Question

A 16.0-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 820-N firefighter has climbed 4.30 m along the ladder from the bottom.

Horizontal Force

magnitude	__N
direction	- towards the wall or away from the wall?

Vertical Force

magnitude	N
direction	up or down?

(b) If the ladder is just on the verge of slipping when the firefighter is 9.40 m from the bottom, what is the coefficient of static friction between ladder and ground?

Answer 1

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

Rw[16sin61] - 490[(16/2)cos61] - 820[4.3cos61] = 0

Rw = 258 N

Sum horizontal forces to zero shows that the horizontal floor reaction is 258 N toward the wall

Sum vertical forces to zero to find the floor vertical force

Fv - 490 - 820 = 0

Fv = 1310 N upward

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

Rw[16sin61] - 490[(16/2)cos61] - 820[9.4cos61] = 0

Rw = Fh = 402.8 N

The vertical reaction remains the same

Fv = 1310 N upward

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

? = Fh / Fv

? = 402.8 / 1310

? = 0.30....Answer

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