Equilibrium — Question 10

Question 10

Equilibrium depends on which motions a constraint allows

A bead is constrained to move horizontally along $y=h$ in a potential $U(x,y)=kx^2/2-mgy$, with $k,m,g>0$. The support is ideal and holds the bead at fixed height.

1. Find its constrained equilibrium and the vertical support force.

2. What happens if the vertical constraint is removed at that configuration?

Question 10 - Solution

Equilibrium depends on which motions a constraint allows

(a) Derivation and calculation

The specified potential produces $$\vec F=-\boldsymbol\nabla U=-kx\hat x+mg\hat y.$$ The allowed horizontal coordinate satisfies $$F_x=0\quad\Longrightarrow\quad\boxed{x=0}.$$ Vertical constraint balance requires $$F_y+R_y=0,\qquad\boxed{\vec R=-mg\hat y}.$$

(b) Result and interpretation

Removing the vertical constraint makes $R_y=0$. At the former constrained equilibrium, $$\ddot x=-kx/m=0,\qquad\boxed{\ddot y=g}.$$ For small allowed horizontal displacements, $$\Delta U\big|_{y=h}=\tfrac12kx^2>0,$$ so horizontal stability remains a valid constrained statement, but does not establish full two-dimensional equilibrium.

Physical check

The upward force follows the minus sign in the given $-mgy$ potential; it must not be replaced by conventional downward gravity.