Question

A uniform ladder of 4 m length rests against a vertical wall with which it
makes an angle of 45°. The coefficient of friction between the ladder and the wall is 0.4 and that
between ladder and the floor is 0.5. If a man, whose weight is one-half of that of the ladder ascends
it, how high will it be when the ladder slips?

Answer 1

Answer:

$\\\sqrt{2}$ m

Answer 2

Given,

The coefficient of friction between the ladder and the wall is 0.4 and

between the ladder and the floor is 0.5.

The length of the ladder = 4m.

The ladder rests against a vertical wall, making a 45° angle.

One man's weight is one-half of that of the ladder ascends

it.

To Find,

The distance between the base point and the man, when the ladder is at the point of slipping, or the height when the ladder is at the point of slipping.

Solution,

Let W be the weight of the ladder.

Then the weight of the man is $\frac{W}{2}$=0.5W.

We draw a diagram of the total from our viewpoint to understand it in a better approach.

The first friction $Fr_1$ will work on the ground against the ladder.

The second friction $Fr_2$ will work on the Wall against the ladder.

$Fr_1=0.5R_1 . . . . . . (i)$

$Fr_2=0.4R_2$. . . . (ii)

If we resolve the forces vertically we get,

 $R_1+Fr_2-W-0.5W=0$

⇒  $R_1+0.4R_2=1.5W. . . . (iii)$

If we resolve the forces horizontally we get,

$R_2-Fr_1=0.$

⇒$R_2=0.5R_1. . . .(iv)$

Put the value from (iv) in (iii)

And we get,

$R_2=0.625W,Fr_2=0.25W$.

Assume that X is the distance between A on the diagram the basepoint and the man when the ladder is at the point of slipping.

The moment about the point A,

W×2cos45°+0.5W×2cos45°-$R_2$×4sin45°-$Fr_2$×4cos 45°.

Now put the values of $R_2,Fr_2$ to get the value of X.

W×2cos45°+0.5W×2cos45°-0.625W×4sin45°-0.25W×4cos 45°.

=2W×$\frac{1}{\sqrt{2} }$+W×$\frac{1}{\sqrt{2} }$-2.5×$\frac{1}{\sqrt{2} }$-W×$\frac{1}{\sqrt{2} }$.

=-0.5W.

Hence, those numerics are may be wrong in this problem.

#SPJ3