1. Physics 23 Supplementary Problem
Winter/Spring 2003
Test-Level Problem on Statics (from a Test given in Winter 1996).
A stationary ladder of total length L=13 m and weight W=216 N leans against a vertical smooth wall, while its bottom legs rest on a rough horizontal surface at a distance d=5 m from the base of the wall. Its top is located a distance H=12 m above the floor.
A rope is attached to the ladder at a distance d=2m from its base. A worker pulls horizontally on the rope, producing a tension of T=260 N.

2. In the diagram to the right, sketch all the forces acting on the ladder, and the location where they act. N
Weight.
Weight. At L/2 up the ladder.
Tension, of course.
Normal force exerted by wall.
L/2 W
The floor clearly exerts an upward normal force, to “balance” W.
If TN then a friction force acts on the ladder due to the floor. We can’t tell the direction of this friction force from the data given. T P
The total floor force (call it P) is the sum of the floor normal and friction forces (assumed in + directions).

3. What are the vertical and horiz-ontal components of the force applied on the ladder by the floor if the ladder does not slip? W
L/2 T N P
First think. We have been studying torques. Do I need to use torques?
I will have two sum of force equations (x and y components). I have three unknowns, Px, Py, and N.
I need a third equation, so yes, I will have to use the sum of torques equation.

4. W
L/2 T N P
Think some more. Where is a good place to put the rotation axis?
Answer: at the point of application of one of the forces.
I think force equations are easier than torque equations (don’t need to do RF or RF or RF sin).
Therefore, I choose to put the origin at the floor contact point, where I have two unknown torques.
Therefore, I choose to put the origin at the floor contact point, where I have two unknown torques. I’ll pick clockwise for positive rotations. +

5. OSE: Fy=may W
L/2 T N P +
Ny + Wy + Ty + Py = may
-W + Py = 0
Py = +W
OSE: Fx=max
Nx + Wx + Tx + Px = max
-N + T + Px = 0
Px = N - T
Don’t know N, don’t know Px. Need another equation.

6. OSE: τz= Iz W
L/2 T N P +
τNz + τWz + τTz + τPz = Iz
-NH + (D/2)W + Td = 0
Dang! Forgot to show D/2 and d on the diagram! Better do it now!
D/2
Dang again! d is not a system parameter! Better figure it out now! d
There are several ways to do it. I see two similar triangles. Click twice to see them.
d/d = (H/2) / (L/2)  d = d H/L

7. Two equations, two unknowns, solve for Px.
Continuing… W
L/2 T N P +
D/2 d
-NH + (D/2)W + Td = 0
-NH + (D/2)W + T d H/L = 0
Two equations, two unknowns, solve for Px.
Px = N - T
-NH + (D/2)W + T d H/L = 0
NH = (D/2)W + T d H/L
N = DW/2H + T d /L
Px = DW/2H + T d / L - T
Numerically, Px = -175 N, so it is directed in the –x direction. You are invited to check my algebra.