DRACO 1 Posted June 11, 2020 Report Share 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. Quote Link to post Share on other sites

balagna 5 Posted June 11, 2020 Report Share Posted June 11, 2020 can you provide more context? Quote Link to post Share on other sites

DRACO 1 Posted June 12, 2020 Author Report Share Posted June 12, 2020 can you provide more context?I have provided all that I can. Quote Link to post Share on other sites

DRACO 1 Posted June 12, 2020 Author Report Share Posted June 12, 2020 can you provide more context?I have provided all. Quote Link to post Share on other sites

OceanBreeze 410 Posted June 12, 2020 Report Share Posted June 12, 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. This is a question in basic mechanics, but it still is interesting to think through and clear the rust off a few synapses. What is carrying the ball upwards is its momentum versus the force of gravity pushing it downwards. The interesting question is, since momentum is not a force, how can it oppose a force? To answer that we need to consider the dimensional factors that comprise both momentum and force: Momentum is just mass x velocity Force has factors of mass x acceleration and since acceleration is just how velocity changes over time, we can write this as: force = mass x (velocity/time) or force = (mass x velocity)/time We see that if the both sides of that last expression for force are multiplied by time we have: Force x time = mass x velocity In other words, a force acting over a period of time has the same dimensions as momentum and that is how the downward force of gravity can oppose the momentum that is carrying a ball upward. Maybe a simple numerical example will make this clearer? Suppose the ball is thrown straight upward at an initial velocity of 10 m/s At what time t, will it reach its maximum height, neglecting air resistance and only opposed by the force of gravity? We know it will reach maximum height when force x time = mass x velocity Since force = ma, we can write this as ma x time = mass x velocity Cancelling out mass on both sides: a x t = v Gravitational acceleration a = 9.8 m/s^2 and we know v = 10 m/s So: 9.8 m/s^2 x ts = 10 m/s We can cancel the unit factors out leaving 9.8 x t = 10 and the time t is easily found to be 1.02 sec So, at 1.02 seconds the upward momentum is completely countered by the downward force of gravity The velocity is momentarily 0 at this time but the acceleration due to gravity remains constant at 9.8 m/s^2 meaning this is not a point of equilibrium as the downward force is not balanced by any upward force. I will leave it as an exercise to find what the maximum height H reached will beH = vt - ½ at^2 For the diagram, you are on your own; just Google one up or preferably learn how to draw a free body diagram. DRACO 1 Quote Link to post Share on other sites

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