Shear stress = F_1/A = F_1/(0.80 m)(0.0050 m) = (250 m^-2)F_|| The shear strain

Question

Shear stress = F_1/A = F_1/(0.80 m)(0.0050 m) = (250 m^-2)F_|| The shear strain is Shear strain = x/h = 1.6 times 10^-4 m/0.80 m = 2.0 times 10^-4 SOLVE The shear modules S is S = Shear stress/Shear strain = 0.36 times 10^11 Pa = (250 m^-2)F_1/2.0 times 10^-4 Solving for F_1, we obtain F_1 = 2.9 times 10^4 N REFLECT This is nearly 7000 lb, or roughly twice the weight of a medium-size car Part A We replace the plate with another brass plate that has the same horizontal dimensions (0.80 m on a side) but a different thickness, If the shear forces have magnitude 1.2 times 10^5 N and the shear strain is the same as before, Express your answer in centimeters to two significant figures.

Explanation / Answer

Shear strain = 2.0 * 10^-4
Shear Stress = F/A
Where, F = 1.2 * 10^5 N

S = Shear Stress/Shear strain
0.36 * 10^11 = (1.2 * 10^5)/(A) / (2.0 * 10^-4 )
A = 0.01667 m^2
0.8 * x = 0.01667 m^2
x = 2.1 cm

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.