A ladder (l_L = 7.70 m) of weight W_L = 350 N leans against a smooth vertical wa

Question

A ladder (l_L = 7.70 m) of weight W_L = 350 N leans against a smooth vertical wall. The term "smooth" means that the wall can exert only a normal force directed perpendicular to the wall and cannot exert a frictional force parallel to it. A firefighter, whose weight is 869 N, stands 6.50 m up from the bottom of the ladder (this distance goes along the ladder, it is not the vertical height). Assume that the ladder's weight acts at the ladder's center, and neglect the hose's weight. What is the minimum value for the coefficient of static friction between the ladder and the ground, so that the ladder (with the fireman on it) does not slip? (Assume theta = 52.0 degree.)

Explanation / Answer

Weight of Ladder = 350 N

Weight of the Firefighter = 869 N

Length of Ladder, L = 7.7 m

Distant of Fire fighter , d = 6.50 m

= 52.0°

At the bottom of the ladder, the vertical normal force on the ladder (upwards) is NV.

Resolving vertically:

NV = 350 + 869 = 1219 N

At the top of the ladder the horizontal normal force on the ladder is NH.

Taking moment about the bottom of the ladder:

NH * L * sin(52) = 350 * (L/2) * cos(52) + 869 * d* cos(52)

NH * 7.7 * sin(52) = 350 * (7.7/2) * cos(52) + 869 * 6.50* cos(52)

NH = 710 N

At the bottom of the ladder the horizontal frictional force, F, acts towards the wall.

Resolving horizontally

F = NH = 710 N

For the minimum Value of Coefficient of Friction -

= F/NV
= 710/1219
= 0.582

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