Question #fae13

1 Answer
Feb 2, 2018

29 kg


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

#r_1W - r_2R = 0#

W = weight = mg
#r_1 = (200-10)mm = 190 mm#, 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 =(724-95)mm= 629mm#, distance from the pivot to R.

#R= (190W)/629 ~~ 0.3mg#

To express R in kg, divide it by g

#R/g ~~ 0.3m = 0.3*95kg ~~ 29 kg #

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) kg ~~ 99 kg# 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.