Question

A 0.10 kg meter stick is held perpendicular to a vertical wall by a 2.5 mm string going from the wall to the far end of the stick.
(A) Find the tension in the string.
(B) If a shorter string is used, will its tension be greater than, less than, or the same as that found in part (A)?
(C) Find the tension in a 2.0 m string.

Answer 1

Explanation:

a)

Sum of moments = 0 (Equilibrium)

T . cos (Q)*L = m*g*L/2

$cos Q = \frac{\sqrt{(2.5^2 - L^2) } }{2.5}$

$T*\frac{\sqrt{(2.5^2 - L^2) } }{2.5} * L = 0.1 *9.81*L/2$

$T = \frac{2.4525}{\sqrt{(2.5)^2 - L^2} }$

b) If the String is shorter the Q increases; hence, Cos Q decreases which in turn increases Tension in the string due to inverse relationship!

c)

$T = \frac{1.962}{\sqrt{(2)^2 - L^2} }$