Year 12s: what will be the first way you research your future options? >>

ladders!!! - M2

Hi guys, i am not very good at ladders ;so i was hoping someone could help me on here :redface:

. I can do part a) of this question fine, however i am a bit confused with part b) i can get the right answer, (in the end....!! :rolleyes:

), but i was wondering if someone could take me through how they would do it...just so i can understand it more clearly.......

thanks in advance :smile:

its much appreciated....

"A ladder AB, of weight W and length 4a, has one end A on rough horizontal ground. The
coefficient of friction between the ladder and the ground is . The other end B rests against a
smooth vertical wall. The ladder makes an angle with the horizontal, where tan = 2.
A load of weight 4W is placed at the point C on the ladder, where AC = 3a. The ladder is modelled as a uniform rod which is in a vertical plane perpendicular to
the wall. The load is modelled as a particle. Given that the system is in limiting equilibrium,

(a) show that = 0.35.

A second load of weight kW is now placed on the ladder at A. The load of weight 4W is
removed from C and placed on the ladder at B. The ladder is modelled as a uniform rod which
is in a vertical plane perpendicular to the wall. The loads are modelled as particles. Given that
the ladder and the loads are in equilibrium,

(b) find the range of possible values of k. <---this bit, thanks :smile:

i have attached a diagram also..... :smile:

M2 q2.JPG75.0KB

Reply 1

mizfissy815

Alter the diagram, to fit the new scenario and it should look something like this.

∑Fx =0
S= μR

∑Fy=0
R= (5+k) W

Now take moments about B…

W x 2acosθ + kW (4acos&#952 :wink:

+ μR(4asin&#952 :wink:

= R(4acos&#952 :wink:

Sub. in R= (5+k) W…

W x 2acosθ + kW (4acos&#952 :wink:

+ μ x (5+k) W (4asin&#952 :wink:

= (5+k)W (4acos&#952 :wink:

All the ‘a’s and ‘W’ s cancel out…and divide everything by cosθ. (If you are ever given that tan θ = whatever...then you are most likely going to have to divide by cos θ (so that all the cos terms cancel out and any sin is converted to tan)

2 + 4k + μ(5+k) (4 x tan &#952 :wink:

= 20 + 4k

Sub. in your value of μ = 0.35

2 + 14 + (14/5)k = 20
(14/5)k = 4
k= 10/7

Since this is in limiting equilibrium, this is the minimum value for k. So k must be equal to or greater(which would give it more stability) than this value.

k≥ 10/7

altered.JPG17.4KB

Reply 2

uk6458

thank you very much, mizfissy815 , that really helped!!! .......i will give you rep when i have enough power :smile:

Reply 3

dhokes

I was just about to start a new topic about part b. until I found this post. So for a question like this where it asks for a range of values, is it always >= value ???

Cheers

Reply 4

mizfissy815

No not always. Depends on the question really. But if something is in limiting equilibrium, you know that this is just the amount of force you need to keep it stable. You need to think, whether or not increasing the mass would stabilize the system further, or cause it to fall apart.