The pole in Figure 5.51 is at a 90° bend in a power line and is therefore subjec

Question

The pole in Figure 5.51 is at a 90° bend in a power line and is therefore subjected to more shear force than ones in straight parts of the line. The tension in each line is 5.00 104 N, at the angles shown. The pole is 15.0 m tall, has an 23.0 cm diameter, and can be considered to have half the strength of hardwood. Figure 5.51

(a) Calculate the compression of the pole. .837 Correct: Your answer is correct. mm

(b) Find how much it bends and in what direction. Magnitude .888 Incorrect: Your answer is incorrect. mmDirection to the right to the left Correct: Your answer is correct.

(c) Find the tension in a guy wire used to keep the pole straight if it is attached to the top of the pole at an angle of 30.0° with the vertical. (Clearly, the guy wire must be in the opposite direction of the bend.)

Explanation / Answer

a) in vertical direction

F net = 2*(5*10^4 )*(cos 80) = 17364.817 N

diameter =23 cm =0.23 m , cross section area=A = 0.0415 m^2

stress = F/A =417782.158

stress /strain = 12*10^9 for hardwood elastic modulus

strain = 3.48*10^-5 = compression/15

compression = 0.5222 mm , please check modulus for harwood remaining process is everything same

b)pole will bend exactly in middle of two wires and in their direction

let pole bend and line connecting tip of pole and bottom makes angel theta with vertical

(2*5*10^4 * cos10 *sin45 ) / strain = 5*10^9

strain = 1.39*10^-5 = [ 1- cos theta ] / [costheta]

theta = 0.302 degrees towards in the diection as mentioned

15*tan theta = distance bent = 0.079

c) T sin 30 = 2*5*10^4 * sin45 *cos10

T = 139272.848 N

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