Due to the high tension in the string, the distance the string is pulled
before it slips is a small fraction of the length of the string.
Response:
The given parameters are;
Normal force on the string = 0.5 N
From a similar question posted online, we have;
Length of the violin string, L = 0.33 m
Tension in the string, T = 60 N
Static friction, [tex]\mu_s[/tex] = 0.8
Required:
The distance the string can be pulled before it slips.
Solution:
Friction force, [tex]F_f[/tex] = 0.5 N × 0.8 = 0.4 N
According to Newton's law of motion, we have;
[tex]F_f[/tex] = 2·T × sin(θ)
Which gives;
0.4 N = 2 × 60 N × sin(θ)
Therefore;
[tex]sin(\theta) = \dfrac{0.4}{2 \times 60} = \mathbf{ \dfrac{1}{300}}[/tex]
The triangle formed by the half length of the string and the displacement gives;
Where;
x = The distance the string is pulled before slipping
Which gives;
[tex]\dfrac{1}{300} = \mathbf{\dfrac{2 \cdot x}{0.33 \, m}}[/tex]
0.33 m = 300 × 2 × x = 600·x
[tex]x = \dfrac{0.33 \, m}{600} = 0.00055 \, m = \mathbf{0.55 \, mm}[/tex]
Learn more about friction here:
https://brainly.com/question/11539805
