In order to solve for the tension in the rope and the horizontal and vertical co

Question

In order to solve for the tension in the rope and the horizontal and vertical components acting at the hinge C, you need to break each of the vectors into their vertical and horizontal components. The sum of the forces in the horizontal direction and the sum of the forces in the vertical direction must equal zero. In other words, up has to equal down and left has to equal right. For the following diagram

Part A
Given that A = 60 degrees and B = 35 degrees and P = 600 newtons. The distance d1 = 1.2 meters and d2 = 2.6 meters.
Find the vertical component of P.

Explanation / Answer

let the assumption.... First thing first, that is Geometry. We need to know the angle the cable makes with the vertical (and the bridge), in order to find its magnitude and components. Let's designate some points: A is where the cable attaches to the bridge, O the hinge of the bridge, B where the cable attaches to the castle wall. We have, in the triangle AOB, OB = 12 m, OA = 5 m. Now, OB is vertical and OA is 20 deg below horizontal, that makes angle AOB 110 deg, hence, angles OAB + OBA (call them A and B) is supplement to 110 deg: A + B = 70 or A = 70 - B Also, applying Sine Rule of Triangles: sinA / 12 = sinB / 5 12 sinB = 5 sin(70 - B) = 5 sin 70 cosB - 5 cos70 sinB tanB = ... Solving, B = ... deg, A = ... deg Now that Geometry is out of the way, let's delve into Statics. Moment of a force about a point is the cross (outer) product of the force and radius vector (line from point of moment to point of action of force). If we take moment about the hinge, the moment of the cable tension must balance the moments of the weights of the bridge and the knight and stead. Also if the angle between the bridge and horizontal is 20 deg, then angle between bridge and vertical (direction of weight) must be 70 deg. So here we go T x 5 sin 51.1 = 2000 g x 4 sin 70 + 1200 g x 7 sin 70 where T is cable tension Solving T = ... N Only horizontal force causing a reaction at the hinge is the horizontal component of the tension (the weights are vertical) H = T sin B = ... N Direction: opposite to horizontal component of T. Vertical component of tension = T cos A = ... N That is not the only vertical force on the bridge, there are the weights to consider. So V + T cos B - 2000 g - 1200 g = 0 V = ... N Direction: figure out from sin of V, positive upwards.. A = 51.1, B = 28.9, T = 38852 N, H = 12585 N, compression, V = -5365 N, downwards.

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---

---

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