# Does force of falling object include its own weight?

Tags:
1. Aug 9, 2015

### Yong Chun Hon

1. The problem statement, all variables and given/known data
From what I have seen on internet sources, the average force exerted by an falling object on the ground is equal to the work done by the ground to stop the object. Assume collusion is completely inelastic and object sticks to ground after impact.

F: Net force on ground
m: mass of object
g: acceleration due to gravity
h: height of object
d: distance object travels during deceleration

F = Energy of object/distance which the force is applied = mgh/d

Another perspective would be to calculate by the change in momentum. Assume collusion is completely inelastic and object sticks to ground after impact.

F: Net force on ground
m: mass of object
v1: velocity of object the moment it touches the ground
v2: velocity of object at end (0 in this case)
t: time object takes to come to stop

F = =Change in momentum/Time = (mv1-mv2)/t

My question is should we include the weight of the object for the impact?

2. Relevant equations
F = mgh/d
F = (mv1-mv2)/t

3. The attempt at a solution
Should the solution be be F = Change in momentum/time + weight of object = (mv1-mv2)/t + mg

2. Aug 9, 2015

### Andrew Mason

The force on the ground is the body's weight + its rate of change of momentum. If dp/dt = 0 (when it is stopped) the force is still mg.

AM

Last edited: Aug 9, 2015
3. Aug 9, 2015

### Yong Chun Hon

Ah I see thanks. I was wondering what were the variables for a force of a punch assuming it is constantly accelerating. This indirectly helped answered it.

4. Aug 11, 2015

### Andrew Mason

Welcome to PF, by the way!

To analyse these kinds of thing, I suggest that you start with a free-body diagram. In the case a falling body, the forces are the downward force of gravity and the upward normal force. These forces must sum (as vectors) to "ma" where m is the mass of the falling body and a is its acceleration. Using the "up" direction as positive, this means: N - mg = ma or N = ma + mg.

In the case of a forceful horizontal punch on a body that is fixed to the earth, you have two forces being applied to the fist: the force being applied to the fist by the arm and the force being applied by the body that the fist is striking (ignoring gravity, assuming it is a perfectly horizontal punch). These two forces must add up to "ma" where m is the mass of the fist.

AM

5. Aug 15, 2015

### Yong Chun Hon

Thanks Andrew!

I got roughly how the free body looks like. I posted this question because my A'level homework included the weight of the object for its impact when falling, but answers on the internet did not. So I thought I was missing something.

6. Aug 15, 2015

### haruspex

Please post the homework question exactly as given to you. Otherwise there is a risk we will mislead you.
(You seem to be confused between force, impulse and energy.)

7. Aug 16, 2015

### Yong Chun Hon

This was a question I had in my mind. But I didn't know where to post this. My reference for the force of a falling object was from my a'level tutorial. But comparing against Internet answers, they did not include the weight, hence the question.

8. Aug 16, 2015

### haruspex

OK.
It might depend on what is considered to be the height through which the object descended.
If h is the height it started above the ground, and d is the depth it sank into the ground, then the total height of fall is h+d. This gives you F=mg(h+d)/d = (mgh/d)+mg.
If you only count the energy it had when it hit the ground, mgh, then you should add on the weight, because the force from the ground has to support the weight as well as absorbing the KE it already has: F = (mgh/d)+mg.
Same result.

9. Aug 17, 2015

### Andrew Mason

Just to be clear, the force from a falling body on another body (such as the earth) is equal and opposite to the force that the other body (the earth) exerts on the falling body. The latter force, called the normal force, $\vec{N}$, plus the force of gravity: $m\vec{g}$ results in $m\vec{a}$. So regardless of how you want to analyse the situation $\vec{N} + m\vec{g} = m\vec{a}$ which can be rewritten: $\vec{N} = m\vec{a} - m\vec{g}$. Since the "force from the falling body" on the earth (let's call it $\vec{F}$) is equal and opposite to the normal force, $\vec{N}$: $\vec{F} = m\vec{g} - m\vec{a}$.

AM

Last edited: Aug 17, 2015