# 60 in Examkrackers Physics textbook

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#### [email protected]

##### Member
15+ Year Member

Please refer to number 60 in Examkrackers Physics textbook for helpful diagram

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

#### G1SG2

##### Full Member
10+ Year Member
5+ Year Member
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)
I'm not sure I understand what you did in the above equations...where did you get the 6 from?

How am I supposed to know which is the best point of rotation to choose?
I think EK is right about using the center of gravity of the board (if its not massless) as the axis of rotation, but I usually pick a point where there is an unknown so it would get zero torque. But in this situation, there were two unknowns, so I picked the center as that of the board. You used the left side of the board as your axis of rotation-that would be problematic, as that would only allow you to use the tension and the weight of the board in your torque equation-BOTH of which are unknowns. So try to pick a point for the axis of rotation that would give zero torque to one of your unknowns.

Here is how I solved it:

I used the center of the board as the axis of rotation, which is 0.5 m. Then I set up my torque equation to find the tension:

(Note that the Torque for the board would be zero because I chose the middle of the board as my axis of rotation.)

0.3 T=0.5 x 3 (9.8)
T=49 N

Then set up your forces up equals forces down equation:

49N=(3kg)(9.8) + x
49N =29.4N + x
x=19.6N. Divide by g and you end up with 2 kg.

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