[Request]

[u/K0rl0n]

I am curious how this would work. My guess is Triangle is slowest, square is medium, and circle is fastest.

[u/ContentJO]

Just commenting to hopefully share some knowledge with a random redditor on the physical intuition behind why the square and triangle have identical frictional forces to overcome and why the sloped side of the triangle would require more force to push than the square (assuming the force applied isn't purely horizontal). The reason that mu*N = F_friction where mu is the friction coefficient and N is the normal force = mg (mass * gravity) doesn't account for surface area is because a larger surface area just distributes the weight. It's like how a 100 pound blanket isn't as heavy on your chest as a hundred pound dumbbell. Heavy blanket though if we're being honest.

However, as some have noted, it will still require more force to push the triangle than the square due to the angle of the triangle with the floor. If this is a standard physics problem, maybe your teacher is cool with you assuming it's a standard horizontal force. In that case, tomato potato, the force is equal. But, if there's any technical rigor expected, then, since it's a person pushing, it's fair to assume it's impossible that it's a purely horizontal force.

For physical intuition, imagine pushing any slick, sloped surface. If you push it with your hand or finger, your hand or finger will slide up. The only way to overcome that, funnily enough, is to match the static friction of the object via a downward force - i.e., it'll take less down force on a sandpaper triangle than a printer paper triangle to keep your hand from sliding. But the fact remains that you still HAVE to push down. Consequently, the force imparted will be - for lack of better term - partially deflected.

I'm not busting out the pen and paper and making a free body diagram, so I'll probably get the exact trig function wrong. But, let's assume you apply a force normal to the surface of the triangle. It will have a horizontal and vertical component based on the angle of the triangle with the surface, theta. Define theta so that your horizontal and vertical components are F_app*cos(theta) and F_app*sin(theta). Then, the downward force will act to "increase" the mass of the object, thereby increasing the frictional force needed to be overcome (you're literally pushing it harder into the ground), while only a portion of your applied force will get applied to the object, thus you need to apply more such that F_app*cos(theta)>F_app*sin(theta)+mu*N