- #1
spenc193
- 7
- 1
Thank you in advance for your help!
Consider a lifter performing a bench press. The weight of the bar is 225 lbs. Assume the duration of the entire movement from the lowest point to the highest is 1 sec., and that the bar will move a total of 2 ft. from the lifter's sternum to the height of his reach. After zero pause at the top of the lift, the bar will be lowered at a constant velocity, requiring 2 sec. to return it to the initial position.
It would help me if we used units of foot-pounds; I will use these units and g = 32.17 ft/s^2.
How would I calculate the force produced during BOTH the positive (upward; concentric) phase of this lift and the negative (downward; eccentric)?
I realize that the force is essentially equal to the poundage, but it seems to me there are two components here: (1) the force necessary simply to keep the bar suspended in the gravity field and (2) the force to accelerate the bar upward. So knowing that we accelerate the bar 2 ft/1sec, it would follow that the total force would be → F = (225) + [(225) * (2/32.17)]
So is it 225 lbs. or is it 238.9 lb ?
Next: Work.
I am having difficulty calculating work, especially as the combined quantity for both phases of the lift. For the upward phase, would it simply be:
W = F x d = (either 225 or 238.9 lb)(2 ft)
OR would it be correct to calculate work as the change in PE for the weight:
W = ΔPE = (m)(g)(h) = (225 lb)(32.17)(2 ft)
because then it seems to me we'd still have to add in the component of the bar being accelerated right? So we'd get:
(m)(g)(h) + (m)(a)(ft) BUT, how would we really approximate the acceleration? The
velocity is zero at the initial and final position of the lift. Wouldn't we hav
to know the instantaneous acc. immediately after lift-off and just
before peak height in order to do this?Lastly: Work for downward part of lift.
I understand that the work will be negative on the downward component of the lift. The weight will be doing work on the muscle, and the muscle will be absorbing energy. But I still am not sure how to calculate the work here.
Despite the fact the bar is descending, we are CONTROLLING the descent using muscular force, despite the fact the muscle is elongating. How would we calculate the work being done on the descending weight in order to control that descent for say perhaps 2 seconds? We are still acting with an upward force, but one which must be less than 225 lbs. I'm just not sure how to calculate that and the subsequent work done during that phase. This probably sounds stupid, because I get the work is being done BY the weight, but how is it the lifter is not doing some positive work to keep the weight from simply dropping?THANKS!
Consider a lifter performing a bench press. The weight of the bar is 225 lbs. Assume the duration of the entire movement from the lowest point to the highest is 1 sec., and that the bar will move a total of 2 ft. from the lifter's sternum to the height of his reach. After zero pause at the top of the lift, the bar will be lowered at a constant velocity, requiring 2 sec. to return it to the initial position.
It would help me if we used units of foot-pounds; I will use these units and g = 32.17 ft/s^2.
How would I calculate the force produced during BOTH the positive (upward; concentric) phase of this lift and the negative (downward; eccentric)?
I realize that the force is essentially equal to the poundage, but it seems to me there are two components here: (1) the force necessary simply to keep the bar suspended in the gravity field and (2) the force to accelerate the bar upward. So knowing that we accelerate the bar 2 ft/1sec, it would follow that the total force would be → F = (225) + [(225) * (2/32.17)]
So is it 225 lbs. or is it 238.9 lb ?
Next: Work.
I am having difficulty calculating work, especially as the combined quantity for both phases of the lift. For the upward phase, would it simply be:
W = F x d = (either 225 or 238.9 lb)(2 ft)
OR would it be correct to calculate work as the change in PE for the weight:
W = ΔPE = (m)(g)(h) = (225 lb)(32.17)(2 ft)
because then it seems to me we'd still have to add in the component of the bar being accelerated right? So we'd get:
(m)(g)(h) + (m)(a)(ft) BUT, how would we really approximate the acceleration? The
velocity is zero at the initial and final position of the lift. Wouldn't we hav
to know the instantaneous acc. immediately after lift-off and just
before peak height in order to do this?Lastly: Work for downward part of lift.
I understand that the work will be negative on the downward component of the lift. The weight will be doing work on the muscle, and the muscle will be absorbing energy. But I still am not sure how to calculate the work here.
Despite the fact the bar is descending, we are CONTROLLING the descent using muscular force, despite the fact the muscle is elongating. How would we calculate the work being done on the descending weight in order to control that descent for say perhaps 2 seconds? We are still acting with an upward force, but one which must be less than 225 lbs. I'm just not sure how to calculate that and the subsequent work done during that phase. This probably sounds stupid, because I get the work is being done BY the weight, but how is it the lifter is not doing some positive work to keep the weight from simply dropping?THANKS!