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

In summary: For negative Θ, the tension increases as Θ decreases, while for positive Θ, the tension decreases as Θ decreases.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) eΘ 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.
  • #1
Jzhang27143
38
1

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?
 
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  • #2
Jzhang27143 said:

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?
 
  • #3
CWatters said:
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.
 
  • #4
Ok forget that. I see there is another Problem 8 on page I-13.
 
  • #5
Jzhang27143 said:
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.
 
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  • #6
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?
 
  • #7
For negative Θ, T(0) eμΘ is a lower bound for the tension. For positive Θ, T(0) eμΘ 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) eΘ 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.
 

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What is a disk held up by string?

A disk held up by string is a physical demonstration of equilibrium in which a disk is suspended by two strings, each attached to opposite sides of the disk, and held in place by the tension in the strings.

What is the principle behind a disk held up by string?

The principle behind a disk held up by string is Newton's First Law of Motion, which states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity unless acted upon by an external force.

How does the tension in the strings affect the disk?

The tension in the strings is what keeps the disk in place and prevents it from falling. The strings are pulling in opposite directions with equal force, creating a state of equilibrium where the forces are balanced and the disk remains stationary.

What factors can affect the stability of a disk held up by string?

The stability of a disk held up by string can be affected by the weight and size of the disk, the length and angle of the strings, and any external forces acting on the disk, such as air resistance or vibrations.

What real-world applications does a disk held up by string have?

A disk held up by string is often used as a simple demonstration of equilibrium in physics and can also be used to study the effects of forces and stability in engineering and architecture. It can also be found in various children's toys and games as a fun and interactive way to learn about balance and gravity.

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