# Question #fae13

Feb 2, 2018

29 kg

#### Explanation:

Balance the torques acting on the cabinet above the pivot point. i.e.,

${r}_{1} W - {r}_{2} R = 0$

W = weight = mg
${r}_{1} = \left(200 - 10\right) m m = 190 m m$, distance from the pivot to the center of mass in the horizontal direction as this distance is perpendicular to the weight vector.

R = reaction force exerted by the wall on the cabinet at the anchor. This force is left pointing/

Solving the equation above,

${r}_{2} = \left(724 - 95\right) m m = 629 m m$, distance from the pivot to R.

$R = \frac{190 W}{629} \approx 0.3 m g$

To express R in kg, divide it by g

$\frac{R}{g} \approx 0.3 m = 0.3 \cdot 95 k g \approx 29 k g$

The pivot assumes more than weight, it also takes on the horizontal force exerted on the cabinet by the wall at the anchor. Effective, the pivot experience $\sqrt{{29}^{2} + {95}^{2}} k g \approx 99 k g$ force. The pivot is part the base board, hence the baseboard absorbs all the force. In case the board is removed, then the anchor will take on all the weight.