A 1.0 kg rock is suspended by a massless string from one end of a 1.0 m long measuring stick. What is the mass of the measuring stick if it is balanced by a support at the 0.25 m mark?
Torque=(distance perpendicular to force from the center of mass)(mass of force)(acceleration of force)
Torquesum=0 in static equilibrium
The Attempt at a Solution
Torquesum=0= +(counterclockwise)(.25meters)(9.81m/s^2)(1kg) + (counterclockwise)(?.5meters?)(.25total mass Stick)(9.81m/s^2) - (clockwise)(?.5meters?)(.75TMS)(9.81m/s^2)
solve for TMS and 2.4525=2.4525TMS; TMS=1kg
So, why is ?.5meters? considered to be how far away the meter stick is from the center of mass?
Originally, the individual centers of masses I tried to put as .25meters for the left part of the stick and .75meters for the right part of the stick. How did I make a mistake?