# Easy Physics Question

1. Aug 6, 2010

### ThomasMagnus

I have a very simple question.

Say If I have a mass of 90kg, and I punch a wall. While punching the wall I accelerate at a rate of 10m/s squared (hypothetical). If I wanted to find out the force I applied to the wall, could I use F=MA. So (90)(10)=900N of force? To keep this question very simple, let's ignore air resistance and the like.

Thanks for the help!! :)

2. Aug 7, 2010

### Dr Lots-o'watts

900N is the force that you apply on the floor without thinking about it. And the force the floor has to apply on your feet to prevent you from going it.

The force that seems to interest you would be F=ma, where m is the effective mass of your fist, and a is the "deceleration" of your fist while it interacts with the wall. I say effective mass because it depends on how well you punch, or how much of your body you push it with.

3. Aug 7, 2010

### zeromodz

If you punch the wall with a force of (m)(a)= (90)(10) = 900 N, then the wall will push back with the same force according to newtons third law. So you will accelerate 10m/s^2 backwards if there is no friction. Also, your center of mass needs to hit the wall for this to happen, not your hand.

4. Aug 7, 2010

### Riyuki

Well, it seems that it depends on what you assume is accelerating at 10m/(s^2).

It could be the average acceleration of the person’s center of mass. In which case the average force exerted on the wall would indeed be 900N. And the initial acceleration of their fist would be higher than 10m/(s^2), while that of more distant parts of the body would be zero, and the acceleration of the fist would drop off and that of the rest of the body would increase until they were moving at the same speed.
Of course it’s not like there are only two masses; you can decompose the body into as many smaller masses and corresponding average accelerations as you like (well, up to a quantum mechanical limit anyway, I guess) to arrive at estimates that yields better and better resolution of distance traveled per unit time for various parts of the body.

Or the 10m/(s^2) could be the average acceleration of the fist over a subset of the time interval of the whole body’s acceleration. In that case, depending on how large the time interval in question is, the body’s average acceleration could be more or less or equal to 10m/(s^2).

As a practical matter, it will be easier to measure the average acceleration of your center of mass / your whole body than to measure a shorter-time-interval subset of that acceleration. In the latter case you would need, I suppose, a highly sensitive elastic diaphragm or really high-speed photography or laser or ultrasonic rangefinding to get useful measurements given the especially small time intervals involved.

5. Aug 10, 2010

### GRDixon

I'm not sure, but wouldn't the center of mass need to hit the wall in order for there not to be any rotation? No matter which part of the body hits the wall, the acceleration (of the center of mass) should still be 10m/sec^2.

6. Aug 10, 2010

### Riyuki

You're right that a one-handed push would probably (perhaps necessarily, regardless of how you hold your arm, given the shape of the human body?) result in some net torque and consequent angular acceleration in the process of pushing off the wall.
Which (if I understand things correctly) would result in the linear acceleration of your center of mass being less than 10m/(s^2) and the CoM’s final linear velocity being somewhat less than [ 10m/(s^2) * t seconds], because some of the acceleration contributes instead to providing you with angular momentum.

Of course that wouldn't necessarily preclude the force exerted on the wall from being 900N.

But it's not really the case that “an object's center of mass needs to hit” another object in order for a collision to result in no net torque. Usually a (relatively compact) body’s center of mass won’t be located at the point of the collision because it’ll be inside the object.
I think what you mean is that “the net acceleration needs to be directed radially inward from the point of impact / causative acceleration toward the object’s center of mass, with no tangential component, in order to avoid the exertion of a net torque on the object”.

With respect to the foregoing sentence, I’m imagining the center of mass as being at the center of arbitrarily many concentric spheres (or circles, in the 2-d case) which extend as far out as the outermost member of the many-particle system (in this case, a human body).
Depending on which part(s) of the system directly receive acceleration from the other object, and how much they receive, there will be a torque on the center of mass equivalent to

τ = Σ [ (radius from center of mass to point of impact)_n * sin(θ)_n * F_n ]

, where θ is the angle between the line from the center of mass to the point of impact, and the line of the direction of the force. The “n” subscripts just represent that I mean the sum of (n) forces.

There are various ways in which there could be no net torque. For example if you push equally with both hands, the torques will cancel out (at least in one plane).

My apologies if I've said anything wrong in the above; if so someone should correct me.

Last edited: Aug 10, 2010
7. Aug 13, 2010

### cragar

what about using the change in momentum divided by the impact time ?
dp/dt

8. Aug 14, 2010

### Acut

I think cragar's idea is the best one.