Den Hartog’s Mechanics

A web-based solutions manual for statics and dynamics

Problem 33

Your instructors will likely tell you to always draw free-body diagrams, and they’ll be right. But I’m feeling a bit lazy tonight, and I think we can get through this one by just looking at the drawing in the book.

We’ll start by figuring out the tension in the left cable, T, necessary to balance the weight of the bridge deck. Isolating the bridge deck and taking the moments about its right hinge, we get this equilibrium equation:

\sum M_{H_1} = 4000 \cdot 8 - 16 T = 0

The solution is T = 2000 lbs. Now we isolate the horizontal bar at the top and take moments about its hinge:

\sum M_{H_2} = 2000 \cdot 16 - 3000 \cdot 10 - 10 P = 0

Solving this we get the necessary downward pull on the counter weight rope, P = 200 lbs, which is the answer to part a).

Part b) requires no calculations, just a bit of insight. As the bridge deck is tilted upward, the top horizontal bar will tilt up at the same angle, and the cable will remain vertical. (That is, the rectangle formed by the deck, the cable, the top bar, and the upright will turn into a parallelogram. This type of mechanism is called a four-bar linkage and we’ll be seeing it later in the book.) All of the forces will remain vertical and all of the moment arms will be reduced according to the cosine of the tilt angle. So the new equilibrium equations will be the same as the old equilibrium equations, but with a common term of cos 30° that cancels out. The upshot of all this is that the tension in the left cable will remain 2000 lbs and the downward pull on the counterweight rope will remain 200 lbs.

Problem 34Problem 32

Last modified: January 22, 2009 at 8:32 PM.