The Physics of Force

[Frank_B]

I wrote this all once before but something got lost during the submission process so if this is a repeat post please delete:

We call strength “the ability to produce force against an external resistance,” yet I found myself in a sort of paradoxical conundrum when thinking about force production through the course of a lift.

Suppose we have a lifter who will lift a 500 pound deadlift. Let’s ignore any frictional forces and suppose that the bar path will be perfectly vertical. Let’s also assume that the lift will cover a distance of 12 inches and that upon initial acceleration of the bar over the first 3 inches of travel, the bar travels at a constant velocity the remaining 9 inches until lockout.

It is here where I have encountered conceptual difficulties. Since acceleration is a derivative of velocity, or a rate of change in velocity over time, and if the bar is moving at a constant rate, then the total force applied to the bar while it’s speed is constant is 0. In real life, the velocity won’t be constant and will have some slight perturbations, but will likely still approximate constant through some part of the lift.

Here’s the rub: I’ve pulled a heavy deadlift before and I know I’ve produced force throughout the entirety of the lift. So what gives? How is the force 0 if I’m feeling this through the entirety of the lift?

The best I can come up with is that the lifter need only produce an initial force off the floor that accelerates the bar to a constant speed. After which, the lifter need only produce the same amount of force as the force opposing it. In other words, the initial force off the floor must be greater than 500 pounds, but once a constant speed is reached, the amount of force in the downward vector is -500 pounds (gravity) and the lifter need only apply +500 pounds upward (constantly) to keep the bar moving at a constant rate. The Sum forces acting on the bar are subsequently ~0 regardless of the (constant) bar speed. If the lifter is producing 501 pounds of force, then the sum total would be 1 pound of force in favor of the upward direction (and the speed will be slightly higher).

Once the bar has reached lockout, the bar is once again not moving and the constant velocity is 0. The lifter is still producing +500 pounds of force against gravity, which is producing -500 pounds of force in the opposite direction. Again, since velocity is not changing then acceleration is 0 and the sum total force on the bar is actually 0 regardless of how shitty the lifter feels.

Am I understanding this correctly?

[Mark Rippetoe]

If the bar weighs 500 pounds, the earth is exerting 500 pounds of force on the bar, and the floor is pushing back with 500 pounds (the "normal force"). When you start to pull the bar off the floor, its velocity is zero, so you have to accelerate the bar to some non-zero velocity. This requires that you apply force in excess of 500 pounds -- lets call that "501" with the understanding that the more force you apply the more the bar accelerates and achieves a higher velocity. If you apply 525, it moves faster than 501, and if you apply 725 you can clean it. But since you want to deadlift it, you have to get it started upward and then maintain upward. It has to leave the floor with more than 500 because it has to accelerate from zero velocity to some positive number that will get the bar to lockout within a few seconds. But the first couple of inches is where the acceleration takes place, and the rest of the pull is just you maintaining a velocity, with force in excess of 500. These numbers might be 525 to start it, then "501" to maintain the velocity up to lockout. If your pull drops down to 500, the bar stops. It's 500 at lockout, but it has to be "501" on the way up or it stops.

But anybody with a better explanation is free to chime in.

[Mark Le Comte]

it has to be "501" on the way up or it stops.

if the force drops under 500 on the way up the bar decelerates. If it does this for long enough the bar stops and then starts to go down.

You can also think work = force x distance and work due to gravity is force x height. To get a 500lb bar to rise 3 feet you could pull with 550lb force for 3 inches (accelerating), 500 lb for 2 ft 6 in (constant velocity) then 450lb for final 3 in (decelerating).

Pull with enough force over enough distance and the bar accelerates enough that you can stop pulling and jump under it (clean or snatch).

[asm44]

If your pull drops down to 500, the bar stops.

If the bar is moving, and if there is no friction as OP said (i.e. air resistance), the bar does not stop. It continues at its current velocity.

If the ROM was infinitely long, and you pulled the 500lbs with 500.000001 lbs of force, it would accelerate forever and thus move infinitely fast.

How is the force 0 if I’m feeling this through the entirety of the lift.

Because you're maintaining a force output. the net force on the bar may be zero after lift-off, but the internal forces and torques on your bones are not. Even at the top of of the pull when the velocity is zero and the entire system is static, your musculature is isometrically contracting hard to produce the force necessary to counteract the force of gravity to keep everything in place.

The barbell is doing it too. Not actively like you, but a loaded and racked barbell is under a bunch of tensile stress that it has to resist, even though it's not moving. That's why it's a bit bent (or a lot). But the bar doesn't have feelings and can't complain.

[Subby]

But anybody with a better explanation is free to chime in.

That's a trap if ever I saw one.

It is here where I have encountered conceptual difficulties. Since acceleration is a derivative of velocity, or a rate of change in velocity over time, and if the bar is moving at a constant rate, then the total force applied to the bar while it’s speed is constant is 0. In real life, the velocity won’t be constant and will have some slight perturbations, but will likely still approximate constant through some part of the lift.

Here’s the rub: I’ve pulled a heavy deadlift before and I know I’ve produced force throughout the entirety of the lift. So what gives? How is the force 0 if I’m feeling this through the entirety of the lift?

I'm not sure I'm understanding the problem, but if you apply 0 force to the bar, then it will fall at the rate gravity wants it to fall at, slowed only momentarily by the bar accelerating your face towards the floor. (assuming you're still holding the bar)

I think the underlying issue though is that you are relying on "what you feel" during the lift. That is subjective and unreliable.

[Frank_B]

If the bar weighs 500 pounds, the earth is exerting 500 pounds of force on the bar, and the floor is pushing back with 500 pounds (the "normal force"). When you start to pull the bar off the floor, its velocity is zero, so you have to accelerate the bar to some non-zero velocity. This requires that you apply force in excess of 500 pounds -- lets call that "501" with the understanding that the more force you apply the more the bar accelerates and achieves a higher velocity. If you apply 525, it moves faster than 501, and if you apply 725 you can clean it. But since you want to deadlift it, you have to get it started upward and then maintain upward. It has to leave the floor with more than 500 because it has to accelerate from zero velocity to some positive number that will get the bar to lockout within a few seconds. But the first couple of inches is where the acceleration takes place, and the rest of the pull is just you maintaining a velocity, with force in excess of 500. These numbers might be 525 to start it, then "501" to maintain the velocity up to lockout. If your pull drops down to 500, the bar stops. It's 500 at lockout, but it has to be "501" on the way up or it stops.

But anybody with a better explanation is free to chime in.

OK. I understand it virtually the same, with the exception of needing to exceed 500 pounds once the acceleration phase has completed and the bar is in constant velocity. Assuming there is precisely +500 pounds of force in the upward vector and -500 pounds of force in the downward vector, but the bar is moving at a constant speed, then there is no net change in force and the bar can continue moving upwards.

Excess force in any direction, as I understand it, is only needed to get things moving or stop/reverse them. An object with no net force has constant velocity, not constant position and will remain in equilibrium and/or keep moving.

So wouldn’t that imply that we only need 500 pounds of force to move the lift at a constant rate, after initial acceleration, and not necessarily “501?”

[francesco.decaro]

I wrote this all once before but something got lost during the submission process so if this is a repeat post please delete:

We call strength “the ability to produce force against an external resistance,” yet I found myself in a sort of paradoxical conundrum when thinking about force production through the course of a lift.

Suppose we have a lifter who will lift a 500 pound deadlift. Let’s ignore any frictional forces and suppose that the bar path will be perfectly vertical. Let’s also assume that the lift will cover a distance of 12 inches and that upon initial acceleration of the bar over the first 3 inches of travel, the bar travels at a constant velocity the remaining 9 inches until lockout.

It is here where I have encountered conceptual difficulties. Since acceleration is a derivative of velocity, or a rate of change in velocity over time, and if the bar is moving at a constant rate, then the total force applied to the bar while it’s speed is constant is 0. In real life, the velocity won’t be constant and will have some slight perturbations, but will likely still approximate constant through some part of the lift.

Here’s the rub: I’ve pulled a heavy deadlift before and I know I’ve produced force throughout the entirety of the lift. So what gives? How is the force 0 if I’m feeling this through the entirety of the lift?

The best I can come up with is that the lifter need only produce an initial force off the floor that accelerates the bar to a constant speed. After which, the lifter need only produce the same amount of force as the force opposing it. In other words, the initial force off the floor must be greater than 500 pounds, but once a constant speed is reached, the amount of force in the downward vector is -500 pounds (gravity) and the lifter need only apply +500 pounds upward (constantly) to keep the bar moving at a constant rate. The Sum forces acting on the bar are subsequently ~0 regardless of the (constant) bar speed. If the lifter is producing 501 pounds of force, then the sum total would be 1 pound of force in favor of the upward direction (and the speed will be slightly higher).

Once the bar has reached lockout, the bar is once again not moving and the constant velocity is 0. The lifter is still producing +500 pounds of force against gravity, which is producing -500 pounds of force in the opposite direction. Again, since velocity is not changing then acceleration is 0 and the sum total force on the bar is actually 0 regardless of how shitty the lifter feels.

Am I understanding this correctly?

You're saying that the bar has enough momentum after the first 3 inches off the floor because of the higher force produced, that it can keep going up even if the net force is zero accross the remaining 9 inches? Something like that?

I'm thinking of what happened in Chase's 405lbs Press. The bar accelerated up, stopped, went down a little, and then started going up again.
It seems obvious that what happend in terms of force production is that it was much higher than the weight of the bar at first, then it was the same, then lower, and then higher again, and at the lockout it might've been lower because of all the skeleton helping supporting the bar. I don't know if the bar path was perfectly vertical over the mid-foot but I think it was. But in any case, when the bar stopped, it's obvious that the net force was zero and that it was negative when it dipped.
I don't see why it would be different in a deadlift or any other lift. The net force in the opposite direction of gravity has to be higher than that in the direction of gravity to move the bar up.

If someone can maintain 525lbs of force against 500lbs of weight accross the whole movement, the velocity remains constant anyway, right?
But that probably never happens, so I would say it's safe to assume the bar's velocity is always changing accross the movement and it's at zero only when it's not moving.

[FrankNJ]

Since acceleration is a derivative of velocity, or a rate of change in velocity over time, and if the bar is moving at a constant rate, then the total force applied to the bar while it’s speed is constant is 0. Here’s the rub: I’ve pulled a heavy deadlift before and I know I’ve produced force throughout the entirety of the lift. So what gives? How is the force 0 if I’m feeling this through the entirety of the lift?

You're getting hung up on the definition of acceleration. Even if an object is sitting on the ground with 0 velocity, we still describe the force in terms in terms of acceleration (32.2 feet/sec2) because that is what would happen if the object was in free fall. To show this in your scenario, let's imagine that you have gotten the bar off the ground and it is now moving up at some constant velocity. Let's then imagine that the force of gravity suddenly disappeared while you were moving the bar up at a constant velocity. Your bar would in fact then start accelerating upward at a rate of something slightly above 32.2 feet/sec2.

[VNV]

...
Here’s the rub: ... How is the force 0 if I’m feeling this through the entirety of the lift?...

...

Because you don't feel the net force (~0), you only feel the force you apply, which is about equal to the force of gravity throughout the lift. A little greater at the beginning (for a moment), equal in the middle (for the duration), and a little less at the end (for a moment).

...Am I understanding this correctly?

Yes.

[Barry Charles]

But anybody with a better explanation is free to chime in.

I don’t know about better , but alternatively:

During the lifting up phase, when the force up is greater than the force down, the bar is accelerating up. When the force up is the same as the force down the acceleration is zero and the velocity is constant which could be zero (a stall or top of lift). When the force up is less than the force down the bar is decelerating.

Force Up + Force down = Mass x Acceleration

Fup > Fd , Accel is upward, velocity increasing

Fup = Fd , Accel = 0, velocity constant

Fup < Fd, Deceleration , velocity decreasing