# Can you ever exert more downwards force than your weight?

So, because I'm a hardcore person, I risked all this afternoon by going out in the wind, the rain and the cold to construct a willow den.

Yes, it seems a menial task, but it was actually quite thought-provoking: some pieces of willow were harder to push into the ground than others, and to try and force them in I'd sometimes end up hanging off of them, but other times I'd just push with all my strength, placing my body above the hole and pushing down, my feet still firmly planted on the ground.

Now, I was wondering, which of these two methods is better for sticking the stick into the ground? Ignoring the fact that by hanging from it, it is likely to topple, which method help me be more efficient in my future den-building exploits?

In other words, can you ever exert more downwards force than your weight? If so, then I'm guessing pushing down would be the better method, and if not, then hanging from the stick, putting all your weight into it, would exert the most possible downwards force.

Here's what a willow den looks like:

(source: raisingsparks.com)

• Jump. Or, can a hammer deliver more downwards force than it's weight? – Nick T Apr 25 '15 at 20:43
• – Hot Licks Apr 25 '15 at 22:25
• What one might potentially do, to make practical use of Newton's legislation, is construct a sort of clamp, say out of two short pieces of 2x4 bolted together with long bolts and thumb nuts. Start the willow shoot into the ground, then clamp this gizmo around it and jump up and down on the gizmo. – Hot Licks Apr 26 '15 at 1:01
• Run an experiment. Stand on the scales, then jump. Scales measure the force with which you push down. When you jump, the arrow will shoot up and exceed your weight, proving that it's possible. Disclaimer: use cheap scales in case you damage them. – Roman Starkov Apr 26 '15 at 19:44
• I created an account just to ask what a "willow den" is. Pieces of willow are things you "push into the ground" but you somehow hang off of them... Please explain. – Adam Jensen Apr 27 '15 at 0:12

The force you can exert is your mass times your acceleration. By the equivalence principle, just standing still is equivalent to accelerating at 9.8 m/s2, which is where the force of your weight comes from when you just stand still. But it is easy to accelerate more - like when you jump.

The force is only limited by your ability to push yourself off (transfer force to) the willow shoot. Imagine that you lie down next to the shoot, holding it in both hands. If you now pulled yourself up rapidly (the way some circus acrobats can pull themselves up a rope while appearing to hang horizontally) then you apply all your weight to the willow - and if you are strong enough to accelerate yourself while doing this, you could apply a force greater than your weight.

However, as you probably realize, there are other far more effective means to drive a stick into the ground. The key is to convert momentum into force - the equation is

$$m\Delta v = F \Delta t$$

This equation tells us that the change in momentum ($\Delta(mv) = m\Delta v$) is determined by the integral of force and time ($\int F\cdot dt=F\Delta t$ if F is constant). This is a direct consequence of the equation $F=ma$, which you can integrate with respect to time to get $\int F\cdot dt = \int m\ a\ dt = m \Delta v$.

When you use a hammer etc, you give it momentum during a long swing (small F, large t); but it slows down and comes to a stop in a very short interval, meaning that for that short time the force is much greater. A post driver is the tool people use to try to replicate this on the scale of large sticks being driven into the ground (hard to hammer the top of a tall thin stick). It may not be possible to use in your particular situation - but in general, it will allow you to apply a force much greater than your weight (bot for a shorter time). This is also the principle behind pile drivers etc . All these methods require the object to be driven to be strong enough to support the force you use to drive them into the ground...

• Good answer! Mind explaining the equation, or providing a link? – theonlygusti Apr 25 '15 at 18:44
• @theonlygusti, $F=ma$ is newton's second law, where we have $a=\frac{dv}{dt}$. Thus $m\Delta v=\int_{t_0}^{t_0+\Delta t} F dt$, assuming that $F$ is constant during the hit, we have $m\Delta v=F\Delta t$. – Shay Ben Moshe Apr 25 '15 at 18:57
• @ShayBenMoshe - I was writing an update to the answer just as you wrote your comment... thanks! – Floris Apr 25 '15 at 18:58
• See here for a nice graphical demonstration of this: the peak forces when jumping and landing are significantly higher than when standing. – 2012rcampion Apr 25 '15 at 20:40
• The force you can exert is your mass times your acceleration plus the force of gravity – Brionius Apr 26 '15 at 1:26

Use a lever. For an application like driving branches in the great outdoors, you would need to come prepared, or locate the site next to something heavy. Anchor one end of the lever under something heavy. Attach something to the branch to press against, and put the branch between you, at one end, and the fulcrum at the other. The lever acts as a force multiplier, so you can exert more downward force than your own weight.

• … or any of the other simple machines that trade off force for distance. – 200_success Apr 27 '15 at 8:24

If you think about Newton's third law and standing still vs jumping. When you stand still the ground exerts a reaction force on you which is equal and opposite to your weight by Newton's third law.

If you jump upwards at the point where you begin to drive upwards you are applying a greater force on the ground than the standing still case, this difference is the reason why you accelerate upwards.

If you could somehow stand on the poles, squat down and jump explosively off them you would drive them into the ground with a greater force than your body weight. (If you could land on it again that would be even better!).

• Good answer, but I'll struggle standing atop a willow shoot: It's as thin and flimsy as a blade of grass at the top! – theonlygusti Apr 25 '15 at 18:04
• I did think that might be problem, physically possible but practically pretty tough. – Chris2807 Apr 25 '15 at 18:05
• @theonlygusti Indeed many pile drivers work as descrived in Chris's answer: a heavy anvil atop a sheath slides over the pole to be driven. The anvil has a system of guides keeping it aligned with the sheath. The anvil also has a hollow in it that sheaths a sticking-out piston on the main sheath. Fuel oil is injected into the hollow and detonated repeatedly: the anvil cyclically jumps upwards and falls again. The movement is typically only a few tens of centimeters at the most, and the frequency about 1Hz. – Selene Routley Apr 28 '15 at 11:18

Sure, I can think of a couple of ways you can exert more downwards force than your weight:

1. Brace against something above you with your hands, e.g. a low ceiling. Then, push against that with your hands as you push down with your legs and you will exert a larger force than your weight.
2. Dynamic force (I assume by 'weight' you just mean static weight, i.e. $W=mg$). If you jump up in the air, then when you hit the ground, as well as reacting the acceleration due to gravity, it also has to decelerate you because you are moving. So, this will exert a larger force on the ground than $mg$. As an example, dropping a lump of metal onto concrete can easily generate instantaneous accelerations in excess of 1000g.

(Disclaimer: Option 1 might be better/safer if dealing with thin willow sticks ...)

• I have no ceiling, unless I push against the sky. Does that mean option 2 is the only way? What about the method where I just push down as hard as I can without lifting myself from the ground? – theonlygusti Apr 25 '15 at 18:42
• @theonlygusti: Well, your question was phrased: 'Can you ever ...?' ;-). I don't think you can exert more force than your weight by just pushing without lifting yourself off the ground. By a force balance argument, if you're applying more force than your weight then you have to be accelerating the rest of your body off the ground (Newton's 3rd Law). – Time4Tea Apr 25 '15 at 18:56
• You could brace with something sufficiently solidly anchored into the ground--I'm thinking of a screw-in type anchor. – Loren Pechtel Apr 25 '15 at 20:53

Yes, you can. The amount of force you can exert on an object is limited only by the geometry and strength of your muscles.

However, Newton's 3rd law dictates that however much force you exert on an object, the object will exert the same amount of force on you, in the opposite direction. So, if you exert a force larger than your weight down on a stick, that stick will exert a force larger than your weight up on you!

If the forces pushing you up are stronger than the forces pushing you down, then there will be a net upward force on you, and therefore an acceleration - i.e. you will begin speeding up in the upwards direction.

Specifically, the net force $F_{net}$ in this situation would be given by

$$F_{net} = F_{stick, you} - F_g$$

where $F_{stick,you}$ is the force the stick puts on you, and $F_g$ is your weight. You can see there is no limit on $F_net$ except for how much force the stick can put on you. That could be limited by how much force your muscles are able to put on the stick, or by how much force the stick can withstand before snapping.

Also note that as you move upwards, unless you were prepared to thrust your arms, your muscles may lose the geometry in which they were contracted, which would decrease the force they exert.

• So there can never be a net downwards force greater than your weight? (This is really what I meant my question to be) – theonlygusti Apr 25 '15 at 18:00
• @theonlygusti: no. Whenever you jump, you exert a downward force on the ground that is greater than your weight. – Jerry Schirmer Apr 25 '15 at 18:09

One approach everybody seems to have missed: increase your weight. Put some weights in a strong backpack.

• While true, I don't really think this is in the spirit of the question. – Kyle Kanos Apr 26 '15 at 0:13
• @KyleKanos It seems to me that it would help the standing-on approach. – Loren Pechtel Apr 26 '15 at 2:53