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 distance the string can be pulled before slipping is <u>0.55 mm</u>
<h3>How does friction force determines the distance the string can be pulled?</h3>
The given parameters are;
Normal force on the string = 0.5 N
From a similar question posted online, we have;
<em>Length of the violin string, L = 0.33 m</em>
<em>Tension in the string, T = 60 N</em>
<em>Static friction, </em><em> = 0.8</em>
Required:
The distance the string can be pulled before it slips.
Solution:
Friction force, = 0.5 N × 0.8 = 0.4 N
According to Newton's law of motion, we have;
= 2·T × sin(θ)
Which gives;
0.4 N = 2 × 60 N × sin(θ)
Therefore;
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;
0.33 m = 300 × 2 × x = 600·x
- The distance the string is pulled before slipping is <u>0.55 mm</u>
Learn more about friction here:
brainly.com/question/11539805