Leaning Ladder

[rowjimmy]

Can someone just explain to me why the weight of a ladder leaning against a wall doesn't have to be broken down into x and y components.

I would've thought that as the angle between the ladder and the floor decreases the horizontal component of the weight would increase and thus be mgcos(theta). The y component of the weight would increase with an increasing angle between the ladder and the floor and would be mgsin(theta). The force pushing of the wall would be the normal force of the wall equal to the horizontal component of the ladders weight (mgcos(theta)) in the opposing direction of the ladders x weight. But this doesn't really make sense because it would imply that you would never need friction on the floor to support a leaning object because the "x component of the weight" would balance with the normal force on the wall.

I guess what confuses me is that the weight of an object that has an angle to the horizontal doesnt get it's weight broken down into component vectors whearas objects on surfaces that have some angle to the horizontal (i.e. inclined plane) do get their weight broken down into component vectors. What am I not seeing right? Thanks!

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[MedPR]

Can someone just explain to me why the weight of a ladder leaning against a wall doesn't have to be broken down into x and y components.

I would've thought that as the angle between the ladder and the floor decreases the horizontal component of the weight would increase and thus be mgcos(theta). The y component of the weight would increase with an increasing angle between the ladder and the floor and would be mgsin(theta). The force pushing of the wall would be the normal force of the wall equal to the horizontal component of the ladders weight (mgcos(theta)) in the opposing direction of the ladders x weight. But this doesn't really make sense because it would imply that you would never need friction on the floor to support a leaning object because the "x component of the weight" would balance with the normal force on the wall.

I guess what confuses me is that the weight of an object that has an angle to the horizontal doesnt get it's weight broken down into component vectors whearas objects on surfaces that have some angle to the horizontal (i.e. inclined plane) do get their weight broken down into component vectors. What am I not seeing right? Thanks!

Is there a specific problem you are confused about? Out of context it's hard to answer.

Edit: As far as friction goes, if you placed the ladder on a frictionless ground surface (x plane) and leaned it against a wall, the friction on the wall would have to be twice as much compared to if there was friction on the ground and the wall. The ladder weighs the same no matter what angle you lean it at. In a realistic situation, leaning ladders don't slide down the wall because of friction in the ground and friction on the wall. The ground friction opposes sliding away from the wall, and the friction on the wall opposes the weight of the ladder which is trying to push the ladder down the wall.

 

[milski]

Short answer: because the weight does not have a horizontal component - it's always pointed down, towards the center of Earth. Try to sketch a free body diagram. I don't have anything handy here but I'll try to put one sometime later tonight and it should become a bit clearer if it is not already.

 

[MedPR]

Short answer: because the weight does not have a horizontal component - it's always pointed down, towards the center of Earth. Try to sketch a free body diagram. I don't have anything handy here but I'll try to put one sometime later tonight and it should become a bit clearer if it is not already.

Based on the fact that he mentioned incline planes, I think he is confused about when/why gravity (weight) has an x component. I don't really know how to explain the convention of x and y components of gravity in incline planes without causing more confusion, so I didn't want to touch on that.

 

[milski]

Here is a pretty good FBD of the situation. The weight does not have a horizontal component, and thus does not need to be treated as two separate forces. Changing the angle of the ladder does not change the weight that acts on it.

Here is a reasonable good explanation of the situation: http://www.hep.vanderbilt.edu/~maguirc/Physics116SP08/Chapter_11/lecture15LadderInEquilibrium.pdf

You have three major forces action on the ladder: weight, straight down, does not change with the angle of the ladder, normal from the wall, perpendicular to the the wall, changes as the angle of the ladder changes and a force from the ground, which is up and at an angle towards the wall (thus not really a normal).

 

[Borealis]

If you're still unsure about the component part, let's say you have a block on an angled plane. If you change your x and y so that the angle of the plane is your x, THEN you'd have to take components of the mg. but if your x and y ref aren't changing, m always points directly down.

 

[rowjimmy]

When you change the directional orientation you then breakdown weight--makes sense, thanks.