Answer:
See attachment for complete answer step by step solving
Explanation:
Given that:
The anchor shown is used to tie tower guy cables to the ground and is supported by a distributed force from the soil, which can be approximated as shown. The anchor also is subjected to the loads shown, where P1 = 5 kNand P2 = 3 kN . The anchor is made from aluminum with E = 69 GPa . It is a cylindrical rod with diameter d1 = 4 cm with three segments of length L = 2 m . One end of the rod is linearly tapered to a diameter of d2 = 2 cm.
a. Calculate the intensity of the reaction load, Po
b. Use the reaction found in Part A, po = 13 kN/m , and calculate the total change in length for the anchor.
Answer:
Detailed solution is given in attached image
A2A.
Let the magnitude of the two forces be x and y.
Let resultant at right angle be p(= √15N) and at 60 degrees be q(= √18N.
Now, p = √(x2 + y2) = √15,
q = √(x2 + y2 +2xycos60) = √18.
So x2 + y2 = 15,
x2 + y2 + xy = 18,
therefore xy = 3 or y = 3/x.
Now, x2 + (3/x)2 = 15,
x4 + 9 = 15x2,
x4 - 15x2 + 9 = 0.
Now find the roots of this quadratic equation, the two values of x will correspond to the magnitudes of the two vectors.
Answer:
2000-7000 * 6% = - 0,000126
Explanation:
inside diameter of the hose