A 16.0-m uniform ladder weighing 510 N rests against a frictionless wall. The la

Question

A 16.0-m uniform ladder weighing 510 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 3.70 m along the ladder from the bottom.

Horizontal Force

magnitude

N

direction

towards the wall

Vertical Force

magnitude

N

direction

up

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

magnitude

N

direction

towards the wall

Explanation / Answer

then using sum of forcesacting along x axis

820 * 3.7 cos 61 + 510 * 8 cos 61 - F * 16 sin 61 = 0

then F = 246.46 N directed towards right

---------------------------------------

Vertical Force Fy = 510+ 820 = 1330 N , directed upwards

----------------------------------------------------

part B :

let F be the magnitude of force o the wall

then using sum of forcesacting along x axis

820 * 8.7 cos 61 + 510 * 8 cos 61 - F * 16 sin 61 = 0

then F = 388.5 N directed towards right

--------------------------------------------------

Vertical Force Fy = 510+ 820 = 1330 N , directed upwards

--------------------------

apply F = u N

u = coefficient of friction = 388.5/1330

u = 0.292

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