Force of tipping block

1. Nov 10, 2013

johnnnnyyy

If it takes 30N to tip over a block based on the equation F=mgL/2r but you strike the stationary block with a force of 50N what is the equation for the force of the block as it is tipping?
I know that there will be an Fnet force of 20N but wouldn't a force of gravity or something give the block more of a force as it tips? Does anybody have any equations that will help me understand this?
Thanks

2. Nov 11, 2013

Simon Bridge

Just striking a block is a collision - resolve it by conservation of momentum.

Note: no such thing as a fixed net force in a collision.
The weight of the block helps or hinders the toppling depending on it's line of action.
When the block tips, the weight acts to provide a torque about the pivot ... sketch it out: how far does the block have to tip for the weight to make it tip more?

3. Nov 11, 2013

johnnnnyyy

Couldn't you measure it as a force though? For example if you strike the block near the bottom so that it will slide and you measure the distance the block slides and you know the coefficients of friction couldn't you measure the force applied to the block from the strike?

4. Nov 11, 2013

Simon Bridge

If you strike a block below the center of mass, isn't there is a chance it will tip over backwards?
Ignoring that.

Lets say the entire blow produces horizontal motion.
A mass $m$ block is so struck that it slides a distance $d$ over a surface with coefficient $\mu$
With what force was it struck?

5. Nov 11, 2013

johnnnnyyy

That's what I cant figure out, I found the force of static friction, the force of kinetic friction, the deceleration of the block, the initial velocity but I cant figure out the force of the strike. How would you go about solving this?

6. Nov 11, 2013

Simon Bridge

From the definition of force: $$F=\frac{\Delta p}{\Delta t}$$ The initial velocity gives you the momentum of the block $p=mv$ right after the collision. Since the block is stationary before the collision, then $\Delta p = mv$ ... what is $\Delta t$?

You are probably thinking that the collision happens in zero time right?
What does that do to the "force" in the equation?

I'll let you off the hook.
There is no such thing as the "force of a collision".
Please allow a while for this to sink in - the idea that a collision has a force is hard to shake.

Collisions take a finite amount of time.
If you put a force-transducer at the impact point, you can measure the force as a function of time.
In simple collisions, the force vs time graph looks like an inverted parabola.
The area under the force-time graph is called "specific impulse". You should look it up.

So... back to the start: you have to handle your problem using conservation of momentum and energy. I know that you are used to thinking in terms of force but that has only limited use. Arguments based on force, in this case, will be misleading.

Last edited: Nov 11, 2013
7. Nov 11, 2013

Simon Bridge

Consider - you have a uniform rectangular block mass m, height h, and a square base length b.
The volume is $V=b^2h$ and the density is $\rho = m/V$. Easy right?
The center of mass is the geometric center of the block.
This means that the line of action of the weight $w=mg$ is down the axis, and right through the center of the base.

To tip it over, you have to do some work.
The minimum amount of work is that needed to shift the center of mass so the line of action of the weight shifts onto the pivot point.

If you tip the block about one edge - you'll see that involves lifting the center of mass (as well as shifting it horizontally) so, the minimum work you need to do is the amount of gravitational PE gained at the start of the process. Once PE is a maximum, the block will be perfectly balanced on one edge: the slightest nudge will topple it.

You should be able to work those equations.