The Student Room Group

(Should be) straightforward mechanics question, but...

for whatever reason (I'm rusty/an idiot..) I can't seem to do it.

Anyway, here we go...taken from a uni 'synoptic' paper but most people who've done some sort of mechanics at A-level should be able to have a crack at it I'd have thought...

Problem: A skier of mass 75 kg skis over a hemispherical mound of snow of radius 15m. At the top of the mound, the skiers velocity vector is horizontal with a magnitude of 40 km/h. Assuming the snow to be frictionless, calculate the magnitude and direction of the force exerted by the skier on the snow at the top of the mound.

General thoughts
: So we've got the skiers weight of (75*g) N vertically downwards and he's on a hemispherical mound, so I was thinking circular motion with the velocity tangential to the top of the mound of 40 km/h. Unless there's a force or something that I'm missing, the only force worth worrying about is the skier's weight, which would then be providing the centripetal force mv^2/r...plugging the numbers in gives me 617.25 N

The answer is, apparently, 118 N. (This is the problem with getting bottom line answers as opposed to solutions when I'm fairly sure I've found a wrong answer elsewhere, it makes me much less trusting of the other answers...)

So anyway, I'd be very grateful if someone can spot where I've gone wrong and will probably reward with some rep or something.

Cheers! :smile:
Reply 1
Hi

you need to know what part of the gravitational force covers the centripetal force and what part is exerted on the snow

F=mg-mv^2/r=m(g-v^2/r)

plug the numbers in :

F=75(9.81-(40/3.6)^2/15)=118.47 rounds to 118

don't forget to change the speed from km/h to m/s (divide by 3.6)

The force is going down btw (I suppose you know that but the original question asks for the direction)
Reply 2
mf2004
Hi

you need to know what part of the gravitational force covers the centripetal force and what part is exerted on the snow

F=mg-mv^2/r=m(g-v^2/r)

plug the numbers in :

F=75(9.81-(40/3.6)^2/15)=118.47 rounds to 118

don't forget to change the speed from km/h to m/s (divide by 3.6)

The force is going down btw (I suppose you know that but the original question asks for the direction)


Thanks :smile: I'd missed out the reaction force on my diagram, always a bad idea. I'm ridiculously rusty at mechanics, lol.