Maths&physics
Badges: 15
Rep:
?
#1
Report Thread starter 1 year ago
#1
all forces are in perfect equilibrium because it's on the verge of slipping.

Therefore, the forces: friction is acting towards the wall, and the ladders mass is acting in the centre?

how do I work out the horizontal force of the ladder and the particle?
Last edited by Maths&physics; 1 year ago
0
reply
Maths&physics
Badges: 15
Rep:
?
#2
Report Thread starter 1 year ago
#2
here is the attachment.
Attached files
0
reply
old_engineer
Badges: 11
Rep:
?
#3
Report 1 year ago
#3
Resolve vertically to find the normal reaction at A (noting that there cannot be any vertical components of forces at B). Then use F = (mu)R at A.
0
reply
Maths&physics
Badges: 15
Rep:
?
#4
Report Thread starter 1 year ago
#4
(Original post by old_engineer)
Resolve vertically to find the normal reaction at A (noting that there cannot be any vertical components of forces at B). Then use F = (mu)R at A.
theres a reaction force from the ladder at the ground (R1) and a reaction force agasint the wall (R2).
because friction is acting in the opposite direction and is parrallel to R2, is friction: mu(R2)?
0
reply
old_engineer
Badges: 11
Rep:
?
#5
Report 1 year ago
#5
(Original post by Maths&physics)
theres a reaction force from the ladder at the ground (R1) and a reaction force agasint the wall (R2).
because friction is acting in the opposite direction and is parrallel to R2, is friction: mu(R2)?
No. F = (mu)R1. Find F first. Then R2 must be equal and opposite to F (as they are the only horizontal forces and the net horizontal force must be zero).
0
reply
Maths&physics
Badges: 15
Rep:
?
#6
Report Thread starter 1 year ago
#6
(Original post by old_engineer)
No. F = (mu)R1. Find F first. Then R2 must be equal and opposite to F (as they are the only horizontal forces and the net horizontal force must be zero).
ok, but why do we use R1 for friction, when ffriction is acting in a perpendicular direction?
0
reply
old_engineer
Badges: 11
Rep:
?
#7
Report 1 year ago
#7
(Original post by Maths&physics)
ok, but why do we use R1 for friction, when ffriction is acting in a perpendicular direction?
That’s always the case when an object is on the point of slipping. F is the friction and R is the normal reaction at the same point.
0
reply
Maths&physics
Badges: 15
Rep:
?
#8
Report Thread starter 1 year ago
#8
(Original post by old_engineer)
That’s always the case when an object is on the point of slipping. F is the friction and R is the normal reaction at the same point.
for part b and working out moments about A, do i take the reaction for the centre of the ladder and the reactioon force (reece) at point c?
0
reply
old_engineer
Badges: 11
Rep:
?
#9
Report 1 year ago
#9
(Original post by Maths&physics)
for part b and working out moments about A, do i take the reaction for the centre of the ladder and the reactioon force (reece) at point c?
There’s nothing in the question to say that the ladder is not uniform, so I think we can assume it is uniform, in which case it’s weight acts vertically downwards from its mid-point. The reaction at C acts horizontally in the direction away from the wall.
0
reply
Maths&physics
Badges: 15
Rep:
?
#10
Report Thread starter 1 year ago
#10
(Original post by old_engineer)
There’s nothing in the question to say that the ladder is not uniform, so I think we can assume it is uniform, in which case it’s weight acts vertically downwards from its mid-point. The reaction at C acts horizontally in the direction away from the wall.
so, do we ignore it's reaction in taking moments?

also, i don't understand what they're asking in c.
0
reply
old_engineer
Badges: 11
Rep:
?
#11
Report 1 year ago
#11
(Original post by Maths&physics)
so, do we ignore it's reaction in taking moments?

also, i don't understand what they're asking in c.
If taking moments about A you can ignore the forces acting at A ( because the distance is zero) but you must include any force acting at C, including the horizontal reaction of the wall on the ladder.

Part c is probably looking for you say that treating the person as a particle allows you to identify accurately where the weight of the person is acting.
0
reply
Notnek
  • Study Helper
Badges: 20
Rep:
?
#12
Report 1 year ago
#12
(Original post by Maths&physics)
x
Why do you make a separate post for your attachments out of interest? You’re not the only one who does this and I’ve always wondered why.
0
reply
Maths&physics
Badges: 15
Rep:
?
#13
Report Thread starter 1 year ago
#13
(Original post by Notnek)
Why do you make a separate post for your attachments out of interest? You’re not the only one who does this and I’ve always wondered why.
probably to save time but what you've said would probably would be easier. I'll do that from now on.
0
reply
Maths&physics
Badges: 15
Rep:
?
#14
Report Thread starter 1 year ago
#14
(Original post by old_engineer)
If taking moments about A you can ignore the forces acting at A ( because the distance is zero) but you must include any force acting at C, including the horizontal reaction of the wall on the ladder.

Part c is probably looking for you say that treating the person as a particle allows you to identify accurately where the weight of the person is acting.
thank you.

is the horizontal reaction of the wall a summation of the horizontal reaction of the ladder and the person?
0
reply
old_engineer
Badges: 11
Rep:
?
#15
Report 1 year ago
#15
(Original post by Maths&physics)
thank you.

is the horizontal reaction of the wall a summation of the horizontal reaction of the ladder and the person?
That terminology is not correct. If taking moments about A it would be correct to say that the sum of the clockwise moments of the ladder and the person must equal the anticlockwise moment of the normal reaction at B.
0
reply
Maths&physics
Badges: 15
Rep:
?
#16
Report Thread starter 1 year ago
#16
(Original post by old_engineer)
That terminology is not correct. If taking moments about A it would be correct to say that the sum of the clockwise moments of the ladder and the person must equal the anticlockwise moment of the normal reaction at B.
ok, im confused. what happens to the reaction force of the person standing on the ladder, why are we ignoring it? if a car is on a slope, we have a reaation force of that person on the ladder.
0
reply
old_engineer
Badges: 11
Rep:
?
#17
Report 1 year ago
#17
(Original post by Maths&physics)
ok, im confused. what happens to the reaction force of the person standing on the ladder, why are we ignoring it? if a car is on a slope, we have a reaation force of that person on the ladder.
If you were trying to determine whether or not the person would slip down the ladder, you would have to find the normal reaction of the ladder to the person’s weight. However that is not what has been asked for in this question.
0
reply
Maths&physics
Badges: 15
Rep:
?
#18
Report Thread starter 1 year ago
#18
(Original post by old_engineer)
If you were trying to determine whether or not the person would slip down the ladder, you would have to find the normal reaction of the ladder to the person’s weight. However that is not what has been asked for in this question.
oh ok, i got you
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 (326)
55.82%
I don't have everything I need (258)
44.18%

Watched Threads

View All
Latest
My Feed