Original post by Darcie_J06
how do you show this lol 😭
I can only see part a here, is that the only part you're stuck on?
For these types of questions I'd always start by sketching a free body diagram of the particle, labelling each force clearly. Here you have a reaction force acting perpendicular to the wall, the weight of the particle and the centripetal force.
Try resolving the forces to create two equations and see what you get.
Original post by Darcie_J06
how do you show this lol 😭
Similar to Skiwi, but rather than resolving its worth noting that when you add the two forces (weight and centripetal), the resultant must be normal to the wall. It would be similar to resolving vertically.
(edited 7 months ago)
Original post by mqb2766
Similar to Skiwi, but rather than resolving its worth noting that when you add the two forces (weight and centripetal), the resultant must be normal to the wall. It would be similar to resolving vertically.
Unless I'm missing something, you don't know the centripetal force as you don't know the radius of the circle.
[I'm guessing the later parts actually ask you to find the radius/height by equating the centripetal force to the horizontal component of the normal reaction].
Original post by DFranklin
Unless I'm missing something, you don't know the centripetal force as you don't know the radius of the circle.
[I'm guessing the later parts actually ask you to find the radius/height by equating the centripetal force to the horizontal component of the normal reaction].
[I'm guessing the later parts actually ask you to find the radius/height by equating the centripetal force to the horizontal component of the normal reaction].
You dont need to know it, so the normal must be weight+centripetal so it is a 5-12-13 right triangle so ...
Or resolve vertically which ignores the centripetal and note the parallel resultant force is zero so ...
(edited 7 months ago)
Original post by mqb2766
You dont need to know it, so the normal must be centripetal+weight so it is a 5-12-13 right triangle so ...
Given that as a hint and not resolving, I think I'd be very likely to swap the 5 and the 12. (Basically I don't think I'd get them the right way round without doing something that's essentially resolving the normal into horizontal/vertical components, even if it wasn't a formal "resolving forces").
[Edit: basically the reason I thought you'd misread and thought we knew the centripetal acceleration was that without knowing the horizontal component I couldn't see how your suggestion was any easier than just resolving vertically (whereas if we did know it your method would be clearly better)]
Or resolve vertically which ignores the centripetal and note the parallel resultant force is zero so ...
Agreed - to me this is the obvious (and I would think intended) approach.
(edited 7 months ago)
Original post by DFranklin
Given that as a hint and not resolving, I think I'd be very likely to swap the 5 and the 12. (Basically I don't think I'd get them the right way round without doing something that's essentially resolving the normal into horizontal/vertical components, even if it wasn't a formal "resolving forces").
[Edit: basically the reason I thought you'd misread and thought we knew the centripetal acceleration was that without knowing the horizontal component I couldn't see how your suggestion was any easier than just resolving vertically (whereas if we did know it your method would be clearly better)]
Agreed - to me this is the obvious (and I would think intended) approach.
[Edit: basically the reason I thought you'd misread and thought we knew the centripetal acceleration was that without knowing the horizontal component I couldn't see how your suggestion was any easier than just resolving vertically (whereas if we did know it your method would be clearly better)]
Agreed - to me this is the obvious (and I would think intended) approach.
I was just replying, but Ive swapped the forces round. The 5-12-13 is just the usual way to say it.
Theres not much in it whether resolving vertically or noting the weight-centripetal-normal must be 5-12-13. To me, its easier to say the parallel resultant must be zero so the weight+centripetal must equal the normal so they must be 5-12-13, but its not the usual way to do these questions (which itself was one reasons for the post). Naively doing the horizontal/vertical might be confusing if you dont spot the parallel resultant is zero and you may be thrown by not knowing the centripetal. Hence it was an addition/alternative to Skiwi's comment.
(edited 7 months ago)
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