UPSC 2015 Maths Optional Paper 1 Q6b — Step-by-Step Solution

13 marks · Section B

Friction (limiting friction) · Dynamics & Statics · asked 5× in 13 yrs · Read the full method →

Question

Two equal ladders of weight 4 kg each are placed so as to lean at $A$ against each other with their ends resting on a rough floor, given the coefficient of friction is $\mu$ . The ladders at $A$ make an angle $60^\circ$ with each other. Find what weight on the top would cause them to slip.

Technique

Free-body diagram of one ladder; horizontal, vertical, and moment equilibrium; symmetric apex reaction is horizontal; impose limiting friction $f=\mu N$ for impending slip; solve for $w$ .

Solution

Setup. Two equal ladders, each of weight $W=4$ kg-wt, lean against each other at the apex $A$ , with the lower ends on a rough horizontal floor (coefficient $\mu$ ). The angle at $A$ between the ladders is $60^\circ$ . By symmetry each ladder makes angle $30^\circ$ with the vertical, equivalently $60^\circ$ with the horizontal floor.

Place an additional weight $w$ at the apex $A$ . Find $w$ such that the system is on the verge of slipping.

Let each ladder have length $2L$ (work with half-length $L$ for cleanness). The apex contact between ladders is smooth (the problem doesn’t say rough at $A$ ; standard assumption).

Each ladder makes angle $\alpha=30^\circ$ with the vertical, so $60^\circ$ with the floor.

Step 1 — Forces on one ladder (say the right one)

Let the right ladder $BA$ have its base at $B$ (on floor) and apex at $A$ .

Forces:

Weight $W$ at midpoint of ladder, vertically downward.
Half of the added weight $w$ at $A$ acts on this ladder (by symmetry, the apex weight distributes equally) — so an additional vertical downward force $w/2$ at $A$ . (Or equivalently, model the apex weight $w$ as a vertical load applied symmetrically to both ladders.)
At $A$ : contact reaction from the other ladder, horizontal (smooth contact, normal to the other ladder’s surface — actually for two leaning ladders at $A$ , the contact force lies along the bisector, but let’s think more carefully).
At $B$ : normal reaction $N$ from floor (vertical, up) and friction $f$ (horizontal, pointing inward toward symmetric centre — preventing slip outward).

For two ladders leaning at $A$ symmetrically, the interaction at $A$ is along the horizontal (by symmetry of the configuration: the line of contact at $A$ is vertical, and the contact reaction is perpendicular to that, i.e. horizontal). Call this horizontal reaction at $A$ on the right ladder $R$ , pointing leftward (i.e. toward the other ladder, providing the horizontal force that supports the apex against sliding outward).

Wait — the right ladder pushes on the left ladder; by Newton’s third law, the left ladder pushes the right ladder with a horizontal force $R$ pointing rightward? No: if both leans outward at the base, then at the apex each ladder is being pushed inward by the other. So on the right ladder, the apex reaction from the left ladder is horizontal, pointing rightward — pushing the right ladder outward to the right.

This horizontal push at the top is what tries to slide the bottom outward (to the right). Friction at $B$ acts inward (to the left) to oppose impending slip.

Step 2 — Equations of equilibrium for the right ladder

Coordinates: $B$ at origin, $A$ at $(2L\sin\alpha,\,2L\cos\alpha)=(2L\sin 30^\circ,2L\cos 30^\circ)=(L,L\sqrt 3)$ . Midpoint $G=(L/2,L\sqrt 3/2)$ .

Forces on the right ladder:

At $G$ : $(0,-W)$ (rod weight).
At $A$ : $(R, -w/2)$ where $R>0$ is the horizontal push from the left ladder, and $-w/2$ is half the apex weight acting vertically downward. (The apex weight sits on both ladders.)
At $B$ : $(-f, N)$ where $f>0$ is friction acting leftward (inward), $N$ is the normal up.

Horizontal balance: $R-f=0\Rightarrow R=f$ .

Vertical balance: $N-W-w/2=0\Rightarrow N=W+w/2$ .

Moment about $B$ (take anticlockwise positive):

Weight $W$ at $G=(L/2,L\sqrt 3/2)$ , force $(0,-W)$ : moment $=L/2\cdot(-W)-L\sqrt 3/2\cdot 0=-WL/2$ (clockwise, i.e. tending to drop the right side of $G$ ).

Wait let me recompute. Moment of force $(F_x,F_y)$ applied at $(x,y)$ about origin: $\tau=xF_y-yF_x$ .

Weight at $G$ : $\tau_W=(L/2)(-W)-(L\sqrt 3/2)(0)=-WL/2$ .
Apex reaction at $A=(L,L\sqrt 3)$ , force $(R,-w/2)$ : $\tau_A=L(-w/2)-L\sqrt 3(R)=-Lw/2-L\sqrt 3 R$ .

Sum of moments about $B$ = 0:

-\dfrac{WL}{2}-\dfrac{Lw}{2}-L\sqrt 3 R=0,

\dfrac{W}{2}+\dfrac{w}{2}+\sqrt 3 R=0.

This gives $R<0$ , which contradicts our sign convention. Let me recheck the direction of $R$ .

Direction recheck. Both ladders have bases spread apart and meet at the top. The right ladder leans from lower-right to upper-left toward the apex. So the apex of the right ladder is to the left of the base $B$ , not to the right. Let me redo coordinates.

Set the symmetric axis at $x=0$ , apex $A$ on the symmetric axis: $A=(0,h)$ for some height $h$ . The right ladder goes from $B_R=(d,0)$ to $A=(0,h)$ where $d>0$ . Its half-length tip is at $G_R=(d/2,h/2)$ .

Each ladder makes angle $\alpha$ with the vertical. Length $2L$ . So $h=2L\cos\alpha$ and $d=2L\sin\alpha$ . With $\alpha=30^\circ$ : $h=2L\cdot\sqrt 3/2=L\sqrt 3$ , $d=2L\cdot 1/2=L$ .

Forces on right ladder:

At $G_R=(L/2,L\sqrt 3/2)$ : weight $(0,-W)$ .
At $A=(0,L\sqrt 3)$ : apex weight half $(0,-w/2)$ and reaction from other ladder. By symmetry the reaction is purely horizontal. The other (left) ladder is to the left, so when the right ladder leans against it, the reaction on the right ladder is horizontal pointing rightward (pushing right ladder outward). So apex reaction = $(R,0)$ with $R>0$ pointing right (positive $x$ ).

But wait — if the right ladder’s top is being pushed rightward by the left ladder, that’s the reaction. The right ladder pushes left on the left ladder at the apex (Newton’s 3rd law).

At $B_R=(L,0)$ : normal $N$ upward $(0,N)$ , friction $f$ . The base tends to slip rightward (outward, due to the top being pushed rightward), so friction acts leftward (inward): $(-f,0)$ with $f>0$ .

Horizontal balance: $R-f=0\Rightarrow R=f$ .

Vertical balance: $N-W-w/2=0\Rightarrow N=W+w/2$ .

Moment about $B_R=(L,0)$ . For a force $(F_x,F_y)$ at point $(x,y)$ , moment about $(L,0)$ is $(x-L)F_y-(y-0)F_x$ .

Weight at $G_R=(L/2,L\sqrt 3/2)$ , force $(0,-W)$ : $(L/2-L)(-W)-(L\sqrt 3/2)(0)=(-L/2)(-W)=WL/2$ .
Apex weight at $A=(0,L\sqrt 3)$ , force $(0,-w/2)$ : $(0-L)(-w/2)-(L\sqrt 3)(0)=Lw/2$ .
Apex reaction at $A$ , force $(R,0)$ : $(0-L)(0)-(L\sqrt 3)(R)=-L\sqrt 3 R$ .

Sum = 0:

\dfrac{WL}{2}+\dfrac{Lw}{2}-L\sqrt 3 R=0,

R=\dfrac{W+w}{2\sqrt 3}.

Step 3 — Limiting friction condition

At impending slip, $f=\mu N$ . With $f=R$ :

R=\mu N,

\dfrac{W+w}{2\sqrt 3}=\mu(W+w/2).

Solve for $w$ :

W+w=2\sqrt 3\mu(W+w/2)=2\sqrt 3\mu W+\sqrt 3\mu w.

w-\sqrt 3\mu w=2\sqrt 3\mu W-W=W(2\sqrt 3\mu-1).

w(1-\sqrt 3\mu)=W(2\sqrt 3\mu-1).

w=W\cdot\dfrac{2\sqrt 3\mu-1}{1-\sqrt 3\mu}.

With $W=4$ :

Answer

\boxed{\;w=\dfrac{4(2\sqrt 3\mu-1)}{1-\sqrt 3\mu}\text{ kg-wt}.\;}