UPSC Maths 2013 Paper 1 Q7b — Solution

15 marks · Section B

Question

A uniform ladder rests at an angle of $45°$ with the horizontal with its upper extremity against a rough vertical wall and its lower extremity on the ground. If $\mu$ and $\mu'$ are the coefficients of limiting friction between the ladder and the ground and wall respectively, then find the minimum horizontal force required to move the lower end of the ladder towards the wall.

Technique

Limiting-equilibrium force and torque balance; identify friction directions from the prescribed motion tendency.

Solution

Setup. Ladder of length $L$ , weight $W$ (uniform, CG at midpoint). Angle $\theta=45°$ . Apply horizontal force $P$ at the lower end, pushing it toward the wall. At limiting equilibrium just before motion:

Lower end about to slip toward wall (in $+x$ direction).
Upper end about to slip up along wall (consequence — geometry of motion: pushing the foot toward the wall makes the top slide up).

Friction opposes tendency:

Ground friction $f_1=\mu N_1$ points away from wall (opposing $P$ ).
Wall friction $f_2=\mu' N_2$ points downward (opposing upward motion of upper end).

Step 1 — Force balance

Axes: $x$ -horizontal toward wall, $y$ -vertical up. Lower end at origin; upper end at $(L\cos\theta,L\sin\theta)=(L/\sqrt 2,L/\sqrt 2)$ .

Horizontal: $P-\mu N_1-N_2=0\;\Longrightarrow\;P=\mu N_1+N_2$ .

Vertical: $N_1-W-\mu' N_2=0\;\Longrightarrow\;N_1=W+\mu' N_2$ .

Step 2 — Torque about lower end

Lever arms × forces, taking counter-clockwise positive:

Weight $W$ at midpoint $(L/(2\sqrt 2),L/(2\sqrt 2))$ , force $(0,-W)$ : torque $-WL/(2\sqrt 2)$ .
Wall normal $N_2$ at top, force $(-N_2,0)$ : torque $+LN_2/\sqrt 2$ .
Wall friction $\mu' N_2$ at top, force $(0,-\mu' N_2)$ : torque $-L\mu' N_2/\sqrt 2$ .
Forces at the lower end contribute zero torque (zero lever arm).

Sum = 0:

-\frac{WL}{2\sqrt 2}+\frac{LN_2}{\sqrt 2}-\frac{L\mu' N_2}{\sqrt 2}=0\;\Longrightarrow\;N_2(1-\mu')=\frac{W}{2}\;\Longrightarrow\;N_2=\frac{W}{2(1-\mu')}.

(Requires $\mu'<1$ — physically the case unless the wall has unusually high friction.)

Step 3 — Back-substitute

N_1=W+\mu' N_2=W+\frac{\mu' W}{2(1-\mu')}=\frac{2W(1-\mu')+\mu' W}{2(1-\mu')}=\frac{W(2-\mu')}{2(1-\mu')}.

P=\mu N_1+N_2=\frac{\mu W(2-\mu')+W}{2(1-\mu')}=\frac{W\bigl[1+\mu(2-\mu')\bigr]}{2(1-\mu')}.

Expanding:

Answer

\boxed{\;P=\frac{W(1+2\mu-\mu\mu')}{2(1-\mu')}.\;}