The Student Room Group

2 M2 Questions Help Please!!!

They are on page 142 edexcel m2 book numbers 5 and 7!

5) A uniform ladder rests with one end on rough horizontal ground and the other end against a rough vertical wall. The coefficient of friction between the ground and the ladder is 3/5 and the coefficient of friction between the wall and ladder is 1/3. The ladder is on point of slipping when it makes an angle x (theta) with the horizontal. Find tan x (tan theta).


7) A uniform ladder of mass 30kg is placed with its base on the rough horizontal ground. The coefficient of friction between the ladder and the ground is 1/4. The upper end of the ladder rests against a smooth vertical wall, the ladder making an angle of 60 deg to the horizontal. Find the magnitude of the minimum horizontal force that must be applied to the base of the ladder to prevent slipping.


Thanks if anyone has time to help!

Ive reposted the thread cause no one has got it right yet!

ANswers are 5) 2/3
7) 11.4 N

Well cheers is anyone can do it :smile:
Reply 1
first: dont you EVER remake a new only coz no one answered in 15 hours!

7)
force up = force down
Nv = 30g

force left = force right
Nh = f + F

f = 1/4 x Nv
f = 1/4 x 30g

Nh = 30/4 g + F

take moment about where the ladder touches the ground:

Nh x 2l x sin60 = 30g x l x cos60
2Nh tan60 = 30g

Nh = 85.0N

Nh = 30/4 g + F
F = 11.4 N

and by the way: when i added the attachment: i discovered that i had the SAME picture that i drew for the SAME question... so: instead of starting 2 new threads, u could've searched and found that question!
Reply 2
5) A uniform ladder rests with one end on rough horizontal ground and the other end against a rough vertical wall. The coefficient of friction between the ground and the ladder is 3/5 and the coefficient of friction between the wall and ladder is 1/3. The ladder is on point of slipping when it makes an angle x (theta) with the horizontal. Find tan x (tan theta).


R1 + f2 = mg

f1 = R2

R2 = f1 = R1 . 3/5
f2 = R2 . 1/3

R1 + 1/3 R2 = mg
R1 + 1/5 R1 = mg

6/5 R1 = mg
R1 = 5/6 mg

R2 . 2l . sinx + f2 . 2l . cosx = mg . l . cosx
2 f1 tanx + 2 (mg - R1) = mg
2 (3/5 R1) tanx + 2(mg - R1) = mg
6/5 . 5/6 mg tanx + 2mg - 2 . 5/6mg = 1
tanx + 2 - 5/3 = 1

tanx = 1 + 5/3 - 2 = 0.666 = 2/3
Reply 3
Thanks very much! Sorry was getting rather impatient because couldnt do either for some reason so was very frustrating! Thank you havent checked it through yet but I think i see where I went wrong. Thank you :smile: