I need help on this question:

A uniform ladder AB, of mass m and length 2a, has one end 1 on rough horizontal ground. The coefficient of friction between the ladder and the ground is 0.5. The other end B of the ladder rests against a smooth vertical wall. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, and makes an angle of 30° with the wall. A man of mass 5m stands on the ladder which remains in equilibrium. The ladder is modelled as a uniform rod and the man as a particle. The greatest possible distance of the man from A is ka.

Find the value of k.

So far I've drawn it out.

A uniform ladder AB, of mass m and length 2a, has one end 1 on rough horizontal ground. The coefficient of friction between the ladder and the ground is 0.5. The other end B of the ladder rests against a smooth vertical wall. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, and makes an angle of 30° with the wall. A man of mass 5m stands on the ladder which remains in equilibrium. The ladder is modelled as a uniform rod and the man as a particle. The greatest possible distance of the man from A is ka.

Find the value of k.

So far I've drawn it out.

(edited 2 months ago)

Original post by Jopping127

I need help on this question:

A uniform ladder AB, of mass m and length 2a, has one end 1 on rough horizontal ground. The coefficient of friction between the ladder and the ground is 0.5. The other end B of the ladder rests against a smooth vertical wall. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, and makes an angle of 30° with the wall. A man of mass 5m stands on the ladder which remains in equilibrium. The ladder is modelled as a uniform rod and the man as a particle. The greatest possible distance of the man from A is ka.

Find the value of k.

So far I've drawn it out.

A uniform ladder AB, of mass m and length 2a, has one end 1 on rough horizontal ground. The coefficient of friction between the ladder and the ground is 0.5. The other end B of the ladder rests against a smooth vertical wall. The ladder rests in equilibrium in a vertical plane perpendicular to the wall, and makes an angle of 30° with the wall. A man of mass 5m stands on the ladder which remains in equilibrium. The ladder is modelled as a uniform rod and the man as a particle. The greatest possible distance of the man from A is ka.

Find the value of k.

So far I've drawn it out.

The usual thing is to resolve forces vertically and horizontally and take moments about a point. Without working it through Id expect it to work here, so what are you stuck with?

(edited 2 months ago)

Original post by mqb2766

The usual thing is to resolve forces vertically and horizontally and take moments about a point. Without working it through Id expect it to work here, so what are you stuck with??

I've done this working out but idk I feel like everything is wrong

Original post by Jopping127

I've done this working out but idk I feel like everything is wrong

A fair bit

•

Weight (ladder and man) acts vertically so only affects the resolving vertically equation.

•

Resolving horizontally should just involve the wall reaction and friction

•

Id probably have taken moments about A, as its a bit simpler

(edited 2 months ago)

Original post by mqb2766

A fair bit

A fair bit

•

Weight (ladder and man) acts vertically so only affects the resolving vertically equation.

•

Resolving horizontally should just involve the wall reaction and friction

•

Id probably have taken moments about A, as its a bit simpler

at

Original post by mqb2766

A fair bit

•

Weight (ladder and man) acts vertically so only affects the resolving vertically equation.

•

Resolving horizontally should just involve the wall reaction and friction

•

Id probably have taken moments about A, as its a bit simpler

i've done this now idk if its the right way to solve it tho

Original post by Jopping127

i've done this now idk if its the right way to solve it tho

Getting there, though not sure where the 24 comes from. Note k must be < 2.

Id also get in the habit of resolving horizontally (even if its simple as here) and explicitly saying why limiting friction applies.

(edited 2 months ago)

Original post by mqb2766

Getting there, though not sure where the 24 comes from. Note k must be < 2.

Id also get in the habit of resolving horizontally (even if its simple as here) and explicitly saying why limiting friction applies.

Id also get in the habit of resolving horizontally (even if its simple as here) and explicitly saying why limiting friction applies.

ohh yh thank you, not sure why I did 24 it was meant to be 2, I changed that and now i got k as 1.19

Original post by Jopping127

ohh yh thank you, not sure why I did 24 it was meant to be 2, I changed that and now i got k as 1.19

Ive not worked it through but you should be able to check by subbing it into the moment equation you start with and verify it balances.

Edit - Im getting k a bit larger so post your working if necessary.

(edited 2 months ago)

Original post by mqb2766

Ive not worked it through but you should be able to check by subbing it into the moment equation you start with and verify it balances.

Edit - Im getting k a bit larger so post your working if necessary.

Edit - Im getting k a bit larger so post your working if necessary.

i did https://docs.google.com/drawings/d/e/2PACX-1vRv_XuJA17AKJ6OoWfed74ePgu88CoBvLorwFOYxfnyobRirpGXC3xHXiWrLSm1yp4t8akBV5AJvTRK/pub?w=960&h=720

(edited 2 months ago)

for some reason your 6 becomes 4?

Edit - note the previous comment about explicitly writing out the resolving horizontally. If you did this and put it in your moment equation at the start, the moment section would be much simpler as you could divide through by mga at the start and there would be less chance of an error. Also, rather than subbing both sin and cos, you could have divided through by sin or cos at the start and there would be a single tan or cot substitution which again means fewer errors

(edited 2 months ago)

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