ball. If the angle between the string and the wall is µ, what is the minimum coefficient of static friction between the ball and the wall, if the ball is not to fall?

 

[TSny]

The wording of problem 8 appears to be different. It says..

I believe it's problem 8 in section 1.4 rather than 1.3.

 

[CWatters]

Ok forget that. I see there is another Problem 8 on page I-13.

 

[TSny]

However, I wanted to see what would happen when theta is 0 at the right end and -pi/2 at the lowest point since it shouldn't matter. As in the solutions, the equation for tension as a function of theta should still be T(theta) <= T(0) e^(u*theta).

Are you sure that the direction of the inequality is correct for negative values of θ? A graph of $T(0) e^{\mu \theta}$ for both positive and negative θ might help.

 

[Jzhang27143]

Ok I see that the direction should be flipped since theta is negative. That would give me T(theta) >= T(0) e^(u*theta). So is this the lower bound for T(theta) and the solution's T(theta) is the upper bound?

 

[TSny]

For negative Θ, T(0) e^μΘ is a lower bound for the tension. For positive Θ, T(0) e^μΘ is an upper bound for the tension.

See the plot of T vs Θ shown below for the case where T(0) = 1 unit. The red curve is a plot of T(0) e^Θ for μ = 1 and represents the tension when the static friction is at its maximum. The blue horizontal line is the tension as a function of Θ when there is no friction. If you now add another curve that represents friction acting at less than its maximum it would need to lie between the red and blue graphs, like the dotted curve. Note the difference for positive and negative Θ.