Which Ramp Requires Less Work Taking Friction into Account?

[teggenspiller]

Homework Statement

Two ramps, one steeper than the other, are available to move a heavy crate from the ground onto the bed of a truck. In the absence of friction, it takes more force to push the crate up the steeper ramp. Does this mean it also takes more work to raise the crate onto the bed of the truck along the steeper ramp? </b>Taking friction into account, which ramp requires less work for pushing the crate on the truck? </b>

Homework Equations

The Attempt at a Solution

IT all depends. The angle of the inclination actually has nothing to do with the amount of work needed to lift an object onto it, only the direction of the force that is causing the displacement has an effect.

However, assuming that more force is required to push an object up the steeper hill, and since force, displacement and the angle between the force and displacement vector are directly proportional, the more force would cause the entire work (J) to be higher. The angle, however, would not have an effect.

Taking friction into account, the less steep hill requires less work

and that's where I am stuck..

 

[pelmel92]

actually ignoring friction, as long as you're raising both the same height (in this case, from the ground to the truck bed) the **work will be the same regardless of the path** (i.e. the steepness of the ramp). work is force times distance, but while the force required to lift something up a steeper (or maximally, a vertical) ramp/path, the distance is shorter so it balances out.

In a case where friction cannot be ignored, the path length matters... for example, in your problem, friction only affects the object while it is in contact with the ramp. more friction means more work, so you want to minimize the distance you have to spend on the ramp... since steep ramps are shorter than relatively long flat ramps, an object moving up a steep ramp will be less affected by friction, and thus need less work to raise a given height.

 

[supratim1]

please note that the friction force also depends on the Normal reaction, which depends on cosine of the angle of inclination.

f = mg.cos(theta). So steeper the slope, lesser is the frictional force.