EK Physics 1001 #627

[BrianK0220]

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[Jepstein30]

Is the answer explanation to this question correct? If so, I must be having trouble with the algebra. Can someone solve for delta x in a step by step fashion? Thanks!

post the problem and we can probably help

 

[BrianK0220]

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[Jepstein30]

A force is applied to the top of a copper block (10^6N). If the bottom of the block is fixed to the ground, how far will the top of the block shift? shear modulus of copper = 40x10^9 N/m^2.

A. 6.25x10^-6 m
B. 1.25x10^-5 m
C. 2.50x10^-5 m
D. 12.5 m

F=10^6N
A=4m^2
h=2 m

Solve for delta x

Modulus = stress/strain

stress = F/A

strain = deltax/x

modulus = F/A * x/deltax = Fx/A*deltax

deltax = Fx/A*modulus = (10^6)*(2)/(4*40*10^9) = 1/(80*10^3) = 1/(8*10^4) = 1.25*10^-5

Answer is B?

 

[BrianK0220]

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[BrianK0220]

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[Jepstein30]

How does (10^6)*(2)/(4*40*10^9) = 1/(80*10^3)? Clearly, my algebra is rusty. I understand everything else. It's a simple plug and chug. Thanks a lot guys.

Yes, B is correct.

handle exponents separately.. 10^6/10^9 leaves you with 10^-3 or 1/10^3

then you have 2/(4*40) which I reduced to 1/(2*20).. = 1/80

1/(80*10^3).. or 1/8*10^4 ... 1/8 is .125 so .125*10^-4 or 1.25*10^-5

 

[sciencebooks]

Modulus = stress/strain

stress = F/A

strain = deltax/x

modulus = F/A * x/deltax = Fx/A*deltax

deltax = Fx/A*modulus = (10^6)*(2)/(4*40*10^9) = 1/(80*10^3) = 1/(8*10^4) = 1.25*10^-5

Answer is B?

NVM 😛

 

[DL011092]

Does anyone know how the area is 4?
The force is applying from the right side of the rectangular box; shouldn't it be 4 x 2=8m2?

 

[Jengreef]

Does anyone know how the area is 4?
The force is applying from the right side of the rectangular box; shouldn't it be 4 x 2=8m2?

area is given as 4m^2. you don't have to calculate it. If you multiply 4*2, and 2 is the given height, you get volume. 4m^2 * 2m = 8m^3

And the force is applied from the top down, not from the right to the left.