I can't understand these kind of problems, when we use arrows etc. and find the overall force..
I pictured the car, with two forces vertically downwards: mg and cf force (mv^2/r)
and one force upwards, which is the reaction force, however I am kind of confused about the reaction force right now, I thought it was the force of same magnitude but the opposite direction, so I pictured another mg force coming vertically upwards.
The answer to this is D and I don't have a good idea why.
I can't understand these kind of problems, when we use arrows etc. and find the overall force..
I pictured the car, with two forces vertically downwards: mg and cf force (mv^2/r)
and one force upwards, which is the reaction force, however I am kind of confused about the reaction force right now, I thought it was the force of same magnitude but the opposite direction, so I pictured another mg force coming vertically upwards.
The answer to this is D and I don't have a good idea why.
There are two forces on the car * its weight, mg (downwards) * reaction from the road, R (upwards)
Together, the resultant of these two must equal the centripetal force on the car (downwards) mv2/r
Just to clarify- for circular motion the resultant force equals mv^2/r. So you know the weight is mg, and you know the weight plus the reaction force=mv^2/r. Therefore the reaction force is the difference between them, or mg-mv^2/r.
Just to clarify as above, the force given by rmv2 is the centripetal force, which is directed towards the center of the circle. You mentioned 'cf', which makes me think you are referring to centrifugal force, which is wrong, it is a common misconception. Check these out:
Just to clarify even further. The OP stated that the force (which he referred to as "cf") was "vertically downwards" and equal to rmv2 The infamous and non existent centrifugal force would have been directed upwards.
lerjj You haven't clarified this for the OP at all You stated that ".. the weight plus the reaction force=mv^2/r"
This gives mg + R = mv2/r
and does not produce answer D It actually gives answer C.
Just to clarify even further. The OP stated that the force (which he referred to as "cf") was "vertically downwards" and equal to rmv2 The infamous and non existent centrifugal force would have been directed upwards.
lerjj You haven't clarified this for the OP at all You stated that ".. the weight plus the reaction force=mv^2/r"
This gives mg + R = mv2/r
and does not produce answer D It actually gives answer C.
Sorry, can't do vector arrows... I did intend that statement to take signs into account. So (-mg)+(+N)=(-mv^2/r), taking +ve to mean up. Adding +mg to both sides gives D, which was my point when I said resultant force. I shouldn't have been lazy with the signs though, that probably wasn't helpful