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Hi, can anyone figure this out, I can almost do it, but not quite. An answer was given to this by sumone here, but I think they showed the converse statement, that if the ladder remains at rest, then theta is greater than or equal to 45 degs, and I'm not sure how to turn it around.

A uniform ladder, of mass m, leans against a vertical wall with its base on horizontal ground. The length of the ladder is 6metres. Assume that the wall is smooth and that ground is rough, with the coefficient of friction between the ladder and the ground equal to 0.5.

a)If the angle between the ladder and the ground is Theta, show that the ladder remains at rest if Theta is greater than or equal to 45 degs.

Thanks for your help

A uniform ladder, of mass m, leans against a vertical wall with its base on horizontal ground. The length of the ladder is 6metres. Assume that the wall is smooth and that ground is rough, with the coefficient of friction between the ladder and the ground equal to 0.5.

a)If the angle between the ladder and the ground is Theta, show that the ladder remains at rest if Theta is greater than or equal to 45 degs.

Thanks for your help

cms271828

A uniform ladder, of mass m, leans against a vertical wall with its base on horizontal ground. The length of the ladder is 6metres. Assume that the wall is smooth and that ground is rough, with the coefficient of friction between the ladder and the ground equal to 0.5.

a)If the angle between the ladder and the ground is Theta, show that the ladder remains at rest if Theta is greater than or equal to 45 degs.

a)If the angle between the ladder and the ground is Theta, show that the ladder remains at rest if Theta is greater than or equal to 45 degs.

For ladder to remain at rest i.e.) Prevent slipping: F <= uR -> F <= 0.5R

For equilibrium:

Resolve Vert: R (Ground) = mg | RG - Notation Later On.

Resolve Horiz: Friction = R (Wall) | RW - Notation Later On.

For equilibrium:

Taking moments about base:

-> 3mgcos(theta) = 6.RW.sin(theta)

-> 6RW = 3mg/tan(theta)

-> RW = mg/[2tan(theta)]

To prevent slipping:

F <= 0.5RG

-> RW <= 0.5mg

-> mg/[2tan(theta)] <= 0.5mg

-> 1/[2tan(theta)] <= 1/2

-> 1/[tan(theta)] <= 1

-> tan(theta) >= 1/1

-> tan(theta) >= 1

-> theta >= 45 Deg

Hi, thanks for your efforts, you got the same answer I did before.

Maybe I'm just being petty/over-analytical, but the question says...

If theta>=45 ,show ladder remains at rest.

But you have shown the converse (if ladder remains at rest, then theta>=45).

From a pure maths point of view this is wrong,

basically (P=>Q) =/=> (Q=>P) but do u think its true in this type of question??

I'll give you a repping for your efforts anyway.

Thanks

PS. A tiny error at the end of yur answer, with the direction of inequality sign

Maybe I'm just being petty/over-analytical, but the question says...

If theta>=45 ,show ladder remains at rest.

But you have shown the converse (if ladder remains at rest, then theta>=45).

From a pure maths point of view this is wrong,

basically (P=>Q) =/=> (Q=>P) but do u think its true in this type of question??

I'll give you a repping for your efforts anyway.

Thanks

PS. A tiny error at the end of yur answer, with the direction of inequality sign

cms271828

Hi, thanks for your efforts, you got the same answer I did before.

Maybe I'm just being petty/over-analytical, but the question says...

If theta>=45 ,show ladder remains at rest.

But you have shown the converse (if ladder remains at rest, then theta>=45).

From a pure maths point of view this is wrong,

basically (P=>Q) =/=> (Q=>P) but do u think its true in this type of question??

I'll give you a repping for your efforts anyway.

Thanks

PS. A tiny error at the end of yur answer, with the direction of inequality sign

Maybe I'm just being petty/over-analytical, but the question says...

If theta>=45 ,show ladder remains at rest.

But you have shown the converse (if ladder remains at rest, then theta>=45).

From a pure maths point of view this is wrong,

basically (P=>Q) =/=> (Q=>P) but do u think its true in this type of question??

I'll give you a repping for your efforts anyway.

Thanks

PS. A tiny error at the end of yur answer, with the direction of inequality sign

The condition is that if theta is 45 Deg or greater, then the ladder is at rest. From the pure maths point of view: If the ladder is at rest, then theta must be 45 deg or greater or else this would be impossible as it would slip due to friction.

Error was noted, just a slight slip up with re-arrangement. Read again.

Nima

The condition is that if theta is 45 Deg or greater, then the ladder is at rest. From the pure maths point of view: If the ladder is at rest, then theta must be 45 deg or greater or else this would be impossible as it would slip due to friction.

Error was noted, just a slight slip up with re-arrangement. Read again.

Error was noted, just a slight slip up with re-arrangement. Read again.

So your saying in this case that..

Ladder remains at rest <==> theta>=45

I think thats true, I thought to much about this, and I'm in a daze and can't think straight, I'll give u another repping!!

I know your answers right from before, I was just saying u forgot to turn

your inequality around when you divided.

cms271828

So your saying in this case that..

Ladder remains at rest <==> theta>=45

I think thats true, I thought to much about this, and I'm in a daze and can't think straight, I'll give u another repping!!

Ladder remains at rest <==> theta>=45

I think thats true, I thought to much about this, and I'm in a daze and can't think straight, I'll give u another repping!!

Let's say that theta is less than 45 deg and the ladder is at rest.

If this was so:

mg/[2tan(theta)] >= 0.5mg -> F >= uR -> Slipping -> Contradiction. (We required state of rest)

Therefore when the ladder is at rest, theta >= 45 (QED by answering question) and when theta is >= 45 degrees, the ladder is at rest. (QED by my proof by contradiction above).

Yes?

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