Ladder Problem

[nakb]

Not sure if this is the right thread, but I can re-post in another one if necessary.

Any thoughts on this problem - It is a practice question for Cambridge STEP but I don't have a solution to check my answer against:

Joe is raising a 10m ladder (AB=10m) which weighs 20kg to the vertical position. The foot of the ladder (A) rests against a fixed wall. Joe initially holds the ladder at its other end (B), having raised it to a height of 2m, and the point vertically below B (X1) is therefore sqrt(10^2 - 2^2)m from A (Pythagoras’ theorem).

Joe proceeds to raise the ladder by walking towards the wall, moving his hands along the ladder so that he continues to grip it at a constant height of 2m as it rises, until it reaches the vertical position (pivoting around Point A). Joe's starting position is therefore X1 and his finish position is A. See diagram below.

(a) Define the function f(x), to determine the total force in Newtons exerted on Joe by the ladder at any point Xn (where x is the distance in meters from A to Xn).

(b) Sketch the graph of the function f(x).

(c) Determine the maximum force in Newtons exerted on Joe by the ladder as he walks towards the wall, and the distance Xn from the wall at that point of maximum force.

You may assume that the ladder is fixed at Point A (i.e. it will not tend to slide up the wall once Joe moves beyond the center of gravity of the ladder and Joe does not need to exert a force to keep the foot of the ladder at Point A).


Thanks

Nigel

 

[HallsofIvy]

You are getting no response because you have not shown anything you have done on this problem. We cannot know what hints will help without know what you understand about this problem, what you have tried, and where you have difficulty.

(You say "check my answer" but you have shown no answer to check nor any attempt to give an answer.)

 

[nakb]

OK, here's how we approached it.

Firstly, just considering what is going on before starting on the maths:

1. At the start (Joe at X1, immediately blow B) the weight will be approx equally divided between Joe at X1 and the ground at A.

2. As the ladder becomes more vertical the weight distribution will change, with more on the ground until when Joe reaches A all the weight is on the ground.

3. As Joe's point of contact with the ladder moves from the end (B) at the start to a point 2 metres from A at the end the force on him by the ladder will tend to increase (principle of the lever).

4. (2) and (3) act in opposite directions so will tend to compensate for each other - although at this stage we can't intuitively tell what the net effect will be.

5. It is assumed the force exerted on Joe by the ladder is perpendicular to the ladder.

Turning to the maths:

a). The force by moments about A at any point X is: F(sqrt(x^2 + 4) - 20g(5) * (x / sqrt(x^2 + 4)

[where (x / sqrt(x^2 + 4) is cos(acute angle at A).

b). Rearranged gives: F(x^2) - 100gx + 4F = 0

c). Differentiated in terms of x gives: 2Fx - 100g = 0

d). So x = 50g/F

[substituting this back into (b) gives:

e). F((50g/F)^2) - 100g(50g/F) + 4F = 0

f). Rearranging gives: F = sqrt((5000(g^2) - (50g)^2) / 4) = 245

g). So x at point of maximum F is: 50g/245 = 2

So answers are:

(a) f = 100gx/((x^2) + 4)

(b) Graph not included here but would be graph of (a).

(c) Force is 245N and distance is 2m.

Are we right?

 

[nakb]

Bump.

Could anyone tell me if we're on the right lines with this one?

Many thanks !

 

[wjm11]

OK, here's how we approached it.

Firstly, just considering what is going on before starting on the maths:

1. At the start (Joe at X1, immediately blow B) the weight will be approx equally divided between Joe at X1 and the ground at A.

2. As the ladder becomes more vertical the weight distribution will change, with more on the ground until when Joe reaches A all the weight is on the ground.

3. As Joe's point of contact with the ladder moves from the end (B) at the start to a point 2 metres from A at the end the force on him by the ladder will tend to increase (principle of the lever).

4. (2) and (3) act in opposite directions so will tend to compensate for each other - although at this stage we can't intuitively tell what the net effect will be.

5. It is assumed the force exerted on Joe by the ladder is perpendicular to the ladder.

Turning to the maths:

a). The force by moments about A at any point X is: F(sqrt(x^2 + 4) - 20g(5) * (x / sqrt(x^2 + 4)

[where (x / sqrt(x^2 + 4) is cos(acute angle at A).

b). Rearranged gives: F(x^2) - 100gx + 4F = 0

c). Differentiated in terms of x gives: 2Fx - 100g = 0

d). So x = 50g/F

[substituting this back into (b) gives:

e). F((50g/F)^2) - 100g(50g/F) + 4F = 0

f). Rearranging gives: F = sqrt((5000(g^2) - (50g)^2) / 4) = 245

g). So x at point of maximum F is: 50g/245 = 2

So answers are:

(a) f = 100gx/((x^2) + 4)

(b) Graph not included here but would be graph of (a).

(c) Force is 245N and distance is 2m.

Are we right?

Hello, Nigel,

I see no one has gotten back to you, so even though I haven't solved yet, I'll make some comments. One reason you may not have gotten a more timely response is that you may have confused some people looking at your work in the very first line: "The force by moments about A at any point X is: F(sqrt(x^2 + 4) - 20g(5) * (x / sqrt(x^2 + 4)". If you count all your left and right parentheses, you'll find they are unbalanced -- a sure way to confuse people. Please edit those for clarity.

First, a quick check: length = 10m; mass = 20 kg; g = 10 m/s^2 (rough approx.);

Force at 10 meters should be half the ladder weight (as you mentioned), W/2 = mg/2 = (20)(10)/2 = 100 Newtons, approx. Secondly, Force at 0 meters should be 0 Newtons.

Make sure your force formula reflects (or closely approximates) those values. Your force equation: f = 100gx/((x^2) + 4) does appear to reflect both values, so we can gain some comfort from that quick check.

I'll try to get back to you later. Sorry for the delay.

 

[nakb]

Yes sorry, balancing the brackets I should have said:

The force by moments about A at any point X is:

F(sqrt(x^2 + 4)) - 20g(5) * (x / sqrt(x^2 + 4))

Nigel

 

[wjm11]

Yes sorry, balancing the brackets I should have said:

The force by moments about A at any point X is:

F(sqrt(x^2 + 4)) - 20g(5) * (x / sqrt(x^2 + 4))

Nigel

Thank you.

b). Rearranged gives: F(x^2) - 100gx + 4F = 0

c). Differentiated in terms of x gives: 2Fx - 100g = 0

I agree with you up through step b); however, when you differentiated, you treated F as a constant. It is not a constant; it is a function of x.

Anyway -- sorry -- I have to run. I'll try to get back later if no one else has jumped in.

 

[nakb]

There is also a follow-up question to this problem which I haven't yet attempted. In the follow-up, instead of the foot of the ladder being against a wall there is a second person, Fred, who keeps it in position by pushing on the foot of the ladder at A with his feet.

This second question asks me to identify and sketch the components of the force that Fred would need to exert on the ladder at A to ensure the foot remains in position at A while Joe erects it and define functions for those forces (assuming a smooth surface at A so no friction is involved and also assuming that Joe's force is as defined by the function in the answer to the first question and continues to be perpendicular to the ladder).

My initial thinking on this second question is that there are two elements involved:

a) Fred will need to stop the ladder sliding horizontally towards him as it is raised.

b) Once Joe gets past the center of gravity of the ladder Fred may have to stop his end lifting.

c) Because the components of Fred's force will change (for both of these elements) as Joe moves towards him I guess this would actually require two functions - one for the horizontal component and one for the vertical component.

I probably won't attempt this second question. I just mention it in case someone is interested in trying it.

 

[wjm11]

There is also a follow-up question to this problem which I haven't yet attempted. In the follow-up, instead of the foot of the ladder being against a wall there is a second person, Fred, who keeps it in position by pushing on the foot of the ladder at A with his feet.

This second question asks me to identify and sketch the components of the force that Fred would need to exert on the ladder at A to ensure the foot remains in position at A while Joe erects it and define functions for those forces (assuming a smooth surface at A so no friction is involved and also assuming that Joe's force is as defined by the function in the answer to the first question and continues to be perpendicular to the ladder).

My initial thinking on this second question is that there are two elements involved:

a) Fred will need to stop the ladder sliding horizontally towards him as it is raised.

b) Once Joe gets past the center of gravity of the ladder Fred may have to stop his end lifting.

c) Because the components of Fred's force will change (for both of these elements) as Joe moves towards him I guess this would actually require two functions - one for the horizontal component and one for the vertical component.

I probably won't attempt this second question. I just mention it in case someone is interested in trying it.

You just have to add a couple more constraint equations: The sum of the forces in both the x and y directions must be zero. For example, however much force Joe applies to the left must be counteracted by Fred pushing to the right. Fred's vertical component of force must add to Jeff's upward force and the ladder's weight to equal zero.

 

[Deleted member 4993]

OK, here's how we approached it.

Firstly, just considering what is going on before starting on the maths:

1. At the start (Joe at X1, immediately blow B) the weight will be approx equally divided between Joe at X1 and the ground at A.

2. As the ladder becomes more vertical the weight distribution will change, with more on the ground until when Joe reaches A all the weight is on the ground.

3. As Joe's point of contact with the ladder moves from the end (B) at the start to a point 2 metres from A at the end the force on him by the ladder will tend to increase (principle of the lever).

4. (2) and (3) act in opposite directions so will tend to compensate for each other - although at this stage we can't intuitively tell what the net effect will be.

5. It is assumed the force exerted on Joe by the ladder is perpendicular to the ladder.

This assumption is complicated. Much better assumption is that Joe's hands are vertical while pushing the ladder and the force applied by the ladder onto Joe (and the force applied by Joe onto the ladder) is vertical.


Turning to the maths:

a). The force by moments about A at any point X is: F(sqrt(x^2 + 4) - 20g(5) * (x / sqrt(x^2 + 4)

[where (x / sqrt(x^2 + 4) is cos(acute angle at A).

b). Rearranged gives: F(x^2) - 100gx + 4F = 0

c). Differentiated in terms of x gives: 2Fx - 100g = 0

d). So x = 50g/F

[substituting this back into (b) gives:

e). F((50g/F)^2) - 100g(50g/F) + 4F = 0

f). Rearranging gives: F = sqrt((5000(g^2) - (50g)^2) / 4) = 245

g). So x at point of maximum F is: 50g/245 = 2

So answers are:

(a) f = 100gx/((x^2) + 4)

(b) Graph not included here but would be graph of (a).

(c) Force is 245N and distance is 2m.

Are we right?

With the modified assumption above, the moment equation becomes:

\(\displaystyle F(x) * x \ - \ \dfrac{WL}{2} \ * \ \dfrac{x}{\sqrt{x^2+4}} \ = \ 0\)