What force would you have to exert on a 324 N trunk up a 19 degree inclined plane so that it would slide down the plane with a constant velocity? What would be the direction of the force? The coefficient of friction between the plane and the trunk is .322.
Break the weight of the trunk into two components, one down the plane, and one normal to it.
Forcedownplane=mg*sinTheta
Forcenormal=mg*cosTheta
Then, the force of friction is down, as motion is upward mu*mg*cosTheta
Summing forces up the plane:
F-mu*mg*cosTheta - mg*sinTheta=ma=0
solve for F. ma is zero because of zero acceleration (constant velocity)
To determine the force required to slide the trunk up the inclined plane with constant velocity, we need to consider the forces at play.
First, let's break down the forces acting on the trunk:
1. The weight of the trunk (mg) acts straight downward and can be calculated as W = m * g, where m is the mass of the trunk and g is the acceleration due to gravity (approximately 9.8 m/s²).
2. The normal force (N) acts perpendicular to the inclined plane and prevents the trunk from sinking into the plane. When the trunk is on an inclined plane, the normal force can be calculated as N = m * g * cos(θ), where θ is the angle of the inclined plane (19 degrees in this case).
3. The force of friction (Ff) opposes the motion of the trunk and acts parallel to the inclined plane. The force of friction can be calculated as Ff = coefficient of friction * N, where the coefficient of friction (μ) is given as 0.322.
4. The force we need to calculate is the force we exert on the trunk, which we'll call F. Since the trunk is sliding with constant velocity, the force we exert (F) must be equal in magnitude and opposite in direction to the force of friction (Ff).
Now, let's calculate the forces:
Weight of the trunk (W) = 324 N
Normal force (N) = m * g * cos(θ)
Force of friction (Ff) = μ * N
Since the trunk is sliding with constant velocity, we know that F (the force we exert) must be equal to Ff (the force of friction), but in the opposite direction.
Thus, F = -Ff = -μ * N
Now, let's substitute the values and calculate:
Weight of the trunk (W) = 324 N
m * g * cos(θ) = m * 9.8 * cos(19 degrees)
Force of friction (Ff) = 0.322 * m * 9.8 * cos(19 degrees)
Therefore, the force required to slide the trunk up the inclined plane with constant velocity is F = -0.322 * m * 9.8 * cos(19 degrees), and the direction of the force is opposite to the direction of motion, which is up the incline.