A uniform ladder of length l rests against a smooth, vertical wall. The mass of

Question

A uniform ladder of length l rests against a smooth, vertical wall. The mass of the ladder is m, and the coefficient of static friction between the ladder and the ground is ?s = 0.41. If the coefficient of static friction is 0.55, and the same ladder makes a 50 degree angle with respect to the horizontal, how far along the length of the ladder can a 56.0 kg person climb before the ladder begins to slip? The ladder is 8.6 m in length and has a mass of 21 kg. distance along the length of the ladder: m

Explanation / Answer

let ? = angle of the ladder with the wall
there are two horizontal forces:1) from the wall Fw , away from the wall, where the ladder leans at the wall, 2) Fr, from the standpoint of the ladder, towards the wall.
There are 2 vertical forces: 1) Fn, the normal force of the ground on the foot of the ladder, upwards 2) Fl, the weight force of ladder, from the gravity center of the ladder at L/2, down.

The ladder is in equilibrium, when Fw = Fr, and Fn = W
and the moments at A are
Fw*L*cos? = W*(L/2)*sin? ............(a)

The ladder starts to slip, when Fr = ?*Fn

Since Fw = Fr = ?*Fn = ?*W --> plug into (a)

?*W*L*cos? = W*(L/2)*sin?

? = 1/2*sin?/cos?

? = 1/2 * tan? (= minimum value for ?s. Neither weight nor length of the ladder play a role for ?s))

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