There will be two (likely different) reaction forces
a vertical reaction force Rf between the bottom of the ladder and the floor
a perpendicular reaction between the ladder and the surface that it is leaning on
if it is a peg then the reaction force Rp will be perpendicular to the peg
if it is a vertical wall then the reaction force Rw will be horizontal
There could be frictional forces keeping the ladder in equilibrium
The frictional force on the floor Ff will act horizontally towards the wall
The frictional force on a vertical wall will act vertically upwards
The coefficients of friction between the ladder and the floor/wall might be different
How do I solve problems involving ladders?
Include the weight of the ladder
If it is modelled as being uniform then the weight will act at the midpoint of the ladder
Include any extra weights for any additional objects on the ladder such as a person standing on it
Form two equations using the horizontal forces and the vertical forces
If the coefficient of friction is involved then:
use F ≤ µR if the ladder is in equilibrium
if the ladder is said to be on the point of slipping it is in limiting equilibrium and so use F = µR
Form moment equations
Useful pivots include the points of contact between the ladder and the floor, wall or peg
You will have to use right-angled trigonometry to find the perpendicular distances
Solve the equations to find any unknown distances, angles, masses or forces
Worked Example
the magnitude of the normal reaction force at the wall,
the magnitude of the normal reaction force at the ground,
the magnitude of the frictional force at the ground.
Exam Tip
If the ladder is resting on the peg then it might be a good idea to resolve forces parallel and perpendicular to the ladder instead of horizontal and vertical. Your answer will be the same, it just depends on what you find easier.
Moments - Hinges
How do I deal with hinges?
If one end of a rod is hinged to a surface then that end is fixed to that point but the rod is free to rotate about that point
There will be a resultant force acting on the rod from the hinge
This force will not be perpendicular to the surface – it will be acting at an angle
Taking moments about the hinge is a good way to avoid including the force from the hinge
If you are asked to find the magnitude and direction of the force from the hinge:
Step 1: Find the horizontal and vertical components of the force
Step 2: Find the magnitude by using Pythagoras
Step 3: Find the direction by using right-angled trigonometry
There are two common ways a rod will be held horizontally at a wall:
The rod could be without a hinge so that there is contact with the wall which means there is a normal reaction force and a frictional force to stop the rod sliding down the wall
The rod could be with a hinge so that there is a resultant force from the hinge acting at an angle of α° to the horizontal which stops the rod rotating about the hinge
Splitting the resultant force from the hinge into horizontal and vertical components makes both scenarios mathematically similar
Worked Example
(a) Find the magnitude of the tension in the cable.
(b) Find the magnitude and direction of the resultant force exerted on the rod by the hinge at A.
Exam Tip
Try not to let hinges scare you! Just remember there will be vertical and horizontal components. If you draw them in the wrong direction (up instead of down or left instead of right) then don’t panic, it just means your answer for that direction will be a negative.