wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

A ladder of mass m is leaning against a wall. lt is in static equilibrium making an angle θ with the horizontal floor. The coefficient of friction between the wall and the ladder is μ1 and that between the floor and the ladder is μ2. The normal reaction of the wall on the ladder is N1 and that of the floor is N2. If the ladder is about to slip, then

A
μ1=0;μ20 and N2tanθ=mg2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
μ10;μ2=0 and N1tanθ=mg2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
μ10;μ20 and N2=mg1+μ1μ2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
μ1=0;μ20 and N1tanθ=mg2
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct options are
C μ10;μ20 and N2=mg1+μ1μ2
D μ1=0;μ20 and N1tanθ=mg2

Balancing forces in horizontal direction,
N1=μN2
Balancing forces in vertical direction,
N2+μN1=mg

Solving we get,
N2=mg1+μ1μ2
N1=μ2mg1+μ1μ2

Applying rotational equilibrium,
mgl2cosθ=N1lsinθ+μN1lcosθ
Solving, tanθ=1μ1μ22μ2

If μ1=0,μ20
tanθ=12μ2
N1=μ2mg
or N1tanθ=mg2

565018_130968_ans_f226fae7214e435584da8c25fed79381.png

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
PHYSICS
Watch in App
Join BYJU'S Learning Program
CrossIcon