**The Physics of Squatting**Kasey Halsey ’18

This past month, a powerlifter
named Ray Williams set a world record for the “raw” back squat. He lifted an
impressive 1,0005 pounds, what makes this feat even more amazing is that he did
so without the assistance of knee wraps or a squatting suit. Traditional knee
wraps and squat suits store elastic energy which aids the athlete in lifting
more weight than they would be able to without the use of such equipment (Blatnik
et al., 2012 and Lake et al., 2012). This means that Ray Williams had to exert
a larger amount of force upwards to lift the weight than he would have had to while
equipped with knee wraps and a squat suit. We can use physics to determine
exactly how much force Ray Williams had to exert during this astonishing
display of strength. First it should be noted that he lifted exactly 456 kg
during the lift. Using F=ma, we can say that the total downwards force exerted
by the weight would be (456 kg)*(9.8 m/s^2)=4470 N. Then, using kinematics we
can calculate how much Ray is able to accelerate the weight when he takes if
off of the rack. Let’s say that the final velocity of the weight if 0.25 m/s,
and he’s moving it from rest, and the lift off of the rack takes 0.3 seconds.

Vf=Vo+at > 0.25 m/s= 0m/s +a(0.3 s) > a=0.833
m/s^2

Now, we know that the net force
upwards must be F=ma=(456 kg)(0.833 m/s^2)=380 N for this lift to occur. This
means that Ray has to exert 4470 N + 330 N= 4850 N of force upwards to un-rack
the weight. Now, during the actual squat, we can calculate the force exerted on
the weight during the concentric upwards portion of the lift. During the
upwards motion, he accelerates constantly and hits a sticking point, so to
simplify the calculation we will only consider the lift up until this sticking
point during which there is constant upwards acceleration. This time, we can
say that the lift took 0.8 seconds, and that his final velocity was 1.2 m/s. So
we can perform a similar calculation for acceleration.

Vf=Vo+at > 1.2 m/s= 0m/s +a(0.8 s) > a=1.5 m/s^2

So, if the weight had to be accelerated
at 1.5 m/s^2, then the forced exerted during the lift by Ray was (F=ma=(456
kg)(1.5 m/s^2)=684) + 4470 N= 5150 N. If the distance traveled during this
portion of the lift was 0.7 meters, then he did W=Fd=(5150 N)(0.7 m)=3605
Joules of work during the lift.

We can also roughly calculate the
torque needed to be exerted by the leg to move the weight by examining its
connection to the knee joint. If he squatted so that his knees were approximately
parallel to the ground, then the torque would needed to rotate the leg about
the knee joint would be T=F*r, and assuming the thigh was 0.5 meters long this
calculation would give us T=(5150 N)*0.5 meters=2575 Nm.

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