A block is sitting on a rough ramp. It is found that a force of F = 200.0 N (directed up the ramp) is required to keep the box from sliding back down the ramp. The angle 𝜃 is 35.0o and the mass of the box is 150.0 kg. What is the coefficient of static friction between the box and the ramp?

Answer 1

Fk=F

Fk=200N

Fk=µk*mg*Cos(theta)

200N=µk*(150kg)*(9.8m/s^2)*Cos(35)

200N=µk*1470N*0.819

200N=µk*1204N

200N/1204N=µk

µk=0.166

Answer 2

µk=coefficient of static friction

Answer 3

Ignore my initial post, I forgot to include one of the components for the frictional force.

Fk=F

Fk=200N

Fk=µk*mg*Cos(theta)+ µk*mg*Cos(theta)

200N=µk*(150kg)*(9.8m/s^2)*[Cos(35)+ Sin(35)]

200N=µk*159.8N*[0.819 + 0.574]

200N=µk*159.8N*[1.393]

200N=µk*222

µk=0.898

200N/1204N=µk

µk=0.166
Physics-additional detail - Devron, Tuesday, May 13, 2014 at 4:12pm
µk=coefficient of static friction

Answer 4

�k=coefficient of static friction