A weightless ladder....but why move the pivot point?

[mistermill]

A weightless ladder 7.0 m long rests against a frictionless wall at an angle of 65° ablove the horizontal. A 72 kg person is 1.2m from the top of the lass. What horizontal force at the bottom of the ladder is required to keep it from slipping?

Στ=0
τ=FsinθdSolutions for this question always have the pivot point at the bottom of the ladder, solving for the horizontal frictional force at the bottom of the ladder as being equal and opposite to the force at the top.Στ=0

τ person = mgsin25°(5.8)
τ friction at the top= Fsin65°(7.0)

F = 272N

My question is WHY doesn't the answer work if we place the pivot point at the top?


Στ=0

τ person = mgsin25°(1.2)
τ friction at the bottom = Fsin65°(7.0)

F = 56N

What is missing in this solution that would make it correct?

 

[TSny]

My question is WHY doesn't the answer work if we place the pivot point at the top?

Στ=0

τ person = mgsin25°(1.2)
τ friction at the bottom = Fsin65°(7.0)

F = 56N

What is missing in this solution that would make it correct?

Are you missing one of torques?

 

[Orodruin]

Note that there is no mention whatsoever of any friction in this problem. There is only a mention of an applied horizontal force at the bottom, which may or not be friction. What is certain is that the problem explicitly states that the contact with the wall is frictionless.

 

[mistermill]

Are you missing one of torques?

Not according to the answer keys I have come across. But what about the normal forces at the top and bottom of the ladder? they also don't show up in any of the answer keys. Perhaps that is the 'applied force' at the bottom?

 

[mistermill]

Note that there is no mention whatsoever of any friction in this problem. There is only a mention of an applied horizontal force at the bottom, which may or not be friction. What is certain is that the problem explicitly states that the contact with the wall is frictionless.

Right, so if the contact with the wall is frictionless, does that mean it can't be used as the pivot point?

 

[Orodruin]

Nope.

The question was clearly rhetorical to make you think about it. The correct answer is yes.

 

[mistermill]

The question was clearly rhetorical to make you think about it. The correct answer is yes.

So there is a torque missing when the pivot point is at the top, which accounts for the different answers.

 

[Orodruin]

Yes. Can you figure out what force it is?

Edit: Also note that the torques you have taken into account both go in the same direction ... This means that whatever force you are missing must give a torque in the opposite direction with the top of the ladder as pivot.

 

[mistermill]

Yes. Can you figure out what force it is?

Is it the Normal force acting perpendicular to the floor?

 

[Orodruin]

Is it the Normal force acting perpendicular to the floor?

You tell me. Does it give a torque relative to the top of the ladder?

 

[mistermill]

You tell me. Does it give a torque relative to the top of the ladder?

Well.

I drew my torque from the person straight down.

I drew the applied horizontal force at the bottom as straight out from the wall. So those two torques do act opposite one another.

When I also include a normal force at the bottom of the ladder, it has a torque opposite the torque from the person. Can one have a normal force for a weightless ladder?

 

[Orodruin]

I drew the applied horizontal force at the bottom as straight out from the wall. So those two torques do act opposite one another.

Are you sure? In which direction does the force from the wall go? Did you do your force equilibrium in the horizontal direction correctly? Also, if you think about experience in real life, will pulling the ladder away from the wall make it more or less stable?

Note that your drawing being in one way does not make it so, you need to do the math.

When I also include a normal force at the bottom of the ladder, it has a torque opposite the torque from the person. Can one have a normal force for a weightless ladder?

What does force equilibrium for the ladder in the vertical direction tell you? Normal forces have nothing to do with weight apart from very often being what balances the weight of an object or other forces acting on that object.

 

[mistermill]

Are you sure? In which direction does the force from the wall go? Did you do your force equilibrium in the horizontal direction correctly? Also, if you think about experience in real life, will pulling the ladder away from the wall make it more or less stable?

Note that your drawing being in one way does not make it so, you need to do the math.What does force equilibrium for the ladder in the vertical direction tell you? Normal forces have nothing to do with weight apart from very often being what balances the weight of an object or other forces acting on that object.

Gotcha.

Fapplied at the bottom goes toward the wall, causing a torque in the same direction as the torque from the person.

Now we have two unknowns! Because we don't know F normal. A good reason to move the pivot point.

 

[Orodruin]

Because we don't know F normal.

Don't you? What does force equilibrium for the ladder in the vertical tell you?