Answer to Question #108373 in Mechanics | Relativity for Abhishek singh

Question

A cricket shoe has 10 cleats (nails) in sole, each have area of 1.65 x 10 -2cm 2in contact with the floor. During walking there is one instant when all 10 cleats support the entire weight of 80 kg person. What is pressure exerted by the nails on the floor?

Answer 1

Solution: The pressure of a body on the ground is determined by the ratio of the force of the weight of this body to the area of its support

(1) $P=\frac{F}{S}$ .

In our case, the weight of the body is equal to $F=mg=80 kg\cdot 9.8 ms^{-2} =784 N$. The support Area coincides with the total cross-section area of the ten cleats  $S=10\cdot1.65\cdot10^{-2} cm^2=1.65 \cdot 10^{-1}\cdot 10^{-4} m^2=1.65\cdot 10^{-5} m^2$ . Substitute this quantities to (1) we find

(2) $P=\frac{784N}{1.65\cdot 10^{-5} m^2}=47.5\cdot 10^6 Pa=47.5 MPa$ This is quite a lot of pressure. Therefore, the cleats enter the ground and hold the player in place well.

Answer: The pressure exerted by the nails on the floor is $47.5 MPa$