A 100-kg box is placed on a ramp. As one end of the ramp is raised, the box begins to move downward just as the angle of inclination reaches 15. What is the coefficient of static friction between box and ramp?

The coefficient of static friction is 0.27

Explanation:

Frictional force Ff is a force of opposition that exist between two surfaces in contact. This force opposes the moving force acting on the object. It is directly proportional to the reaction force acting on the object and this is expressed mathematically as Ff = nR where;

n is the constant of proportionality known as the coefficient of static friction.

R is the normal reaction

For a body on an inclined plane, the moving force which acts downwards and along the incline plane is equal to the frictional force acting upward the plane i.e Fm= Ff

The expression then becomes;

Fm = nR

n = Fm/R

For a body on an inclined plane;

Fm = wsin(theta) and R = wcos(theta) where;

w is the weight of the object

theta is angle of inclination = 15°

n = wsin(theta)/wcos(theta)

n = sintheta/cos(theta)

n = tan(theta)

n = tan15°

n = 0.27

The coefficient of static friction is 0.27

i think if it were me i whould divide 300 dide 15

step-by-step explanation: