Question

A 15.5 m uniform ladder weighing 480 N rests against a frictionless wall. The ladder makes a 56.0° angle with the horizontal.

(a) Find the horizontal and vertical forces the ground exerts on the base of the ladder when an 800 N firefighter is 4.00 m from the bottom.
Magnitude of the horizontal force
Direction

away from the wall

towards the wall    

Magnitude of the vertical force
Direction

up

down    

(b) If the ladder is just on the verge of slipping when the firefighter is 8.90 m up, what is the coefficient of static friction between ladder and ground?

Answer 1

Given,

L = 15 m ; w = 480 N ; = 56 deg

(a)W = 800 N ; d = 4m

Let F be the magnitude of force of the wall.Let Fh be the horizontal and Fv be the vertical forces.

W x d cos + w x L/2 cos - F x L sin = 0

800 x 4 cos56 + 480 x 15/2 cos56 - F x 15 sin56 = 0

F = (1789.4 + 2013.1 )/12.44 = 305.7 N

F = 305.7 N (towards the left)

Horizontal component = Fh = F = 305.7 N (towards the wall)

Vertical component = Fy = W + w = 480 + 800 = 1280 N (directed up)

(b) Now; d = 8.9 m

Using the same equation derived in a

W x d cos + w x L/2 cos - F x L sin = 0

800 x 8.9 x cos56 + 480 x 7.5 x cos56 - F x 15 sin56 = 0

F = 449.7 N

we know that frictional force is Ff = N

1280 x = 449.7

= 449.7/1280 = 0.351

Hence, coefficient of static friction between ladder and ground = = 0.351.