1. A uniform 9.5m long ladder of mass 18.0kg leans against a smooth wall (so the

Question

1. A uniform 9.5m long ladder of mass 18.0kg leans against a smooth wall (so the force exerted by the wall is perpendicular to the wall). The ladder makes a 20degree angle with the vertical wall and the ground is rough.
a) Calculate the normal force and the force of friction that must be exerted by the ground in order to prevent the ladder from moving
b) Determine the coefficient of friction at the base of the ladder if the ladder is not to slip when a 75kg person stands 3/4ths of the way up the ladder.

Attempted solution: A diagram (assuming the wall is to the left). The forces go from the left hand side of the diagram to right.

Fw to the right

Fg downwards

Fn upwards to the left at an angle

Fny (components of Fn) upwards

Fnx to the left

Ff to the right... but where would it be? would be be directly below the Fw, the 'x' distance away from where the ladder touches the ground? Or the same place as Fnx? Or does Fnx provide the frictional force?

The 'x' distance from the point where the ladder touches the ground to the wall, 3.25m

The 'y' distance of the ladder= 8.93m

Vertical Forces Fny= mg Fny= 176N

Horizontal Forces Fw= Ff (is this right????)

Torque: cw=ccw

mg(1.63) + Ff(8.93)= Fny(3.25)--> are these right distances?

mg(1.63) + mg(3.25) = mg(8.93)

1.63 + (8.93)= 3.25

= .18

When I solve for Ff is it .18(Fny) or am I supposed to find the x component of Fn and then solve for Fn?

I would like to solve the first part correctly before I proceed with the next! Help will be GREATLY appreciated!!! I am so close!

Explanation / Answer

First off, i'm not sure how your diagrams are but your Fn's are a little weird. Your force should just be pointing up, with no x and y components. Your frictional force would be pointing towards the wall. This would make sense since the normal force on the wall would be pushing it away from the wall, that would mean something would have to be pushing it into the wall for equilibrium, hence the direction of the friction force. Both these forces are acting at the base of the ladder.

For your torque equation, it seems you are using the wrong distances. You need to pick a pivot point on the ladder and find how far each of the forces are away from the the pivot point you picked. From your work it seems you picked the point where the ladder meets the wall which works out. The formula you have is right but the distances are wrong.

We have three forces putting torque on the ladder (other than the normal force of the wall which the torque would be zero anyway since that's our pivot point.) We have the normal force on the floor, friction between the floor and ladder, and the weight of the ladder. Viewing the normal force of the floor we see that that force is acting on the complete opposite end of the ladder, this means it is 9.5 meters away (the length of the ladder). Same with he frictional force. We know the weight is acting on the center of gravity of the object so this tells us the force is enacting on the middle of the ladder (since the question states it's uniform.) This gives us half the distance of the ladder (4.75m). This gives us all of our distances that we need but we are still forgetting one thing. Let's look at the equation for torque.

=r*F*cos()

This means the torque is the perpendicular force relative to the object. Looking at the diagram, we can see that in fact none of these forces are perpendicular to the ladder itself. Doing some geometry we can find the angle needed to find the perpendicular magnitude for each force. (all of these forces have to be perpendicular to the ladder.

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