A uniform 250 N ladder rests against a perfectly smooth wall, making a 35 degree angle with the wall. (a) find the normal forces that the wall and the floor exert on the ladder. B) what is the friction force on the ladder at the floor?

Answer 1

Try to sketch a diagram as you read through this post. An image is worth a thousand words.

There are four forces acting on the ladder of length L and making θ=35 degrees with the vertical smooth wall and on a horizontal rough floor.

1. the weight W=250N acting downwards through the middle of the ladder
2. vertical reaction on the floor acting upwards, equal to W.

These two forces form a couple of magnitude
W(L/2)sinθ

3. Normal reaction N on the wall (horizontal) at the top end of the ladder, equal in magnitude to
4. Frictional force F acting horizontally at the bottom of the ladder.

These two forces form another couple equal to FLcosθ.

Since the ladder is in equilibrium, the two couples must be equal, thus

W(L/2)sinθ = FLcosθ
From which we can solve for F
= (W/2)tanθ
after cancelling L on each side and replacing sinθ/cosθ by tanθ.

Answer 2

so it would be 125 (tan 35) = 87.5 Awesome that is what it says in the back of the book THANK YOU!

Answer 3

Glad that it works out well!

Answer 4

In case you're still on, just want to let you know that I finish for the night. I can answer any question you may have tomorrow, unless another teacher is still on tonight.