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?
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
µk=coefficient of static friction
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
�k=coefficient of static friction
To find the coefficient of static friction between the box and the ramp, we can use the relationship between the force of friction and the normal force exerted on the box.
The normal force is the force perpendicular to the surface that supports the box, which in this case is the weight of the box. We can find the normal force by multiplying the mass of the box by the acceleration due to gravity (9.8 m/s^2):
Normal force = mass * acceleration due to gravity
Normal force = 150.0 kg * 9.8 m/s^2 = 1470 N
The force of static friction can be calculated using the equation:
Force of static friction = coefficient of static friction * normal force
We can rearrange this equation to solve for the coefficient of static friction:
Coefficient of static friction = Force of static friction / normal force
Now, we can substitute the values given in the problem:
Force of static friction = 200.0 N
Normal force = 1470 N
Coefficient of static friction = 200.0 N / 1470 N ≈ 0.136
Therefore, the coefficient of static friction between the box and the ramp is approximately 0.136.