Answer
A) The tension in the string at the 0-cm mark is about 0.78N
B) The tension in the string at the 90-cm mark is about 0.98N
Work Step by Step
A) We sum the torques about the string on the right to find the tension in the string at the 0-cm mark. Therefore;
$(+ \circlearrowleft) \sum \tau=0$
$.180kg*9.8m/s^2*.40m-T*.90m=0$
$T=.180kg*9.8m/s^2*.40m/(.90m)$
$T\approx0.78N$
The tension in the string at the 0-cm mark is about 0.78N
B) We sum the torques about the string on the left to find the tension in the string at the 90-cm mark. Therefore;
$(+ \circlearrowleft) \sum \tau=0$
$-.180kg*9.8m/s^2*.50m+T*.90m=0$
$T=.180kg*9.8m/s^2*.50m/(.90m)$
$T\approx0.98N$
The tension in the string at the 90-cm mark is about 0.98N
