The chain AB is holding an access door partially open as shown in Fig. 1 below.

Question

The chain AB is holding an access door partially open as shown in Fig. 1 below. The tension in the chain AB is 150 N For the door shown in Fig.1, determine the component of the chain tension vector perpendicular to the door. Also, find the component parallel to the door. The chain AB is holding an access door partially open as shown in Fig. 1 below. The tension in the chain AB is 150 N For the door shown in Fig.1, determine the component of the chain tension vector perpendicular to the door. Also, find the component parallel to the door. 900 900 mm A D 1200 mm 30 Fig. 1

Explanation / Answer

We Can Calculate distance BD
Let the point of contact of door with wall be O.
As the door is elevated by 30o ,the angle Between Door and wall = 90 -30 = 600
Which makes BOD triangle an equilateral triangle.

BD = BO = OD = 900mm

Given, AD = 1200 mm

AB = sqrt(AD^2 +BD^2)
AB = sqrt(1200^2 + 900^2) mm
AB = 1500 mm

Sin(theta) = BD/AB
Sin(theta) = 900/1500 = 9/15 = 3/5

Cos(theta) = AD/AB
Cos(theta) = 1200/1500 = 12/15 = 4/5

Component of chain tension vector perpendicular to the door = AB sin(theta) = 150 * 3/5 = 90N
Component of chain tension vector parallel to the door = AB cos(theta) = 150 * 4/5 = 120N

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

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