Find Force to Tilt a Refrigerator

[logan3]

Homework Statement

You are trying to tilt a very tall refrigerator (2.0 m high, 1.0 m deep, 1.4 m wide, and 100 kg) so that your friend can put a blanket underneath to slide it out of the kitchen. Determine the force that you need to exert on the front of the refrigerator at the start of its tipping. You push horizontally 1.4 m above the floor.

$L_{height} = 2.0 m$
$L_{depth} = 1.0 m$
$L_{width} = 1.4 m$
$L_{applied} = 1.4 m$
$m = 100 kg$
$g = 9.8 m/s^2$

Homework Equations

$\tau_{ccw} = \tau_{cw} \Rightarrow \vec F_w \frac {L_{depth}}{2} = \vec F_a L_{applied} \Rightarrow \vec F_a = \vec F_w \frac {L_{depth}}{2L_{applied}}$

The Attempt at a Solution

The bottom right corner will be our axis of rotation. In order for the refrigerator to tilt the applied, clockwise torque by the person must be larger than the weight, counterclockwise torque by the refrigerator. The applied force of the person is applied perpendicular to the vertical surface (height), while the weight force of the refrigerator is applied perpendicular to the horizontal surface (base/depth). The problem says I am pushing horizontally from the front of the fridge, therefore the depth is the horizontal length.

$\vec F_a = (100 kg)(9.8 m/s^2) \frac {(1.0m)}{2(1.4 m)} = 350 N$

Therefore, the moment when the fridge starts to tilt is when the applied force is > 350 N.

Thank-you

 

[Jilang]

Looks good to me.