Question

Give your answer in SI units and to three significant figures.

A stick which is 1.56 meters long is leaned against a wall at an angle. If the coefficient of static friction between the stick and the wall and floor is 0.310, determine the furthest distance from the wall that the bottom of the stick may be placed without slipping.

Answer 1

Length of stick= 1.56 M Coefficient of friction = 0.310

mg = N1 + 0.310 x N2 ——- (1), where normal N1 is reaction of floor on ladder

N2 is normal reaction of wall on ladder , F1 and F2 are are friction forces on ladder by floor and wall respectively.

F1 = N2

0.310 x N1 = N2 ——— (2)

mg = N1 +( 0.310)^2 N1
mg = N1 ( 1+ 0.310^2 )

‘N1 = mg/ ( 1+ 0.310^2)

let x be angle between floor and ladder

[(mg)L /2]cos x = N2 L sin  x + F2 L cos x

[mg/2] cos x = N2 Sin x + 0.310x N2 x cos x

{mg/2} = N2 [ tan x + 0.310]

mg/2 = 0.310 N1 [ tan x + 0.310] = 0.310 x [tan x + 0.310] x [ mg/ (1+0.310^2)]

0.5 = 0.310[ (tan x + 0.310)/ (1+0.310^2)]

[0.5 x (1+0.310^2) / 0.310] = tan x + 0.310

solving angle x = 55.6 deg

therefore , the distance between wall and tipping point = L cos x = 1.56 m x cos 55.6 = 0.88 m