# A uniform crate with a mass of 29 kg rests on a floor with a coefficient of static friction equal to 0.53. The crate is a uniform cube with sides 1.0 m in length. What horizontal force applied to...

A uniform crate with a mass of 29 kg rests on a floor with a coefficient of static friction equal to 0.53. The crate is a uniform cube with sides 1.0 m in length. What horizontal force applied to the top of the crate will initiate tipping?

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The crate is a cube with sides 1 m and has a mass of 29 kg. It rests on the floor with the coefficient of static friction between the floor and the crate being 0.53.

A horizontal force is being applied to the top of the crate at a height 1 m above the floor. The net vertical force acting on the box is the sum of the normal force N and the the weight of the box. The two add up to 0.

If a horizontal force F is applied on the box at the top a moment is created about the opposite lower edge. This is equal to `F*sqrt(1^2 + 1^2) = sqrt2*F` . There is a moment due to the weight of the box in the opposite direction equal to `29*9.8*sqrt(0.5^2 + 0.5^2)` = `29*9.8*0.5*sqrt 2` . If the net moment `F*sqrt 2 - 29*9.8*0.5*sqrt 2 > 0` , the box tips over. Also, the force F should be less than the static friction between the box and the floor, else the box would slide. This gives `F < 29*9.8*0.53`

Solving `F*sqrt 2 - 29*9.8*0.5*sqrt 2 > 0` gives

`F > 29*9.8*0.5 = 142.1` N

The frictional force is equal to `29*9.8*53 = 150.626` N

As the force required to tip over the box is less than the force of static friction, the box does not slide but tips over. The force required for this to happen is 142.1 N

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