A one meter board with uniform density hangs in static equilibrium from a rope with tension T (0.2 m from the left end of the board). A weight hangs from the left end of the board as shown (3kg at the left end of the board). What is the mass of the board?

A. 1kg

B. 2kg

C. 3kg

D. 4kg

This is an equilibrium problem so I set up the equation as such with the counter-c lockwise forces on one side of the equation and clockwise forces on the other side. In addition, examkrackers audio osmosis stated that when the board is not massless you are supposed to use the center of gravity as the point in which is creates torque in the torque equation. (In this case, it would be the center of the meter board.) I chose the left of the board as my point of rotation. Therefore, my equation would become

Tx=m1g + m2gd2

T(0.2)=(30)+ m2(5)

T=m1g + m2g ? so T=3(10)+ m2(10)

Then T(0.2)= 6+ m2(5) so,

6+ m2(5) =(30)+ m2(5) and when you solve for m2=12kg (which is not one of the answers)

EK solved it with choosing where the string attaches to the board to be the point of rotation, which lead them to the answer of 2kg which is B. How am I supposed to know which is the best point of rotation to choose? In addition, where did I go wrong in trying to solve the equation my way?

Thank you,

Veronica