Consider a pair of ladders leaning against each other making a 90º angle at the

Question

Consider a pair of ladders leaning against each other making a 90º angle at the top. The ladders are 3 m and 4 m long and their masses are 360 N and 480 N respectively. Assume that the floor is frictionless and that there is a thin horizontal bar of negligible mass connecting the ladders 0.8 m from the floor. Assume all contacts are made at points.

a) Determine the normal forces at the bases of the ladders, points A and B.

b) Determine the force along the thin connecting bar. Is this bar under compression or under tension?

c) Determine the force that each ladder exerts on the other at the top joint.

d) Repeat assuming that a 70-kg person is standing 3/4 up the smaller ladder.

Explanation / Answer

a)

Determine the torques about point A (or about B). Only the weight and the normal force at the opposite end produce torques about this point:

360(0.9) +480(3.4) =NB(5) à NB=391 N and NA=449 N.

b)

The torques on each ladder add up to zero. So place the axis at the contact point at the top and consider the torques on one ladder only.

For the smaller ladder: 369(0.9) +T(1.6) =449(1.8)à T=303 N.

You can conclude that the connecting bar is pulling on the ladder so the bar is under tension.

c)

Now look at the forces on one of the ladders (the other ladder will give the same answers). The forces on the top should balance all the other forces on that ladder.

On the larger ladder: Ftop=303i +89j.

On the smaller ladder the forces are equal and opposite, of course.

d)

The method is the same as before except you need to include an extra torque and downward force on the smaller ladder.

Answers here turn out to be:

NB=580 N; NA=960 N; T=520 N.

On the larger ladder: Ftop=520i +100(-j).

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