Mark H. answered • 12/08/19
Tutoring in Math and Science at all levels
Draw a picture...
The forces acting on the box are gravity, friction, and the push by the mover.
Start by resolving the force of gravity into components:
--"normal" to the surface of the ramp (i.e. at 90 degrees)
--tangential to the ramp surface.
The normal force is m*g*cos(30), and the tangential force is m*g*sin(30) (Use your drawing to make sure everything makes sense.
Since the box does not move perpendicular to the ramp, the normal force and the reaction force from the ramp are in balance. The friction force, which is tangential, is the product of the normal force and the coefficient of friction.
Along the ramp, we have 3 forces: The tangential force (from gravity), the friction force, and the push supplied by the mover. The "push" is the sum of the tangential and friction forces.
Work can be found from the product of force and distance. The total work done by the mover is the "push force" times the distance (2.1 meters)
The work done by the mover is the SUM of the work done by friction, and by gravity. First, get the "gravity work" by finding the vertical distance (2.1 * sin (30)) and multiplying with the force of gravity found earlier. The work done by friction is the difference between the work done by the mover and the work done by gravity.
ADDENDUM Dec 10
Using 10 for the acceleration of gravity, the weight of the box is 200N.
This resolves into the normal force, Fn = 200 cos(30) = 173
and the tangential force Ff = 200 sin (30) = 100
The friction force is the normal force time the coefficient of friction: Ff = 0.3 * 173 = 52
the applied force to push the box up the ramp is Ft + Ff = 152
Bob M.
Could Fnet,x be represented as: Fa-Fgx-Ff,k? Thus, Fa=ukFn+fgx which is Fa=(0.3*200)+200sin30=160N12/09/19