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 times 10^2 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. 0.317 Your response differs from the correct answer by more than 10%. Double check your calculations. m

Explanation / Answer

As the object is in equilibrium, net force and net torque acting on it must be zero.

Apply, Fnety = 0

N - m*g = 0

N = m*g

let x is the required distance.

Apply net torque about left bottom = 0

Fp*1.3*sin(90) + m*g*(0.4/2)*sin(90) - N*x*sin(90) = 0

Fp*1.3 + m*g*(0.4/2) - m*g*x = 0

==> x = (Fp*1.3 + m*g*0.2)/(m*g)

= (55*1.3 + 870*0.2)/870

= 0.282 m

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