Samus, pressure, and forcePressure is different from force, since pressure is force over the surface area. Taking this

image again, I will simply address pressure because I was using Brinell's hardness test, which I know was incorrect, since this is not the way to determine pressure. Taking normal force, which is mass times gravity, I will then be able to find the pressure. Gravity is usually 9.81 m/s^2, but this is equal to 9.81 N/kg. We know Samus is 90 kg., so 9.81 N/kg times 90 kg. is 882.9 N. The surface area of the spike that I calculated was 2.753623188405796 cm. 882.9 N / 2.753623188405796 cm^2 is 3.2063 MPa.

I know walking on a bed of nails won't cause any physical damage, since the pressure is evenly distributed over the surface area, this might explain why the nails that are in motion harm Samus, since the nails are no longer even, as the image in the link demonstrates. The image also shows more spikes in a different form that drop down on Samus and slowly rise up. These spikes also have the same surface area. Samus won't take damage from these unless they come down on her. So we could say a pressure of 3.2063 MPa won't harm Samus, especially since 882.9 N doesn't harm Samus, as dashing at Mach 2 would produce 61,740 N.

Edit: So, it turns out Samus' height in pixels in

this image is actually 86 px., not 69 px. I don't know how I miscounted. 1 px. / 86 px. is 0.0116279069767442. 198 cm. * 0.0116279069767442 is 2.3023255813953516 cm. Samus' normal force is 882.9 N and 882.9 N / 2.3023255813953516 cm^2 is 3.8348 MPa.

Edit 2: I learned that the spikes are 2 px., not 1 px. like the other calculation. 2 px. / 86 px. is 0.0232558139534884. 198 cm. * 0.0232558139534884 is 4.6046511627907032 cm. 882.9 N / 4.6046511627907032 cm^2 is 1.9174 MPa. Well, now I want to take the gravitational potential energy, but only find the newtons, not joules, and see how many pascals Samus can withstand. Though the distance between the ceiling and Samus is about 3 Samus' high, I'll ignore the third Samus because if her head is coming in contact with the ceiling, then I'm only going to be concerned with the distance from her feet down.

This means the distance is 2 Samus' high with an additional pixels, or 396 cm. The rest of the pixels add up only to 3 px., so 3 px. / 86 px. is 0.0348837209302326 * 198 cm. is 6.9069767441860548 cm. So the distance is 402.9069767441860548 cm. Using gravitational potential energy, which is U = mgh, where m is mass, g is gravity, and h is height, we end up with U = (90 kg.)(9.81 N/kg.)(402.9069767441860548 cm.). This is equal to 355,726.56976744186778292 N cm. Samus tends to bend her knees as she lands, so taking this into consideration, let's say she bends her knees so that she's half her height. 355,726.56976744186778292 N cm. / 50 cm. is 7,114.5313953488373556584 N, and 7,114.5313953488373556584 N / 4.6046511627907032 cm^2 is 15.450749999999982 MPa.

Maybe Samus can withstand more pressure. I can't say for certain, unless there's another area like the one in the image where Samus shows no damage.