Incline Weight
Incline Weight
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How is static friction calculated with respect to an object's weight and its incline of rest?
A loaded penguin sled weighing 90 N rests on a plane inclined at 20° to the horizontal. Between the sled and the plane the coefficient of static friction is 0.27, and the coefficient of kinetic friction is 0.11. What is the minimum magnitude of the force F, parallel to the plane, that will prevent the sled from slipping down the plane?
Friction force, which always acts in the direction opposite to the direction of motion, is F = kN; where N = mg cos(theta) is the so-called "normal" weight of the object with mass m.
Notice that W = mg is the weight, period, of the object and W always... always... points downward toward the center of the planet. If theta is the angle of surface incline, the weight W always points perpendicular to the surface when theta = 0. Unless the mass, gravity field g, or both change, weight remains the same no matter what the incline. Not so of the normal force N = W cos(theta).
The normal force always... always... points perpendicular into the surface, even when the surface is inclined where theta > 0. This means that friction force along the surface is proportional to the normal force pushing into that same surface. And the constant of proportionality is k, the so-called friction coefficient. So we have F = kN = kW cos(theta) = kmg cos(theta) as the friction force opposing the direction of the sliding sled. So F is trying to stop the motion; what's trying to make the sled move?
Gravity. But gravity, like weight, works one direction only... perpendicular towards the center of the Earth. In fact the force of gravity is W = mg, the weight of the object... your sled. But that's not the force along the surface when the surface is inclined. That would be w = W sin(theta).
As you can see, where the surface is level and theta = 0, there is no surface force of gravity. Gravity will not pull that sled horizontally because it works downward into the Earth, not along the surface of the Earth. On the other hand, were the surface perpendicular to Earth's surface, where theta = 90 degrees, w = W sin(90) = W and all the weight of the sled would be pulling along the incline's surface. But look at this.
F = kW cos(90) = kW 0 = 0 at 90 degrees incline and none of the friction force is opposing that weight W. So the net force acting on the sled in that extreme case is f = W - F = mg; so the sled would be accelerating g = W/m the acceleration due to gravity while in free fall. That sled would be accelerating and moving like a bat our of you know where. Which brings us to your problem.
The net force on the sled at theta = 20 deg is f = w - F = W sin(theta) - k W cos(theta) = 0 when the sled is not yet moving. So k = .27 as this is a static case. From the net force f = 0 = w - F, we see that F = w = W sin(theta) = 90 sin(20) = 30.8 Newtons ANS is the requisite friction force to keep the sled from moving. And now you know why.