Mechanics 2 Help: Forces at an angle

Watch
sjh96
Badges: 2
Rep:
?
#1
Report Thread starter 7 years ago
#1
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.
0
reply
Genesis2703
Badges: 5
Rep:
?
#2
Report 7 years ago
#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!
0
reply
sjh96
Badges: 2
Rep:
?
#3
Report Thread starter 7 years ago
#3
(Original post by 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!
Name:  2013-09-15 21.01.13.jpg
Views: 620
Size:  497.7 KB

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?
0
reply
Genesis2703
Badges: 5
Rep:
?
#4
Report 7 years ago
#4
(Original post by sjh96)
Name:  2013-09-15 21.01.13.jpg
Views: 620
Size:  497.7 KB

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?
A nicer way of expressing the angle is to think of Soh,Cah,Toa. Cosine of the bottom left angle is adjacent/hyp which is 2.5/6.5 = 5/13 which is nicer than some angle in degrees. Note that is is Cos(x) = 5/13 not x=5/13, so if you need sin(x) later on you need to be careful with what value that represents.

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?
0
reply
brianeverit
  • Study Helper
Badges: 9
Rep:
?
#5
Report 7 years ago
#5
(Original post by sjh96)
Name:  2013-09-15 21.01.13.jpg
Views: 620
Size:  497.7 KB

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?
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..
1
reply
IDontKnowReally
Badges: 11
Rep:
?
#6
Report 4 years ago
#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?
0
reply
RDKGames
Badges: 20
Rep:
?
#7
Report 4 years ago
#7
(Original post by 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?
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.
0
reply
IDontKnowReally
Badges: 11
Rep:
?
#8
Report 4 years ago
#8
(Original post by 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.
So if the clockwise and anticlockwise moments cancel each other out, how do i find the co-efficient of the friction?
0
reply
RDKGames
Badges: 20
Rep:
?
#9
Report 4 years ago
#9
(Original post by IDontKnowReally)
So if the clockwise and anticlockwise moments cancel each other out, how do i find the co-efficient of the friction?
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.
0
reply
IDontKnowReally
Badges: 11
Rep:
?
#10
Report 4 years ago
#10
(Original post by 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.
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
0
reply
RDKGames
Badges: 20
Rep:
?
#11
Report 4 years ago
#11
(Original post by 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
Yes it will be irrelevant, but when taking moments about the contact point between the ladder and the wall, then the m stays.
0
reply
IDontKnowReally
Badges: 11
Rep:
?
#12
Report 4 years ago
#12
(Original post by RDKGames)
Yes it will be irrelevant, but when taking moments about the contact point between the ladder and the wall, then the m stays.
But what is m representing?
0
reply
RDKGames
Badges: 20
Rep:
?
#13
Report 4 years ago
#13
(Original post by IDontKnowReally)
But what is m representing?
The weight of the ladder. So just say the ladder weighs m kg, so the gravity force on it is mg.
0
reply
IDontKnowReally
Badges: 11
Rep:
?
#14
Report 4 years ago
#14
(Original post by RDKGames)
The weight of the ladder. So just say the ladder weighs m kg, so the gravity force on it is mg.
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 + 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)
0
reply
RDKGames
Badges: 20
Rep:
?
#15
Report 4 years ago
#15
(Original post by 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)
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 A be the point of contact between the ladder and the ground, let B be the point of contact between the ladder and the wall, let M=mg where m is the mass of the ladder, and let the centre of mass be at point C. You can find the vertical height h between the bottom of the ladder to the top of it against the wall using trigonometry (or Pythagoras). Now let S and R be the reaction forces at B and A respectively. Also let \mu R be the frictional force at point A.

Taking moments about B gives us: M(B): 1.25M+6\mu R=2.5R therefore:
\frac{1.25M}{2.5-6\mu}=R \longrightarrow \mu R=\frac{1.25M\mu}{2.5-6\mu}

(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 S=\mu R because there are the only two horizontal forces, therefore they must balance.

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

I got \mu=\frac{5}{24} 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.
1
reply
IDontKnowReally
Badges: 11
Rep:
?
#16
Report 4 years ago
#16
(Original post by 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 A be the point of contact between the ladder and the ground, let B be the point of contact between the ladder and the wall, let M=mg where m is the mass of the ladder, and let the centre of mass be at point C. You can find the vertical height h between the bottom of the ladder to the top of it against the wall using trigonometry (or Pythagoras). Now let S and R be the reaction forces at B and A respectively. Also let \mu R be the frictional force at point A.

Taking moments about B gives us: M(B): 1.25M+6\mu R=2.5R therefore:
\frac{1.25M}{2.5-6\mu}=R \longrightarrow \mu R=\frac{1.25M\mu}{2.5-6\mu}

(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 S=\mu R because there are the only two horizontal forces, therefore they must balance.

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

I got \mu=\frac{5}{24} 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.
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?
0
reply
the bear
Badges: 20
Rep:
?
#17
Report 4 years ago
#17
(Original post by 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..
this makes it really simple... you get the same answer as RDK but in one line. *
0
reply
RDKGames
Badges: 20
Rep:
?
#18
Report 4 years ago
#18
(Original post by 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?
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.
0
reply
IDontKnowReally
Badges: 11
Rep:
?
#19
Report 4 years ago
#19
(Original post by 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.
I see. Following your method also gave me 5/24. Thanks for all your help
0
reply
Neveyy
Badges: 2
Rep:
?
#20
Report 1 year ago
#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!!
0
reply
X

Quick Reply

Attached files
Write a reply...
Reply
new posts
Back
to top
Latest
My Feed

See more of what you like on
The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

Personalise

Do you have the space and resources you need to succeed in home learning?

Yes I have everything I need (402)
56.38%
I don't have everything I need (311)
43.62%

Watched Threads

View All
Latest
My Feed