Normal force acting on a ladder lying against a wall

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SUMMARY

The discussion centers on the concept of normal force as it applies to a ladder leaning against a wall. Participants clarify that the normal force exerted by the wall on the ladder is always perpendicular to the wall, regardless of the presence of friction. They emphasize that while the normal force on the ground is also perpendicular, the total force may appear slanted due to the vector sum of forces, including friction. The conversation highlights the importance of accurately representing forces in free body diagrams, particularly in physics problems involving static equilibrium.

PREREQUISITES
  • Understanding of normal force and its definition in physics
  • Knowledge of free body diagrams and their components
  • Familiarity with static equilibrium concepts
  • Basic principles of friction and its role in force interactions
NEXT STEPS
  • Study the principles of static equilibrium in physics problems
  • Learn how to construct accurate free body diagrams for various scenarios
  • Explore the effects of friction on normal forces in inclined planes
  • Investigate the implications of vector components in force analysis
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Students of physics, educators teaching mechanics, and anyone interested in understanding the dynamics of forces acting on objects in static equilibrium.

starwars32
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Hi.. I am just confused normal forces. The definition of normal force is always perpendicular to the surface. But when we draw a free body diagram for a ladder lying against the wall. The normal force is like slanted towards the wall. Please explain thanks.
 
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starwars32 said:
But when we draw a free body diagram for a ladder lying against the wall. The normal force is like slanted towards the wall.
No, the normal force between the wall and ladder is always perpendicular to the wall. By definition! But if there's friction on that wall, the wall may also exert a vertical force on the ladder.
 
Hi there. Here is my two diagrams. The correct answer is the first one. Why can't it be the 2nd one? the normal force acting on the ladder on the ground,shouldnt it be perpendicular to the ground ( normal force definition)? for the normal force on the wall due to the wall also shouldn't it be perpendicular too?Thanks
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starwars32 said:
The correct answer is the first one.
Who says?

I'd say that the second diagram is the correct one. (Assuming no friction between ladder and wall.) The first diagram doesn't make sense.
 
The qn given was that it has friction between the ladder and the ground and also between the ladder and the wall cos to prevent the ladder from slipping. So if we taken into account of friction then the normal force will not be perpendicular? y is it so?
 
starwars32 said:
The qn given was that it has friction between the ladder and the ground and also between the ladder and the wall cos to prevent the ladder from slipping.
Are you sure? You don't need friction between ladder and wall to prevent slipping, only between ladder and floor.

Why don't you post the exact problem that those diagrams are supposed to describe.
So if we taken into account of friction then the normal force will not be perpendicular? y is it so?
No. The normal force is always perpendicular to the wall.

(If I were drawing a diagram, I would always show the forces at wall and floor in terms of vertical and horizontal components. Much cleaner that way.)
 
QN: A ladder of weight W rests against a vertical wall. Friction between the ladder and the ground and also between the ladder and the wall prevents the ladder from slipping. Which diagram shows the directions of the forces on the ladder? The answer given was C. So the forces that is shown is not the normal forces? both on the ground and wall?
Physics.png
 
anyway if we like draw the net force, wouldn't we be drawing the net force instead of the forces that is acting on it? for free body diagrams we are always drawing the indidivual forces? am i right?
 
starwars32 said:
So the forces that is shown is not the normal forces? both on the ground and wall?
Correct. They are showing the total force of wall on ladder and floor on ladder, not just the normal force.

Do this. Show the components (normal force and friction), then add the components to see how the total force points at some angle.
 
  • #10
Ok thanks. Here is another qn on free body diagram. A crane is made from a uniform rigid boom of length 30m and weight 400N, as shown. The boom is supported at its lower end by a frictionless hinge on the ground. ITs upper end is attached to a support case that is fastened on the ground behind the boom. The support cable has variable length, hence allowing the boom to elevate the load. In a particular application, the crane is used to support a crate of weight 2.0 kN. The boom makes and angle of 45 degrees with the horizontal and the support cable makes an angle of 30 degrees with the horizontal. Complete a free body digram of the boom showing all forces acting on it. Label each force clearly. My qn: I thought they are merely asking for the forces not NET force. how come the reaction force on the floor is the first one and not the 2nd one? thanks
Physics 2.png


Physics 3.png
 
  • #11
starwars32 said:
anyway if we like draw the net force, wouldn't we be drawing the net force instead of the forces that is acting on it? for free body diagrams we are always drawing the indidivual forces? am i right?
Yes. In a free body diagram, one shows all the forces acting on the ladder: gravity, force of wall, force of floor.

But you can represent the force of wall on ladder by its components (normal force and friction) or by a single force vector that includes everything (like they do in this example). If you were actually solving a numerical problem, most likely using the components would be more useful. In this particular problem, they just want to test if you know which way the forces are acting.
 
  • #12
starwars32 said:
My qn: I thought they are merely asking for the forces not NET force.
They are just asking for the forces acting on the boom, not the components of those forces.
how come the reaction force on the floor is the first one and not the 2nd one?
The floor is attached to the boom. Why would you assume that it can only exert a vertical force? (There'd better be a horizontal component, otherwise what will balance the horizontal component of the cable tension?)
 
  • #13
ok thanks! appreciate it!
 
  • #14
woahhh thanks for this! i was getting stuck in my own physics and found this online :) anw just to clarify, in real life there isn't such thing as a frictionless wall so the correct resultant force on the ladder from the wall is never perpendicular to the wall right?
 

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