The uniform 10 kg ladder rests against the smooth wall at B and the end A rests on the rough horizontal plane for which the coefficient of static friction is µs = 0.3. Determine the angle of inclination θ of the ladder and the normal reaction at B if the ladder is on the verge of slipping.

Angle of inclination = 16.7°; normal reaction = 93.96N

Step-by-step explanation:

Frictional force is known as force of opposition that tends to oppose the motion of a body (moving force Fm) placed on a rough surface. Frictional force Ff is expressed as;

Ff = nR where;

n is the coefficient of static friction

R is the normal reaction

From the formula;

n = Ff/R

Where Ff = Fm = Weight of the object × sinθ i.e Wsinθ

and R = Wcosθ

n = wsinθ/wcosθ

n = sinθ/cosθ

n = tanθ

Given µs = 0.3

0.3 = tanθ

θ = arctan0.3

θ = 16.7°

Angle of inclination is 16.7°

The normal reaction at B if the ladder is on the verge of slipping can be gotten using the expression;

R = wcosθ

R = mgcosθ where

m is the mass of the ladder = 10kg

g is the acceleration due to gravity = 9.81m/s

R = 10×9.81×cos16.7°

R = 93.96N

42

step-by-step explanation:

256/6=42

there is no question content here, but you for the free points! : d