Question

A massless rope is tossed over a wooden dowel of radius r in order to

lift a heavy object of weight W off of the floor, as shown in the figure. The coefficient of

sliding friction between the rope and the dowel is µ. Show that the minimum downward

pull on the rope necessary to lift the object is

Fdown = We^πµ

Answer 1

Note that keeping in mind the end goal to lift the question, the greatness of the strain T in the rope must be more than or equivalent to the heaviness of the protest. (That is, T ≥ W T≥W.) So the rope is pulled toward one side by a power of greatness T and on the other by a power of extent F down Fdown. The dowel applies an ordinary power ⃗ N N→ on the rope, and the extent of the grating between the rope and the dowel is given by f = μ N f=μN. The issue is that I have no clue how to manage this typical power. On the off chance that I draw a graph where ⃗ f f→ contradicts the movement of the rope, I wind up with ⃗ f f→ and ⃗ T T→ pointing the other way as ⃗ F down F→down, yet ⃗ N N→ is opposite to those powers. The rope is clearly not moving toward ⃗ N N→, so it appears that some obscure power is adjusting the ordinary power out.