King E Posted June 11, 2020 Posted June 11, 2020 When a ball is thrown upward it becomes at rest at maximum height, at this it is not in equilibrium although it is at rest. It is not at equilibrium because force of gravity is acting on it? Still I cannot find good explanation from exam point of view.I also cannot find the figure/diagram.
MigL Posted June 11, 2020 Posted June 11, 2020 (edited) It goes through zero velocity as it changes direction, but does it really come to rest ? I was thinking more along the lines of placing a ball on top of a steep hill. It is at rest, but unstable. IOW, not in equilibrium, as defined by a lowest energy state. edit: is this homework ? Edited June 11, 2020 by MigL
swansont Posted June 11, 2020 Posted June 11, 2020 If you define “at rest” to mean v=0. So why not just say v=0? It avoids the ambiguity of language. What kind of diagram do you need? It’s a ball that is momentarily not moving. There’s a downward force on it.
joigus Posted June 11, 2020 Posted June 11, 2020 At maximum height \[\boldsymbol{v}=0\] but \[\boldsymbol{a}\neq0\] Is that what's confusing you? 2
studiot Posted June 11, 2020 Posted June 11, 2020 1 hour ago, joigus said: At maximum height v=0 but a≠0 Is that what's confusing you? Isn't that just so clear ? +1
King E Posted June 11, 2020 Author Posted June 11, 2020 3 hours ago, joigus said: At maximum height v=0 but a≠0 Is that what's confusing you? Yes. How is it at rest if a force is acting on it?
joigus Posted June 11, 2020 Posted June 11, 2020 (edited) 15 minutes ago, King E said: Yes. How is it at rest if a force is acting on it? Ok. I think it's what MigL was trying to tell you. It doesn't come to rest. It is not at rest. It goes through rest for an instant, so to speak. I'm still looking for a graph. Maybe Studiot can help with the graph, which is after all something you were asking for. Here: 3 hours ago, MigL said: It goes through zero velocity as it changes direction, but does it really come to rest ? Edited June 11, 2020 by joigus added emphasis 1
King E Posted June 11, 2020 Author Posted June 11, 2020 1 minute ago, joigus said: Ok. I think it's what MigL was trying to tell you. It doesn't come to rest. It is not at rest. It goes through rest for an instant, so to speak. I'm still looking for a graph. Maybe Studiot can help with the graph, which is after all something you were asking for. Here: Yeah a graph could help! 3 hours ago, swansont said: If you define “at rest” to mean v=0. So why not just say v=0? It avoids the ambiguity of language. What kind of diagram do you need? It’s a ball that is momentarily not moving. There’s a downward force on it. by figure, I mean graph
joigus Posted June 11, 2020 Posted June 11, 2020 (edited) 32 minutes ago, King E said: Yeah a graph could help! Does that help? v = 0 is when the velocity graph crosses the axis. Just "one instant." Let's say the particle is at rest but doesn't have time to rest. Words are tricky in physics. They sometimes project "mirages" in our minds... Edited June 11, 2020 by joigus mistyped
King E Posted June 11, 2020 Author Posted June 11, 2020 8 minutes ago, joigus said: Does that help? v = 0 is when the velocity graph crosses the axis. Just "one instant." Let's say the particle is at rest but doesn't have time to rest. Word are tricky in physics. They sometimes project "mirages" in our minds... So the object passes through the rest but isn’t continuously at rest because of the force of gravity. This obeys Newton’s law doesn’t it?
joigus Posted June 11, 2020 Posted June 11, 2020 Just now, King E said: So the object passes through the rest but isn’t continuously at rest because of the force of gravity. This obeys Newton’s law doesn’t it? Bingo! 1
J.C.MacSwell Posted June 15, 2020 Posted June 15, 2020 (edited) If a body is in equilibrium, it's in equilibrium in every (inertial) reference frame. It can't be at rest in every frame. But it's always at rest in some frame. So to answer think of a body at rest but not in equilibrium...every body not in equilibrium is an example... Edited June 15, 2020 by J.C.MacSwell
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