Question

A 4 m ladder weighing 250 N is placed against a smooth vertical wall with its lower end

1.5 m away from the wall. If the coefficient of friction between the ladder and the floor is

0.3, show that the ladder will remain in equilibrium in this position.​

Answer 1

Given:

A 4 m ladder weighing 250 N is placed against a smooth vertical wall

Lower end of the ladder is 1.5 m away from the wall.

Coefficient of friction between the ladder and floor is 0.3

To Find:

Show that the ladder will remain in equilibrium position.

Solution:

Let the normal reaction on ladder due to ground = $N_{1}$

Normal reaction on ladder due to wall = $N_{2}$

Net forces and momentum on the ladder should be equal to 0 .

Force balance

vertical direction.

$N_{1} = mg$

Horizontal direction

$N_{2} =F_{\mu} \\N_{2} = \mu (N_{1} ) \\N_{2}= \mu (mg)$

Torque balance about point on ladder in contact with ground.

First we need to calculate the height of point of contact of ladder with wall.

 Use pythogarus theorem,

$h = \sqrt{4^{2} - 1.5^{2} }\\ h = 3.7$

Torque of normal reaction due to wall

T = $N_{2}(h ) = \mu mg (3.7)$ = $3.7(0.3)(250)= 277.5 Nm$

Torque due to mass mg

T = $T = mg (0.75)=187.5 Nm$

Torque is not balanced

Hence the ladder is not in equilibrium .

Answer 2

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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