Question

Two identical ladders are arranged as shown in figure. Mass of each ladder is M and length is L. A load of mass m has been attached at the touching point (O) of the ladders, as shown in figure. The system is in equilibrium. Find the magnitude of static frictional force acting at either point A and B.

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Solution

The correct option is **B** (M+m2)gcotθ

Let N be the normal force acting at point A and B (same by symmetry)

Let N1 and N2 be the horizontal and vertical components of normal reaction exerted by each ladder on the other. (equal and opposite force will act on the two ladders at point O according to Newton's 3rd law)

From symmetry of the ladder arrangement, both ladders will share the load mg equally, hence mg2 will be the load on each ladder.

Let f be the friction acting between each ladder and the ground and due to symmetry, it will be same for both.

As the ladders are in equilibrium,

τnet=0, Fnet=0

For translational equilibrium,

∑Fx=0

∑Fy=0

For right ladder,

N2+mg2+Mg=N ...(i)

and N1=f ...(ii)

For left ladder,

N2+N=Mg+mg2 ...(iii)

and N1=f

From equation (i) and (iii),

N−Mg−mg2+N=Mg+mg2

⇒2N=2Mg+mg

∴N=Mg+mg2 ...(iv)

For rotational equilibrium, the net torque about O should be zero on any ladder,

τnet=0 ...(v) for right ladder.

Taking anticlockwise sense of rotation for torque as +ve

& τ=F×r⊥

∵ Forces N1,N2 & mg2 pass through reference point O, hence their torque about point O is zero

Substituting torques with proper sign in eq (v):

−Mg(L2cosθ)−f(Lsinθ)+N(Lcosθ)=0

Putting value of N from Eq (iv),

⇒Mgcosθ2+fsinθ=(Mg+mg2)cosθ

⇒fsinθ=Mgcosθ−Mgcosθ2+mg2cosθ

⇒fsinθ=(M+m2)gcosθ

∴f=(M+m2)gcotθ

Let N be the normal force acting at point A and B (same by symmetry)

Let N1 and N2 be the horizontal and vertical components of normal reaction exerted by each ladder on the other. (equal and opposite force will act on the two ladders at point O according to Newton's 3rd law)

From symmetry of the ladder arrangement, both ladders will share the load mg equally, hence mg2 will be the load on each ladder.

Let f be the friction acting between each ladder and the ground and due to symmetry, it will be same for both.

As the ladders are in equilibrium,

τnet=0, Fnet=0

For translational equilibrium,

∑Fx=0

∑Fy=0

For right ladder,

N2+mg2+Mg=N ...(i)

and N1=f ...(ii)

For left ladder,

N2+N=Mg+mg2 ...(iii)

and N1=f

From equation (i) and (iii),

N−Mg−mg2+N=Mg+mg2

⇒2N=2Mg+mg

∴N=Mg+mg2 ...(iv)

For rotational equilibrium, the net torque about O should be zero on any ladder,

τnet=0 ...(v) for right ladder.

Taking anticlockwise sense of rotation for torque as +ve

& τ=F×r⊥

∵ Forces N1,N2 & mg2 pass through reference point O, hence their torque about point O is zero

Substituting torques with proper sign in eq (v):

−Mg(L2cosθ)−f(Lsinθ)+N(Lcosθ)=0

Putting value of N from Eq (iv),

⇒Mgcosθ2+fsinθ=(Mg+mg2)cosθ

⇒fsinθ=Mgcosθ−Mgcosθ2+mg2cosθ

⇒fsinθ=(M+m2)gcosθ

∴f=(M+m2)gcotθ

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