A brass cube 2cm on each edge is held in the jaw of a wire with a force 1600N. By aw much is the cube compressed. (Y for brass is 9.1×10^10pa)
I have difficulty getting the meaning "in the jaw of a wire with a force" on each edge.
Length =2cm=0.02m
Area of the brass cube =0.02*0.02=0.0004m2
F=1600N, Y=91000000000pa From Y=Fl/Ae,e=Fl/YA
e=1600*0.02/0.0004*91000000000
e=32/36400000
e=0.000000879m
To find the compression of the brass cube, we can use Hooke's Law, which states that the deformation of an object is directly proportional to the force applied to it.
Hooke's Law can be written as:
F = -kx
Where:
F is the force applied (in this case, 1600N),
k is the spring constant or stiffness of the material (related to its Young's modulus),
x is the deformation or compression.
First, let's find the spring constant (k) using the Young's modulus (Y) provided for brass (9.1x10^10 Pa):
Y = k/A
Where:
Y is the Young's modulus,
k is the spring constant,
A is the cross-sectional area of the material.
In this case, the brass cube has a square cross-section, so the area (A) is obtained by squaring the edge length (a):
A = a^2 = (2cm)^2 = 4cm^2
Converting to meters (since the derived SI units will be used for consistency):
A = 4cm^2 * (1m/100cm)^2 = 0.0004 m^2
Now, let's substitute the known values into the equation to find the spring constant:
9.1x10^10 Pa = k / 0.0004 m^2
k = (9.1x10^10 Pa) * (0.0004 m^2) = 3.64x10^7 N/m
Now that we have the spring constant (k), we can find the compression (x) using Hooke's Law:
1600N = (-3.64x10^7 N/m) * x
Solving for x:
x = 1600N / (-3.64x10^7 N/m)
x ≈ -4.40x10^(-5) m
The negative sign indicates compression. Therefore, the brass cube is compressed by approximately 4.40x10^(-5) meters.