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.

# Think Of A Body Which Is At Rest But Not In Equilibrium. Give Explanation As Well As Figure/diagram.

### #1

Posted 11 June 2020 - 12:22 PM

### #2

Posted 11 June 2020 - 02:43 PM

can you provide more context?

### #3

Posted 12 June 2020 - 01:22 AM

can you provide more context?

I have provided all that I can.

### #4

Posted 12 June 2020 - 02:18 AM

can you provide more context?

I have provided all.

### #5

Posted 12 June 2020 - 10:31 AM

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 be

H = 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 likes this