# Why are clockwise and anticlockwise moments not balanced in this equilibrium scenario? If we take moments about the peg, which is directly below centre of mass of the cone (which is $r$ cm from its circular base) we get $3rTcos(90-ø) = Trcos(90-ø)$. $T$ = tension in string. Clearly this has no solution unless T is 0. And moments should be balanced, as the cone is in equilibrium.

Is this question wrong, or is there a force - or something else- i’m missing?

The problem is that you are only considering the moments due to the vertical components of the forces. The horizontal components also have a contribution to the moment, which you are currently ignoring.

• Oh yeah, that explains it. My mechanics is quite rusty at the moment – Logan545 Apr 22 '17 at 20:44

The force $T$ acts through the peg, therefore its moment about the peg is zero. Taking components of $T$ vertically (and horizontally) is not going to alter this situation.

You have actually already balanced moments when you assumed that the CG of the cone lies vertically below the peg. The only force acting on the cone which does not necessarily pass through the peg is the weight $W$ of the cone. When this force does pass through the peg then all moments on the cone are zero.

The cone can be balanced with various lengths of string. Ensuring that the CG lies below the peg constrains the relation between the angles $\theta_1, \theta_2$ which the string at the left and right makes with the horizontal. These angles are not necessarily equal. Making them equal is not required for static equilibrium. This is a geometrical constraint.

(a) Can be solved by applying geometry. Assume right hand upper edge is distance $x$ below the peg, then equate expressions for $\tan\theta$ in the right-triangles formed by right and left sections of the string.

(b) Can be solved by balancing forces vertically.