# Why is the normal contact force horizontal on an inclined ladder?

Think about how a ladder stands up in real life. Would the ladder stay in the orientation shown in your image if the ground were ice? No! The reason? Friction.

The friction force, represented by $$\vec{F}_{ff}$$ in the figure acts to prevent the ladder from sliding to the right.

There are actually 5 forces acting on this ladder:

• $$\vec{F}_g$$: the gravitational force (aka the "weight" force), which pushes the ladder toward the ground
• $$\vec{F}_w$$: The normal force of the wall on the ladder, which prevents the ladder from falling into the wall.
• $$\vec{F}_{fw}$$: The friction force of the wall on the ladder, which prevents the ladder from sliding down the wall
• $$\vec{F}_f$$: The normal force of the floor on the ladder, which prevents the ladder from falling through the ground.
• $$\vec{F}_{ff}$$: The friction force of the floor on the ladder, which prevents the ladder from sliding to the right.

There must be a horizontal force acting on the wall to exert a horizontal force on the ladder. What causes the horizontal force on the wall and what is it called?

Actually, there must be a horizontal force acting somewhere on the ladder to require an equal and opposite normal reaction force on the wall for equilibrium. That horizontal force acting on the ladder is the friction force at the base of the ladder. So what your question really boils down to is, why is there a friction force at the base of the ladder? @Bunji has given you an intuitive explanation. The following is in terms of the gravitational force acting on the ladder.

To answer that question note that any force can be resolved into mutually perpendicular components. Therefore $$F_g$$ can be resolved into two components, one acting down and parallel to the ladder, $$F_{g}sinα$$, and one perpendicular to the ladder, $$F_{g}cosα$$. At the base of the ladder, this force down and parallel to the ladder has a vertical downward component acting on the ground and a horizontal component acting on the ground to the right. Per Newton's third law these forces have equal and opposite reaction forces as shown in the free body diagram of the ladder at the base. One of those is the horizontal friction force acting to the left. For equilibrium you then need a horizontal reaction force on the wall for the sum of the horizontal forces on the ladder to be zero.

All of the above is intended only to explain the reason for a normal reaction force at the wall. Given that you now have 4 unknown reaction forces and one known force, $$F_g$$. So solve for the 4 unknowns, you need 4 equations. From here you should be able to identify the needed equations if you realize that the sum of the moments where the ladder contacts the ground and the wall have to be zero for equilibrium.

Hope this helps.

I feel like there is something missing in this diagram, which is torque. In reality, there is a torque on the ladder, due to gravity, which causes it to want to rotate counterclockwise around the point where it touches the floor. This torque is "responsible" in some sense for the force of the top of the ladder against the wall (and the counterbalancing force of the foot of the ladder against the floor's friction.)

I don't see any torques in your free body diagram, although I do see an angle "alpha" at the base of the ladder, which is suggestive that maybe there should be some. If you haven't covered torque yet, this is not a great problem to try to work through.