Question

A 5.0 kg ladder 2.5m long is leaning against a wall at an angle of 40 degrees above the horizontal. How far up the ladder would a 1.0 kg mass be placed before the ladder slides down? the coefficient of friction between the ladder and the ground is 0.5 while between the ladder and the ground is 0.1

Answer 1

"the coefficient of friction between the ladder and the ground is 0.5 while between the ladder and the ground is 0.1" which is the wall and which is the ground?

Answer 2

Oups, the ground is 0.5 and the wall 0.1

Answer 3

Alright. What you have to do is use three equations. Set up the "x" components and "y" components of your FBD and then pick a point A here:
Take the moment about point A and set all our sum of forces equal to zero.

Also, at the beginning find the vertical distance and the horizontal distance of the ladder using the angle.

You should have three equations and three unknowns. From there you can solve for the N(wall), N(floor) and then finally "d" the length up the ladder of the box that the box can be before slipping.

Answer 4

I don't understand how you can add them up like that, I get the following :
\[(1/2) (2.5)mgcos40 + xm _{2}gcos40= (2.5)F _{wall}\cos 40\]
\[F_{fr w} + F_{N} =mg + mg_{2}\]
\[F_{w}=\mu F_{N}\]
What do I do from here exactly?

Answer 5

Let's start off with the "x" and "y" components. We will assign direction vectors for positive and negative directions as so:


from here I'm going to find all the Forces that Act in the "x" direction, could be either negative or positive depending on their direction.

\[\rightarrow \sum F _{x} = (F _{\mu _{s}})_{F} - (N)_{w} = 0\]\[\sum F _{y} = (N)_{F} - W _{L}-W_B + (F _{{\mu}_{s}})_{W} = 0\]

Where the subscript "F" and "W" represents the "floor" and "wall" respectively. The major hint between these two questions is the equation for Friction force which is:
\[F_{\mu} = \mu N\]