Reaction the rod makes with the ground

In part C of this question , I don't get why the "reaction" force ; something I have always learnt as being perpendicular to the ground or surface in contact, is inclusive of friction..

Am I misinterpreting the meaning of reaction force ? To my understanding, it should not include friction as friction only arises as a result of motion , as opposed to arising because of a force pair of the object in contact with the ground.

Thanks.

Well friction doesn't always arise as a result of motion; indeed, there is no motion here, but we have a frictional force.

The reaction is just that - "A reaction". The weight acts down the ground "reacts" with an upward force. The string is trying to pull the rod to the left, the ground "reacts" with a frictional force to the right.

We often work out reaction by considering the component perpendicular to the ground, and friction separately, but they are both reactions and combined form the reactive force.

Great answer. Thank you.

Does this mean that if I have a ball on a rough slope, then the reaction force has 2 components ; the normal force and the friction ?

Yes, the reactive force can be resolved into two components; the normal reaction and the frictional force. Together they combine to give the reactive force.

A block (rather than a ball) resting on a rough inclined plane would be an example you're probably already familiar with.

In the equation F = μ*R , is R the normal component of the reaction force ? Because in all of the examples I have seen it has been, and the ground itself must be rough to have friction in the first place.

Yes, R is the normal component of the reaction force in that equation.

Ahh ok. thank you. You are a great explainer btw.