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3 years ago

Determine the maximum mass of the crate so

Engineering

1 answer:

Tpy6a [65]3 years ago

7 0

Answer:

293 kg

Explanation:

Let's say the tension in each cable is Tb, Tc, and Td.

First, find the length of cable AD:

r = √(2² + 2² + 1²)

r = 3

Using similar triangles:

Tdx = 2/3 Td

Tdy = 2/3 Td

Tdz = 1/3 Td

Sum of the forces in the x direction:

∑F = ma

Tb − 2/3 Td = 0

Td = 3/2 Tb

Sum of the forces in the y direction:

∑F = ma

2/3 Td − Tc = 0

Td = 3/2 Tc

Sum of the forces in the z direction:

∑F = ma

1/3 Td − mg = 0

Td = 3mg

From the first two equations, we know Td is greater than Tb or Tc. So we need to set Td to 8.6 kN, or 8600 N.

8600 N = 3mg

m = 8600 N / (3 × 9.8 m/s²)

m ≈ 292.5 kg

Rounded to three significant figures, the maximum mass of the crate is 293 kg.

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A solid shaft and a hollow shaft of the same material have same length and outer radius R. The inner radius of the hollow shaft

alexandr402 [8]

Answer with Explanation:

By the equation or Torque we have

$\frac{T}{I_{p}}=\frac{\tau }{r}=\frac{G\theta }{L}$

where

T is the torque applied on the shaft

$I_{p}$ is the polar moment of inertia of the shaft

$\tau$ is the shear stress developed at a distance 'r' from the center of the shaft

$\theta$ is the angle of twist of the shaft

'G' is the modulus of rigidity of the shaft

We know that for solid shaft $I_{p}=\frac{\pi R^4}{2}$

For a hollow shaft $I_{p}=\frac{\pi (R_o^4-R_i^4)}{2}$

Since the two shafts are subjected to same torque from the relation of Torque we have

1) For solid shaft

$\frac{2T}{\pi R^4}\times r=\tau _{solid}$

2) For hollow shaft we have

$\tau _{hollow}=\frac{2T}{\pi (R^4-0.7R^4)}\times r=\frac{2T}{\pi 0.76R^4}$

Comparing the above 2 relations we see

$\frac{\tau _{solid}}{\tau _{hollow}}=0.76$

Similarly for angle of twist we can see

$\frac{\theta _{solid}}{\theta _{hollow}}=\frac{\frac{LT}{I_{solid}}}{\frac{LT}{I_{hollow}}}=\frac{I_{hollow}}{I_{solid}}=1.316$

Part b)

Strength of solid shaft = $\tau _{max}=\frac{T\times R}{I_{solid}}$

Weight of solid shaft = $\rho \times \pi R^2\times L$

Strength per unit weight of solid shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{solid}}\times \frac{1}{\rho \times \pi R^2\times L}=\frac{2T}{\rho \pi ^2R^5L}$

Strength of hollow shaft = $\tau '_{max}=\frac{T\times R}{I_{hollow}}$

Weight of hollow shaft = $\rho \times \pi (R^2-0.7R^2)\times L$

Strength per unit weight of hollow shaft = $\frac{\tau _{max}}{W}=\frac{T\times R}{I_{hollow}}\times \frac{1}{\rho \times \pi (R^2-0.7^2)\times L}=\frac{5.16T}{\rho \pi ^2R^5L}$

Thus $\frac{Strength/Weight _{hollow}}{Strength/Weight _{Solid}}=5.16$

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