2) A stepladder consists of two halves that are hinged at the top and connected

Question

2) A stepladder consists of two halves that are hinged at the top and connected by a tie rod which keeps the two halves from spreading apart. In this particular instance, the two halves are 2.5 m long, the tie rod is connected to the center of each half and is 70 cm long. An 800-N person stands 3/5 of the way up the stepladder, as shown in the figure. The ladder is light enough that we can neglect its weight, and it rests on an extremely smooth floor. What is the tension in the tie rod? (Note: To solve this problem, it is helpful to imagine cutting the ladder in half vertically and consider the forces and torques acting on each half of the ladder.) =2.50 m 70.0cm

Explanation / Answer

Draw triangle ABC where A is the apex - the hinged point - and AB = AC = the legs of the ladder.

Let point D be on AB and let point E be on AC where DE is the tie rod.

Let the 800-N person be a point F, which is on AC.

Now for a bit of geometry.

Let the angle BAC = 2. Drop a perpendicular from A to DE, meeting DE in point G.

Consider the triangle AGE.

Angle GAE = , so sin() = GE / AE.

Since GE = half of DE, GE = 0.35 m. AE = 1.25 (AE = half of AC; AC = 2.5 m.).

So, sin() = GE / AE = 0.35 / 1.25 = 0.28 and cos() = 0.96.

The standard procedure in statics for dealing with freely hinged joints is to separate the joints and introduce a force at the hinge to represent the internal force that was at the joint prior to their separation. So, separating the joint at A, we need to introduce a force F acting away from the joint. (There will be a corresponding force F acting on the other half of the ladder at A to maintain equilibrium, but this does not concern us here).

Now draw another diagram and put in the forces that we now know:

We have ladder AC with force F acting horizontally at A (This force must be horizontal since the hinge is freely jointed). At point E we have a force T - the tension in the tie-bar acting horizontally - but in the opposite direction to that of force F. At point F we have a force of 800-N acting vertically downwards. And at point C there is the force R - the reaction of the ground on that half of the ladder - acting vertically upwards.

We know all of the required distances in that diagram:

FC = 1.5 m., EC = 1.25 m., EA = 1.25 m. and the half-angle between the two ladders = 2; we know sin() and cos(). {As calculated above}.

Resolve the horizontal and vertical forces.

We get:

F = T and W = R.

Consider the moments acting around point C

W x FC.sin() + T x EC.cos() = F x AC.cos().

So: 800 x 1.5 x 0.28 + T x 1.25 x 0.96 = F x 2.5 x 0.96

That reduces to:

336 + 1.2T = 2.4F.

But T = F,

so: 336 = 1.2T,

thus: T = 336 ÷ 1.2 = 280-N.

The tension in the tie-bar = 280/2 = 140N

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