A 15.0 m uniform ladder weighing 490 N rests against a frictionless wall. The la

Question

A 15.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 61.0° angle with the horizontal. (a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 840 N firefighter is 3.70 m from the bottom.
Magnitude of the horizontal force
Direction_____

away from the wall or towards the wall     ?

Magnitude of the vertical force
Direction _____

down or up ?

(b) If the ladder is just on the verge of slipping when the firefighter is 8.70 m up, what is the coefficient of static friction between ladder and ground?

Explanation / Answer

Sum moments about the floor contact to find the wall reaction horizontal force.

Rw[15sin 61] - 490[(15/2)cos61] - 840[3.7cos61] = 0

Rw[15sin 61] = 274.887

Rw = 20.953 N

Sum horizontal forces to zero shows that the horizontal floor reaction is 252 N toward the wall

Sum vertical forces to zero to find the floor vertical force

Fv - 490 - 840 = 0

Fv = 1330 N upward

b) Using the same logic to find the horizontal reactions when the firefighter is higher

Rw[15sin61] - 490[(15/2)cos61] - 850[8.7cos61] = 0

Rw = (5366.84)/[15sin61]

= 409.08

Rw = Fh = 409.08 N

The vertical reaction remains the same

Fv = 1330 N upward

coefficient of friction is the ratio of the maximum horizontal force to vertical force

= Fh / Fv

= 409.08 / 1330

= 0.3075

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