Thursday, December 1, 2016

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.

No comments:

Post a Comment

Note: Only a member of this blog may post a comment.