Question

A 4 m ladder weighing 250 N is placed against a smooth vertical wall with its lower end1.5 m away from the wall. If the coefficient of friction between the ladder and the floor is0.3, show that the ladder will remain in equilibrium in this position.​

Answer 1

The ladder will not remain in equilibrium in this position.​

Explanation:

When the ladder is balanced, the condition is:

Torque due to normal reaction = Torque due to mass of ladder

Let the normal reaction on ladder due to ground be N₁

⇒ N₁ = mg

Let the normal reaction on ladder due to wall be N₂

⇒ N₂ = μmg

Torque due to normal reaction:

T₁ = N₂ × h

On applying, we obtain height of the ladder which is h = 3.7 m

T₁ = μmg × 3.7

T₁ = 0.3 × 250 × 3.7 = 277.5 Nm

Torque due to mass of ladder:

T₂ = 0.75 × mg

T₂ = 0.75 × 250 = 187.5 Nm

Since, Torque due to normal reaction ≠ Torque due to mass of ladder

Answer 2

Explanation:

Explanation:

see figure , here a ladder weighing 250 N is placed against a smooth vertical wall having coefficient of friction of 0.3 between it and floor.

here, weight of ladder , W = mg = 250N

coefficient of friction between floor and ladder ,  = 0.3

now, maximum frictional force =  

= 0.3 × 250 = 75 N

hence, maximum force of friction available at the point of contact between the ladder and the floor is 75N

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