IDNLearn.com is committed to providing high-quality answers to your questions. Discover thorough and trustworthy answers from our community of knowledgeable professionals, tailored to meet your specific needs.
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
