Jun-pyo, Jandi, and the art of falling down

Question: Jandi shoves ice cream into Jun-pyo’s face. What is the minimum force Jandi needs to apply so that Jun-pyo falls down?

Answer: ~696 N.


There are four forces acting on Jun Pyo: Fnorm, Ffrict, Fapp, and Fgrav.


Image result for free body diagram with friction and force applied

image from physicsclassroom.com

Let’s first talk about Fnorm and Fgrav.

Fnorm is a contact force. Basically, whenever an object lies on a surface, or is touching a surface, the surface will apply a normal force on the object. In this case, since Jun Pyo is standing on ground, the ground is applying a normal force on him.

Fgrav is the force of gravity. Gravity acts on every object, and it always points downwards.

Jun Pyo is standing on ground, but he isn’t ascending vertically into the air nor is he falling into the earth. Therefore, the two vertical forces cancel each other out: Fnorm=Fgrav.

Jun Pyo is experiencing two other forces: Ffriction and Fapplied.

Fapplied is the force that the ice cream exerts on his body. (If Jun Pyo were simply standing on the ground, without Jandi pushing him, then there would be no Fapplied.) Ffriction is force that opposes any sliding motion. If Fapplied < or = Ffriction, then Jun Pyo won’t move. If Fapplied > Ffriction, the Jun Pyo will move.

Fapplied must exceed Ffriction if Jun Pyo falls.

By definition: Ffriction=μFnorm (*)

And in this problem, Fnorm=Fgravity since Jun Pyo isn’t flying upwards or sinking downwards into the earth.

By definition: Fgravity=mg

Therefore, substituting into (*): Ffriction = μmg

Lee Min Ho weighs 71kg, according to Asianwiki. Gravity is 9.8m/s^2.

So Ffriction =μ(71kg)(9.8m/s^2)

Assuming that static friction is between rubber shoes and concrete, the coefficient is 1.0.

Therefore Ffriction = (1.0)(71)(9.8) = 696 N

Fapplied must be greater than 696 N in order for Lee Min Ho to move.


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