Comic-book superheroes are sometimes able to punch holes through steel walls. Co

Question

Comic-book superheroes are sometimes able to punch holes through steel walls.

Comic-book superheroes are sometimes able to punch holes through steel walls. (a) If the ultimate shear strength of steel is taken to be 2.10 x 10^2 cm^2 and is approximately circular.N (b) Qualitatively, what would happen to the superhero on delivery of the punch? What physical law applies? x 10^9 Pa, what force is required to punch through a steel plate 2.30 cm thick? Assume the superhero's fist has cross-sectional area of 1.17

Explanation / Answer

Part (a).
The equation is Smax = Fmax/Amax.................................. S=F/A
You need to find Fmax....................................... Maximum Force
So first find Amax....................................... Maximum Area
Amax = C x L
C = 2 x ? x R
L = 2.3 cm (given).................................... Thickness of plate
Afist =117 cm^2 (given).................................... Area of fist
R in (C = 2 x ? x R), is ?(Afist / ?)
Put R in C: C = 2?(Afist x ?)
Amax = L x 2?(Afist x ?)
REMEMBER TO CONVERT: cm to m, Pascal is kg/m^2, so convert to meters when using Pascals
Amax = 0.023 x 2?(1.17 x ?) = 0.08819129075
Divide by 10: 0.08819129075/10 =0.008819129075
Smax=2.1*10^9 Pa (given).................................... Maximum Strength
Smax x Amax = 0.008819129075 x 2.1*10^9
Fmax =18520171.06 Newtons or 1.852*10^7 N

.

Part (b). Qualitatively, what would happen to the superhero on delivery of the punch? What physical law applies?
By Newton's third law, the wall would exert a force of equal magnitude in the opposite direction on the superhero, who would be thrown backward at a very high recoil speed

Related Questions

Navigate

• Browse (All)
• Browse C
• Subjects

---