3. 0,661.66 points I Previous Answers SerPSET9 12.P.013.MI. A 16.0 m uniform lad

Question

3. 0,661.66 points I Previous Answers SerPSET9 12.P.013.MI. A 16.0 m uniform ladder weighing 490 N rests against a frictionless wall. The ladder makes a 59.0° 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 has climbed 4.10 m along the ladder from the bottom. Horizontal Force magnitude It may be helpful to think first about the force the wall exerts on the ladder. How is this related to the force that the ground exerts on the ladder? N direction towards the wall Vertical Force magnitude direction b) If the ladder is just on the verge of slipping when the firefighter is 8.90 m from the bottom, what is the coefficient of static friction between ladder and ground Need Help? Read L It L Master It My Notes

Explanation / Answer

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

Rw[16sin59] - 490[(16/2)cos59] - 840[4.1cos59] = 0

Rw = 276.54 N

Sum horizontal forces to zero shows that the horizontal floor reaction is 276.54 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[16sin59] - 490[(16/2)cos59] - 840[8.9cos59] = 0

Rw = Fh = 427.96 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

= 427.96 / 1330

= 0.322

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