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Deriving the expression for average vertical force

  • Thread starter jisbon
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  • #1
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Homework Statement:

Ball (mass m) launched horizontally at height H from ground.
Horizontal displacement at first bounce is 4/3 H and maximum height after the first bounce is 9/16 H.
Duration of contact between ball and ground is t

Homework Equations:

-
Here's what I have done:

Let t1 be the time taken for the ball to reach the ground after the first bounce, and ux be the initial speed in the horizontal direction of ball.
Also, let v be the speed after the bounce.
For horizontal,
$$\dfrac {4}{3}H=U_{x}t_{1} $$
For vertical,
$$H=\dfrac {1}{2}g\left( t_{1}\right) ^{2} $$
Solving these 2, I find:
$$U_{x}=\dfrac {2}{3}gt_{1} $$
Im supposed to find the vertical average force. In this case will it be:
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t} $$
?
Not sure what the question is requiring me to do
 

Answers and Replies

  • #2
PeroK
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What vertical force are you trying to find? gravity, air resistance, contact force with the ground?
 
  • #3
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What vertical force are you trying to find? gravity, air resistance, contact force with the ground?
I will assume contact force. The question simply stated to derive the expression for vertical average force acting on ball
 
  • #4
PeroK
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I will assume contact force. The question simply stated to derive the expression for vertical average force acting on ball
If you don't know what force you are supposed to be calculating then I'd move on.

I'd assume it's the contact force.

Force is rate of change of momentum, I believe.
 
  • #5
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If you don't know what force you are supposed to be calculating then I'd move on.

I'd assume it's the contact force.

Force is rate of change of momentum, I believe.
So this is correct I assume?
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$
 
  • #6
PeroK
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So this is correct I assume?
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$
Looks good. Although, you might need to figure out what they want you to do about gravity.

Is this the contact force or the net force?
 
  • #7
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Looks good. Although, you might need to figure out what they want you to do about gravity.

Is this the contact force or the net force?
The answer provided to me was :
$$\dfrac {7m}{4t}\sqrt {2gH} $$
What I have here is:
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$
Where uy is 0
So I found:
$$\begin{aligned}v_{y}=gt_{1}=g\sqrt {2H}\\ F=\dfrac {m\left( v_{y}-0\right) }{t}=\dfrac {mg\sqrt {2H}}{t}\end{aligned} $$
What's the error here? Thanks
 
  • #8
PeroK
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Why
The answer provided to me was :
$$\dfrac {7m}{4t}\sqrt {2gH} $$
What I have here is:
$$F_{avg}=\dfrac {m\left( v_{y}-u_{y}\right) }{t}$$
Where uy is 0
So I found:
$$\begin{aligned}v_{y}=gt_{1}=g\sqrt {2H}\\ F=\dfrac {m\left( v_{y}-0\right) }{t}=\dfrac {mg\sqrt {2H}}{t}\end{aligned} $$
What's the error here? Thanks
Why do you have ##u_y = 0##?
 
  • #9
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Why

Why do you have ##u_y = 0##?
Because the ball is launched horizontally, so
$$u_y = 0$$
 
  • #10
PeroK
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Because the ball is launched horizontally, so
$$u_y = 0$$
I was hoping you wouldn't say that!

So why not take ##v_y =0##, as the ball reaches the top of the first bounce? That's where the problem statement ends.

Or, should these velocities relate to the phase of the motion you are interested in?
 
  • #11
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I was hoping you wouldn't say that!

So why not take ##v_y =0##, as the ball reaches the top of the first bounce?

Or, should these velocities relate to the phase of the motion you are interested in?
Not sure what you meant.. But wasn't it constrained to the first phase when I initially subbed $$\dfrac {4}{3}H=U_{x}t_{1}$$?
 
  • #12
PeroK
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Not sure what you meant.. But wasn't it constrained to the first phase when I initially subbed $$\dfrac {4}{3}H=U_{x}t_{1}$$?
I'm not sure what you mean by that. If you want the force over a time interval ##t##, then you need the velocities at the start and end of that interval.

If you include an earlier phase of motion, then you need to include the time for that phase too. In this case ##t_1## would be relevant.
 
  • #13
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I'm not sure what you mean by that. If you want the force over a time interval ##t##, then you need the velocities at the start and end of that interval.

If you include an earlier phase of motion, then you need to include the time for that phase too. In this case ##t_1## would be relevant.
So from what I understand, I need to find the velocity JUST before the bounce and JUST after the bounce?
So JUST before the bounce, velocity of y = $$ g\sqrt {2H}$$
How do I find the velocity JUST after the bounce in this case?
 
  • #14
PeroK
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So from what I understand, I need to find the velocity JUST before the bounce and JUST after the bounce?
So JUST before the bounce, velocity of y = $$ g\sqrt {2H}$$
How do I find the velocity JUST after the bounce in this case?
Re your first question: why isn't that obvious?

Can't you deduce ##v_y## from high it bounces?
 
  • #15
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Re your first question: why isn't that obvious?

Can't you deduce ##v_y## from high it bounces?
Sorry, wasn't thinking clearly.
So can I say that $$\begin{aligned}0=v^{2}_{y}+2\left( -g\right) \left( \dfrac {9}{16}H\right) \\ v^{2}_{y}=\dfrac {18}{16}gH\end{aligned} $$
 
  • #16
PeroK
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Yes. Be careful with signs now.
 
  • #17
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Ok.. So force is derived to be:

$$\dfrac {m\left( \sqrt {\dfrac {18}{16}gH}-g\sqrt {2+t}\right) }{t} $$

How does one simplify to the suggested answer? Can't get my head around it :H
 
  • #18
PeroK
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Ok.. So force is derived to be:

$$\dfrac {m\left( \sqrt {\dfrac {18}{16}gH}-g\sqrt {2+t}\right) }{t} $$

How does one simplify to the suggested answer? Can't get my head around it :H
I can't see where you got that second term from. The speed as it leaves the surface can be determined by the same method you got the incident velocity.

In general, it is always ##v^2 = 2gh##. It's only ##h## that changes each time.
 
  • #19
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I can't see where you got that second term from. The speed as it leaves the surface can be determined by the same method you got the incident velocity.

In general, it is always ##v^2 = 2gh##. It's only ##h## that changes each time.
Made a mistake in Latex from my post, I meant 2gH.
Also, I managed to get the solution after realising that it should be negative in the second part since it is bouncing in a different direction.
Thanks for the help.
 

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