Many types of weightlifting, such as curls, are easier for shorter people to do, as shorter people tend to also have shorter arms. When considering the weight as a point mass, the total torque needed for the muscles in the arm to keep the weight still must equal 0. A problem like this already appears on the lecture slides from class 28, so I won't do all the math again. In this problem, the length of the forearm was 38.1cm and the amount of force needed from the biceps was 404N. Now, let's say the arm was 5cm shorter. Using the same formula, the needed force of the biceps can be found.

∑τ = τ biceps +τ bone +τ arm +τ ball = 0

(Fbiceps0.050m(sin(110°))) = (13.3N ×.152m)+(44.5N ×.331m)

Fbiceps = 357N

This shows that by just cutting off just 5cm of your forearm, 50 fewer newtons are required to lift weights! (Don't actually cut off part of your forearm.)

I actually have some experience with this next activity. Being shorter is quite advantageous as a lineman in football, especially when it comes to blocking. Much of this has to do with center of gravity, but not much about this was discussed in this course, so I'll simplify this using concepts we know more about, like basic forces. Let's say that the shorter lineman pushes his opponent with a force of 200N at an 80 degree angle between his opponent and his arms. The taller lineman pushes on the other lineman with the same force, but at a 100 degree angle. By using basic trigonometry, it can be determined that both players push in the x-direction with about 196N of force, but the shorter player pushes the taller player with 35N in the upwards direction, and the taller player pushes with 35N in the downwards direction. These vertical forces are then added onto the normal force of each player, assuming that the downward direction is positive and the upwards direction is negative. Assuming the players both players weigh 80kg and the static coefficient of friction is 0.9, the following frictional forces can be calculated.

Ffr = μFn

Ffr(tall) = (0.9)((80kg)(9.8m/s^2) + (-35N)) = 670N

Ffr(short) = (0.9)((80kg)(9.8m/s^2) + 35N) = 740N

Therefore, the short player has a higher force of friction acting on him than the taller player, meaning that the shorter player will have a much easier time pushing the taller player than the taller player would pushing the shorter player.

The next time anyone makes fun of how short you are, just let them know that you could probably beat them in most strength-based activities.

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