Question

Two identical ladders are arranged as shown in the figure, resting on a horizontal surface. Mass of each ladder is M and length L. A block of mass m hangs from the apex point P. If the system is in equilibrium, find direction and magnitude of friction?

Answer 1

Friction is horizontal, toward the other ladder. Its magnitude is `( M + m )/2 tan alpha, alpha` = the angle between a ladder and the altitude PN to the horizontal surface,

Explanation:

The `triangle`PAN is a right angled `triangle`, formed by a ladder PA and the altitude PN to the horizontal surface.

The vertical forces in equilibrium are equal reactions R balancing the weights of the ladders and the weight at the apex P.
So, 2 R = 2 Mg + mg.
R = `( M + m/2 ) g` ... (1)

Equal horizontal frictions F and F that prevent sliding of the ladders are inward and balance each other,
Note that R and F act at A and, the weight of the ladder PA, Mg acts at the middle if the ladder. The apex weight mg acts at P.

Taking moments about the apex P of the forces on the ladder PA,
F X L cos `alpha+ Mg X L/2 sin alpha = R X L sin alpha`.Use (1).
F -= `((M + m ) / 2 ) g tan alpha`.

If F is the limiting friction and `mu` is the coefficient of friction of the horizontal surface,
F = `mu`R..
`mu=( M + m ) / ( 2 M + m )tan alpha`..