# Mechanics 2 Help: Forces at an angle

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The question I am stuck on is from A2 Mathematics - Mechanics.

"A uniform ladder of length 6.5 m rests with one end against a smooth wall and the other on rough horizontal ground at distance 2.5 m from the wall. If the foot of the ladder is on the point of slipping, find the coefficient of friction between the ladder and the ground."

I know that F=μR, but there are no forces or mass mentioned, which is why I am stuck. It would be great if someone could help me with this question.

"A uniform ladder of length 6.5 m rests with one end against a smooth wall and the other on rough horizontal ground at distance 2.5 m from the wall. If the foot of the ladder is on the point of slipping, find the coefficient of friction between the ladder and the ground."

I know that F=μR, but there are no forces or mass mentioned, which is why I am stuck. It would be great if someone could help me with this question.

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#2

Hmm have you done moments yet? (it was in OCR M2 anyway). If so I would recommend taking moments from the centre of the ladder (since it is uniform you know where the centre of mass of it is). With some small trigonometry and thinking about all the forces etc. you can do some calculations that make you find the coefficient needed. If you want more specific details just ask!

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(Original post by

Hmm have you done moments yet? (it was in OCR M2 anyway). If so I would recommend taking moments from the centre of the ladder (since it is uniform you know where the centre of mass of it is). With some small trigonometry and thinking about all the forces etc. you can do some calculations that make you find the coefficient needed. If you want more specific details just ask!

**Genesis2703**)Hmm have you done moments yet? (it was in OCR M2 anyway). If so I would recommend taking moments from the centre of the ladder (since it is uniform you know where the centre of mass of it is). With some small trigonometry and thinking about all the forces etc. you can do some calculations that make you find the coefficient needed. If you want more specific details just ask!

This is what I have done, but then I realised that what I did probably isn't relevant then I got slightly confused. Any ideas on what I should do next?

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#4

(Original post by

This is what I have done, but then I realised that what I did probably isn't relevant then I got slightly confused. Any ideas on what I should do next?

**sjh96**)This is what I have done, but then I realised that what I did probably isn't relevant then I got slightly confused. Any ideas on what I should do next?

Now note that the three forces in your moments equation are all horizontal or vertical forces. this means you want the vertical and horizontal distance from the centre for the moments equation. Since the centre is half way along do you see it is 3.25 (13/4) along from the wall or the floor. so now you have a small triangle with hypotenuse 13/4, and the angle you found earlier for the bottom left hand corner. Do you see finding the other 2 sides of the triangle will find you the relevant distances for all 3 forces? And can you form the equation from there?

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#5

**sjh96**)

This is what I have done, but then I realised that what I did probably isn't relevant then I got slightly confused. Any ideas on what I should do next?

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#6

I'm also stuck on this!

How do you know if the Reaction force at the point where the ladder is on the ground leads to a clock wise moment or an anticlockwise moment?

How do you know if the Reaction force at the point where the ladder is on the ground leads to a clock wise moment or an anticlockwise moment?

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#7

(Original post by

I'm also stuck on this!

How do you know if the Reaction force at the point where the ladder is on the ground leads to a clock wise moment or an anticlockwise moment?

**IDontKnowReally**)I'm also stuck on this!

How do you know if the Reaction force at the point where the ladder is on the ground leads to a clock wise moment or an anticlockwise moment?

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#8

(Original post by

Depends what you are taking moments about. If you are taking moments about the point where the ladder makes contact with the wall, then the reaction force where the ladder hits the ground creates an anti-clockwise moment (the force goes up therefore anticlockwise) while the weight of the ladder creates a clockwise moment (as it is pushing it down). Both cancel each other out.

**RDKGames**)Depends what you are taking moments about. If you are taking moments about the point where the ladder makes contact with the wall, then the reaction force where the ladder hits the ground creates an anti-clockwise moment (the force goes up therefore anticlockwise) while the weight of the ladder creates a clockwise moment (as it is pushing it down). Both cancel each other out.

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#9

(Original post by

So if the clockwise and anticlockwise moments cancel each other out, how do i find the co-efficient of the friction?

**IDontKnowReally**)So if the clockwise and anticlockwise moments cancel each other out, how do i find the co-efficient of the friction?

First take moments about the point of contact on the wall, this will lead you to the reaction force in terms of m. This can be used to get the frictional force.

Then take moments about the centre of the ladder and use some substitutions for the reaction forces. Things cancel and you get left with mu.

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#10

(Original post by

They don't 'cancel' out per se, but they are equal to each other therefore the net moment is 0. Make the anti-clockwise and clockwise moments equal to each other.

First take moments about the point of contact on the wall, this will lead you to the reaction force in terms of m. This can be used to get the frictional force.

Then take moments about the centre of the ladder and use some substitutions for the reaction forces. Things cancel and you get left with mu.

**RDKGames**)They don't 'cancel' out per se, but they are equal to each other therefore the net moment is 0. Make the anti-clockwise and clockwise moments equal to each other.

First take moments about the point of contact on the wall, this will lead you to the reaction force in terms of m. This can be used to get the frictional force.

Then take moments about the centre of the ladder and use some substitutions for the reaction forces. Things cancel and you get left with mu.

By taking moments about the centre of the ladder, will the weight become irrelevant? If so, wont this leave me with a net anticlockwise moment? :s

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#11

(Original post by

Where did the m come from?

By taking moments about the centre of the ladder, will the weight become irrelevant? If so, wont this leave me with a net anticlockwise moment? :s

**IDontKnowReally**)Where did the m come from?

By taking moments about the centre of the ladder, will the weight become irrelevant? If so, wont this leave me with a net anticlockwise moment? :s

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#12

(Original post by

Yes it will be irrelevant, but when taking moments about the contact point between the ladder and the wall, then the m stays.

**RDKGames**)Yes it will be irrelevant, but when taking moments about the contact point between the ladder and the wall, then the m stays.

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#13

(Original post by

But what is m representing?

**IDontKnowReally**)But what is m representing?

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#14

(Original post by

The weight of the ladder. So just say the ladder weighs kg, so the gravity force on it is mg.

**RDKGames**)The weight of the ladder. So just say the ladder weighs kg, so the gravity force on it is mg.

I tried taking moments about the point of contact between the ladder and the wall, and about the centre of the ladder like you said.

This gave me:

1.25 mg = 2.5R + 6F

0 mg = 1.25R + 3F

Doing simultaneous equations gave an answer of mg = 0, which cant possibly be right? Where have I gone wrong here? (sorry to keep bothering you about this question)

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#15

(Original post by

Ah okay thanks!I tried taking moments about the point of contact between the ladder and the wall, and about the centre of the ladder like you said.This gave me:1.25 mg = 2.5R + 6F0 mg = 1.25R + 3FDoing simultaneous equations gave an answer of mg = 0, which cant possibly be right? Where have I gone wrong here? (sorry to keep bothering you about this question)

**IDontKnowReally**)Ah okay thanks!I tried taking moments about the point of contact between the ladder and the wall, and about the centre of the ladder like you said.This gave me:1.25 mg = 2.5R + 6F0 mg = 1.25R + 3FDoing simultaneous equations gave an answer of mg = 0, which cant possibly be right? Where have I gone wrong here? (sorry to keep bothering you about this question)

Let's label some things; let be the point of contact between the ladder and the ground, let be the point of contact between the ladder and the wall, let where m is the mass of the ladder, and let the centre of mass be at point . You can find the vertical height between the bottom of the ladder to the top of it against the wall using trigonometry (or Pythagoras). Now let and be the reaction forces at B and A respectively. Also let be the frictional force at point A.

Taking moments about gives us: therefore:

(I see you've done something similar there but realise that the mass and the frictional force act clockwise while the reaction force acts anti-clockwise)

We know that because there are the only two horizontal forces, therefore they must balance.

Now take moments again about and see where that gets you, use the substitutions for and in terms of M and mu. Remember that and go clockwise and goes anticlockwise when considering moments about C.

I got but it's been a while since I've done these types and this Q in particular is awkward to work with, so I'm not sure but my working makes sense to me. Let me know if this is the right answer or not so I wouldn't have to go back to my working to check it for any errors.

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#16

(Original post by

Nope. mass cannot be 0, makes no sense. I had a go at the question and here's what I'm thinking:

Let's label some things; let be the point of contact between the ladder and the ground, let be the point of contact between the ladder and the wall, let where m is the mass of the ladder, and let the centre of mass be at point . You can find the vertical height between the bottom of the ladder to the top of it against the wall using trigonometry (or Pythagoras). Now let and be the reaction forces at B and A respectively. Also let be the frictional force at point A.

Taking moments about gives us: therefore:

(I see you've done something similar there but realise that the mass and the frictional force act clockwise while the reaction force acts anti-clockwise)

We know that because there are the only two horizontal forces, therefore they must balance.

Now take moments again about and see where that gets you, use the substitutions for and in terms of M and mu. Remember that and go clockwise and goes anticlockwise when considering moments about C.

I got but it's been a while since I've done these types and this Q in particular is awkward to work with, so I'm not sure but my working makes sense to me. Let me know if this is the right answer or not so I wouldn't have to go back to my working to check it for any errors.

**RDKGames**)Nope. mass cannot be 0, makes no sense. I had a go at the question and here's what I'm thinking:

Let's label some things; let be the point of contact between the ladder and the ground, let be the point of contact between the ladder and the wall, let where m is the mass of the ladder, and let the centre of mass be at point . You can find the vertical height between the bottom of the ladder to the top of it against the wall using trigonometry (or Pythagoras). Now let and be the reaction forces at B and A respectively. Also let be the frictional force at point A.

Taking moments about gives us: therefore:

(I see you've done something similar there but realise that the mass and the frictional force act clockwise while the reaction force acts anti-clockwise)

We know that because there are the only two horizontal forces, therefore they must balance.

Now take moments again about and see where that gets you, use the substitutions for and in terms of M and mu. Remember that and go clockwise and goes anticlockwise when considering moments about C.

I got but it's been a while since I've done these types and this Q in particular is awkward to work with, so I'm not sure but my working makes sense to me. Let me know if this is the right answer or not so I wouldn't have to go back to my working to check it for any errors.

Except I don't understand why friction is clockwise. As shown in the diagram in this thread, the arrow points towards the wall from point A, so shouldn't the moment be anticlockwise?

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#17

(Original post by

Have you come across the angle of friction yet. If so the easiest method is to treat it as a three force problem. i.e. the reaction at the ground passes through the point of intersection of the weight and the reaction at the wall. The coefficient of friction is then just the tangent of the angle that this reaction makes with the vertical..

**brianeverit**)Have you come across the angle of friction yet. If so the easiest method is to treat it as a three force problem. i.e. the reaction at the ground passes through the point of intersection of the weight and the reaction at the wall. The coefficient of friction is then just the tangent of the angle that this reaction makes with the vertical..

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#18

(Original post by

It makes so much more sense now-thank you!

Except I don't understand why friction is clockwise. As shown in the diagram in this thread, the arrow points towards the wall from point A, so shouldn't the moment be anticlockwise?

**IDontKnowReally**)It makes so much more sense now-thank you!

Except I don't understand why friction is clockwise. As shown in the diagram in this thread, the arrow points towards the wall from point A, so shouldn't the moment be anticlockwise?

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#19

(Original post by

I drew my diagram pointing the opposite direction (as if the ladder was put up on the right side of the wall) which is why the direction of moments is different. In the diagram above, weight and friction act anti-clockwise whereas the reaction force at the ground acts clockwise. Still gives the same equation.

**RDKGames**)I drew my diagram pointing the opposite direction (as if the ladder was put up on the right side of the wall) which is why the direction of moments is different. In the diagram above, weight and friction act anti-clockwise whereas the reaction force at the ground acts clockwise. Still gives the same equation.

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#20

Can anyone help with this question please??A cylindrical object with mass 8kg rests on two cylindrical bars of equal radius. The lines connecting the centre of each of the bars to the centre of the object make an angle of 40 degrees to the verticala) draw a diagram showing all the forces acting on the object. Describe each of the forces using wordsI have no clue how to draw this diagram. Please help!!

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