Lift a weight quickly over your head and then lower it at the same speed.
It is much easier to lower it. Why? Because it is in free fall for part of
the time while you are lowering it so no work is done by your muscles during
the free falling time. In contrast. it always requires a force against
gravity to lift it over your head.
Now decrease the speed at which you lift and then lower the weight. The
effort it takes to lift the weight and then lower the weight become more
similar. If done very slowly, the efforts are effectively the same. Try
the experiment by lifting and lowering a barbell weight (3 seconds up and 3
seconds down vs. 15 seconds up and 15 seconds down.)
When you lift a weight, there is no way to cheat gravity. When you lower a
weight, you can easily cheat gravity when the weight is in free fall. In
fact, I suspect that a realistic model for lowering a weight is one in which
the weight free falls a small distance then you muscles slow down the
weight, followed by subsequent free falls and catches. The same holds true
for push ups and pull ups.
Lawrence D. Woolf; General Atomics; 3550 General Atomics Court, San Diego,
CA 92121; Phone:858-455-4475; FAX:858-455-4268; http://www.sci-ed-ga.org
-----Original Message-----
From: Tucker Hiatt
Sent: Monday, July 23, 2001 9:24 AM
Subject: Re: Muscle work
I've explained why this question bugs me. (Although Tim Folkerts
recent posting has helped considerably. Thanks!) Please let me also
explain why it bugs my students.
Often, the muscle motion under examination is repetitive. Push-ups
and pull-ups are particular favorites. I instruct the students to
complete each half-cycle of their exercise at constant speed, if
possible. (And I acknowledge that we are ignoring the work done
during turn-around times.) This means that a student does work
during the downstroke as well as during the upstroke. On the
upstroke of a constant-speed pull-up, the average muscle force
applied is roughly the student's weight. On the constant speed
downstroke, however, the average muscle force is ...??? Mustn't it
also be the student's weight? And therefore the (magnitude of the)
work done is the same? Doesn't that vivify the question: Why do the
upstroke and downstroke FEEL different?