The normal force exerted on an object by a surface is distributed over the area

Question

The normal force exerted on an object by a surface is distributed over the area of contact between the object and the surface. The effective point of application of the normal force must then be located to produce a torque equal to the net torques of the distributed force (much like how we locate the center of mass when determining the torque due to the weight of an object). Consider a filing cabinet of height 1.3 m and width 0.40 m resting on a level, horizontal floor. The cabinet has a weight of 8.7 102 N and is loaded so that its center of mass is at its geometric center. Now suppose someone leans against it, pushing horizontally with a force of 55.0 N at the top edge, and the cabinet remains in static equilibrium (see figure below). Locate the effective point of application of the normal force due to the floor as measured from the left side of the cabinet.
m

Explanation / Answer

center of cabinet is 0.2m in x and 0.65m in y

Now

Weight of cabinet(W)= 8.7 x 102 N=870 N

Leaning force(F)=55N

Height of cabinet(h)=1.3m

Distance of effective normal force from point of turning of cabinet=x

For the body in equilibrium,

Clock wise moments = anti clockwise moments

F * h= W * x

55 * 1.3 = 870 * x

x=0.0822m

Hence from left side it is 0.2+0.0822 = 0.2822m

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