# muscle tension and force I will describe two different scenarios.

Hi,
Sorry I am not that good at explaining and my not too good Grammar, and if you do have time, thank you in advance.

muscle tension and force I will describe two different scenarios.

QUESTION,
which please do you say puts the most overall tension and fatigue on the muscles, the faster or slower of the rep speed models, and which produces the most overall force ???

Let us say we are doing the bench press, but it could be any lift. Say there are three people with identical strengths and everything else. They are using 80% of their 1RM, ROM is say 20inchs. {50cm} Which rep cadence puts the most fatigue thus overall tension on the muscles.

Scenario 1,
First one does his bench press at a 2/4 rep cadence, which is as you know, 2 seconds concentric and 4 seconds eccentric, he does 5 reps = 24 seconds of lifting, moving the weight 100inches {250cm} in all. And immediately hits muscular failure after the fifth rep.

Second one does his bench press at a .5/.5 rep cadence, he does 24 reps = 24 seconds of lifting, moving the weight 480inches {1200cm} in all. And immediately hits muscular failure after the twenty-fourth rep.

Scenario 2,
Bench press at a 2/4, he does 1 rep = 6 seconds of lifting, moving the weight 40inches {100cm} in all.

Bench press at a .5/.5, he does 1 rep = 1 second of lifting, moving the weight 40inches {100cm} in all.

Bench press at a .5/.5, he does 6 reps = 6 seconds of lifting, moving the weight 240inches {600cm} in all.

As I said, we worked out the forces to lift a 100kg at the different speeds up and down, and the average force needed was basically the same for both the 1 rep of each speed reps. However, as you know when a given load is lifted very fast, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by far exceeds the nominal weight of the load. Even thou the average forces were the same, the slower lifting time I would have said, would have put less tension on the muscles, as with the fast lifting model we are lifting the weight up and down 19 more times in the same time frame, or in the second scenario the weight is lifter 5 more times.

And as you know far better than me, that a small force applied for a long time can produce the same momentum change, as a large force applied briefly, but as I said, with the fast lifting model we are lifting the weight up and down 5 more times in the same time frame, or in the first scenario the weight is lifted 19 more times, thus I would have thought 19 more average forces used by the muscle equals 19 more tensions to the muscles, and more power {work energy} used more overall tension on the muscles.

Because it is the product of the force and the time for which it is applied that is important. As the slower lifting and lowering was a lower force lift, as it was done over a longer time frame, in the 1 rep at .5/.5 against the 1 rep at 2/4.

Thus if 1 rep at .5/.5 and 1 rep at 2/4 have the same average force, they should have the same tensions, however as I am doing 5 more reps in the same time frame, should not I would think I have 5 times more tension on my rep{reps} ???

The below was worked out for a 100kg and 1m ROM each way.

Fast rep, forces required to accelerate the weight up and down, at .5/.5 one time.
1962 kg m/s^2

Slow rep, forces required to accelerate the weight up and down at 2/4 one time.
1999.5 m/s^2

However, we did not take into account and add in the peak forces, that will be at the transition from negative to positive; we call them the MMMTs, {Momentary Maximum Muscle Tensions}could anyone know how to work these out please ??? to add to the fast rep, forces required to accelerate the weight up and down.

Thx for you time again.

Wayne

For all reasonable lifting speeds, it doesn’t make any real difference.

Hi there,

At this moment in time I am not debating which is the best for hypertrophy or strength, just which rep{reps} have the most tension, fatigue and overall force output, between 1 rep at 2/4 and 6 reps at .5/.5 both are completed in the same time frame.

I would say as the 6 reps at .5/.5 have more power {work energy} that they will have far more of tension, fatigue and overall force output, I would suggest 50% more. However, would like f possible some more comments from members please.

Wayne

Hi all,

Ok any clever people here ???

I was trying to work out the force needed to lift a 100kg weight. However the lifter first is lowering this weight at .5 of a second, in 1m, and at the last instant has the accelerate this weight up 1m at .5 of a second.

Someone worked out for me the lifting and lowering of a 100kg for 1m each way at .5/.5 and 2/4. However, as the force need is going to be higher in the transition from the eccentric to the concentric, in each repetition of the lifting and lowering in weightlifting, I was hoping to work this out, as it’s a bit over my head. It’s nothing to do with homework by the way, I have posted similar questions before, it’s just for a debate we are having. Thx in advance for your time if anyone is able to work it out, here are the calculations a friend worked out, hope they are right not sure.

[b][u[Let’s assume we’re comparing 2 sets using the same weight, but different rep speeds. set 1 uses 0.5/0.5 and set 2 uses 2/4. For simplicity’s sake we’ll assume the weight accelerates 100% of the way up and down for both sets.

Known: Mass(m) =100kg (220lbs) Acceleration(a)=?? Distance(d)=1m

First let’s solve for the acceleration required to move the weight 1m in the time frames of the sets.

Calculate a, to travel 1m in 0.5s (Raising and Lowering of set 1):
d=1/2at^2
1m=1/2a(0.5s)^2
a=2(1m)/(.25s^2)
a=8 m/s^2

Calculate a, to travel 1m in 2s (Raising set 2):
d=1/2at^2
1m=1/2a(2s)^2
a=2(1m)/(4s^2)
a=0.5 m/s^2

Calculate a, to travel 1m in 4s (Lowering set 2):
d=1/2at^2
1m=1/2a(4s)^2
a=2(1m)/(16s^2)
a=.125 m/s^2

Now lets solve for the forces required to accelerate the weight

Calculate the force required to raise 100kg, 1m, in 0.5s:
Sum of forces=ma
F1-mg=ma
F1-(100kg)(9.81m/s^2)=(100kg)(8m/s^2)
F1=(981+800) kg m/s^2
F1=1781 kg m/s^2 required to raise the weight 1m in 0.5s

Now compare it to the force required to raise 100kg, 1m, in 2s
Sum of forces=ma
F2-mg=ma
F2-(100kg)(9.81m/s^2)=(100kg)(0.5m/s^2)
F2=(981+50) kg m/s^2
F2=1031 kg m/s^2 required to raise the weight 1m in 2s

***As predicted, it takes more force to raise the weight 1m in 0.5s than it does to raise it in 2s. It takes F1/F2=1.73 times as much force to do so. You’re right about this but nobody is disagreeing with you here

However, now let’s look at what happens on the way down.

Calculate the force required to lower 100kg, 1m, in 0.5s:
Sum of forces=ma
mg-F3=ma
(100kg)(9.81m/s^2)-F3=(100kg)(8m/s^2)
F3=(981-800) kg m/s^2
F3=181 kg m/s^2 is required to lower the weight 1m in 0.5s

Now compare it to the force required to lower 100kg, 1m, in 4s
Sum of forces=ma
mg-F4=ma
(100kg)(9.81m/s^2)-F4=(100kg)(.125m/s^2)
F4=(981-12.5) kg m/s^2
F3=968.5 kg m/s^2 is required to lower the weight 1m in 4s

***Contrary to your belief, it takes MORE force to lower the weight 1m in 4s than it does to lower it in 0.5s. In fact, it take A LOT more. It takes F5/F3=5.35 times as much force to do so! That’s 5.35 times more force to go slow than fast!!![/b][/u]

Wayne

I’ll look through your work later tonight after practice. I have a math degree and have done a fair amount of physics.

Your friend’s calculations are accurate. The assumption about a constant acceleration is an oversimplification, but it doesn’t matter for you questions.

Of course it takes more force to lower a weight slowly: how much force does it take to drop a weight?

I would have thought if you say lowered 100kg slowely as in 8 seconds for 1m, and then lowred the same weight for 2 seconds, that as of the acceleration component, That when a given load is lifted slightely faster, the acceleration component means that the forces exerted on the load (and thereby by the muscles) by will exceed the nominal weight of the load ???

But what I was really trying to work out was the below, if you have time please could you help.

I was trying to work out the force needed to lift a 100kg weight. However the lifter first is lowering this weight at .5 of a second, in 1m, and at the last instant, say the last 20% he has the accelerate this weight up 1m at .5 of a second.

Someone worked out for me the lifting and lowering of a 100kg for 1m each way at .5/.5 and 2/4. However, as the force need is going to be higher in the transition from the eccentric to the concentric, in each repetition of the lifting and lowering in weightlifting, I was hoping to work this out, as it’s a bit over my head.

Wayne