Unstable equilibrium, centre of mass problem

[Kennedy]

Homework Statement

Two masses, A and B, are placed on the edges of a mass of 20 kg which is homogeneously distributed figure such that an unstable equilibrium is achieved. If the mass of A is 12 kg and the mass of B 16 kg, how large is the distance x from the left end of the board to the support point? A) 3 m B) 3.25 m C) 3.4 m D) 3.75 m E) 4 m

Homework Equations

Centre of mass = m1(x1) + (m2)(x2)/(m1 + m2)

The Attempt at a Solution

I attempted this problem but got it wrong, mostly because I don't know what an unstable equilibrium is. I looked it up and it appears to me that any external force that acts on a system with an unstable equilibrium will gravely affect the system. Is this correct? I just went about finding the centre of mass of the system, by doing 12(0) + (16)(6) / (12 + 16) = 3.43 m. So, that was my answer, but it is wrong. I also never used the mass of the board, by I had assumed that since the mass was homogeneously distributed that it wouldn't affect the centre of mass.

 

[PeroK]

Homework Statement

Two masses, A and B, are placed on the edges of a mass of 20 kg which is homogeneously distributed figure such that an unstable equilibrium is achieved. If the mass of A is 12 kg and the mass of B 16 kg, how large is the distance x from the left end of the board to the support point? A) 3 m B) 3.25 m C) 3.4 m D) 3.75 m E) 4 m

Homework Equations

Centre of mass = m1(x1) + (m2)(x2)/(m1 + m2)

The Attempt at a Solution

I attempted this problem but got it wrong, mostly because I don't know what an unstable equilibrium is. I looked it up and it appears to me that any external force that acts on a system with an unstable equilibrium will gravely affect the system. Is this correct? I just went about finding the centre of mass of the system, by doing 12(0) + (16)(6) / (12 + 16) = 3.43 m. So, that was my answer, but it is wrong. I also never used the mass of the board, by I had assumed that since the mass was homogeneously distributed that it wouldn't affect the centre of mass.

Perhaps best not to assume that. Put the 20kg mass into the calculations - it can't do any harm!

 

[Kennedy]

Perhaps best not to assume that. Put the 20kg mass into the calculations - it can't do any harm!

How would I add it into the calculations? I know its mass is 20 kg, but how do I figure out where the mass is acting?

 

[PeroK]

How would I add it into the calculations? I know its mass is 20 kg, but how do I figure out where the mass is acting?

It's homogeneous, so it COM must be in the middle, surely?

 

[Kennedy]

It's homogeneous, so it COM must be in the middle, surely?

You are one hundred percent right! 0(12) + 16(6) + (20)(3) / (20 + 12 +16) = 3.25 m. Thank you. I never thought of adding that in, but I knew that its centre of mass was in the middle.