A 22.0-kg box rests on a frictionless ramp with a 14.1° slope. The mover pulls on a rope attached to the box to pull it up the incline. If the rope makes an angle of 38.4° with the horizontal, what is the smallest force F the mover will have to exert to move the box up the ramp?

Answer 1

Since there is no friction, all you have to do is balance the component of the tow-rope force in the uphill direction against the component of the weight in the downhill direction.

Draw yourself a figure. That should help show you what angle sines and cosines to use.

Further assistance will be provided if you show your work.

Answer 2

say that this young whippersnapper pulls the rope so that it is parallel to the incline; and you are given an acceleration of the box. I'm looking for the tension in this rope. Could someone please help me here by setting this up for me? I'm confused with the acceleration value here for no good reason whatsoever. I seem to get my y components added up from the normal force and the mg force, then I get my x component from the mg but I have nothing to combine it with since I am solving for T. the values I have here for my completeley different yet similar problem are m=168kg, a=.8m/ss, and my theta=30 degrees. If someone could assist me here, I would be more than appreciative.