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I got the following question.

a 12 m ladder, whose mass is 20 kg is leaning agains a wall with its base being 6 m from the wall and the angle between the floor and the ladder is 60 degrees. The wall and the floor are frictionless. So we tie a rope 0.5 m from the base of the ladder to the wall to keep the ladder from sliding. A 72 kg person climbs 3 quarters up the ladder and stps. We need to find the tension in the rope and the reactive force in the floor.

For the force on the floor it is easy, u just get the mass and multiply it by 9.i8 to get the total force up in order for thsi to be static equalibirum.

Now i was trying to find the tension in teh rope, so i did this:

i set the place where the ladder is tied to the rope to be the pivot and did this:

0=-(9-0.5)sin 30 * 72 *9.8 -5.5 sin30 * 20 * 9.8 + sin 30*11.5 Fx

So i solved for Fx adn i get:

600. The correct answer is 380. The reason i used 30 is becuase that is the angle between the force applied and the arm as opposed to use 60 which is the angle between the floor and the ladder.

any ideas where i made the mistake?

Thx

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# Homework Help: Tourque the ladder question

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