2) Determine the tension developed in wire CB. 3) Determine the tension develope

Question

2)Determine the tension developed in wireCB.

3)Determine the tension developed in wireBA.

Explanation / Answer

Let tension in AB = x lb, in BC = y lb and in CD = z lb. For equilibrium of F and E in the horizontal direction, xcos45° = ycos? = zcos30° => x/v2 = (v3/2) z ... ( 1 ) For equilibrium of F and E in the vertical direction, 110 - xsin45° = ysin? = zsin30° - 30 => x/v2 = 140 - z/2 ... ( 2 ) From (1) and (2), (v3/2) z + z/2 = 140 => z = 280/(v3 + 1) = 140(v3 - 1) = 102.487 N Plugging in (1), x = v(3/2) z = v(3/2) * (102.487) = 125.52 N Next, squarring and adding, ycos? = zcos30° and ysin? = zsin30° - 30 => y^2 = z^2 - 30z + 900 => y = v[(102.487)^2 - (30)*(102.487) + 900] = 91.26 N Also, ycos? = zcos30° => cos? = (z/y)cos30° = [(102.487)/(91.26)] cos30° = 0.972 => ? = 13.59 degrees

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.