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In the usual analysis, work needed to lift an object of mass (m) a certain distance (h) is the same whether it is through lifting the object straight up (W=mgh) or pushing it up a ramp with length d and angle A (W = mgsinA x h/sinA = mgh).

However, this does not account for the work needed to move the said object the ground distance (h/tan A), no?

For example, in a 30/60/90 degree triangle scenario with height of 1 meter ramp 2 meter, and ground distance square root of 3 meter, the work needed to move an object of 2 kg horizontally first would be 2gxsquare root of 3 and the work to move the object up 1 m would be 2g. The combined work would be 2g (square root of 3 + 1), no?

The usual analysis of moving that object up the 2m ramp is 2gx(1/2) divided by 1/(1/2) = g x 2 = 2g. These two methods don't add up. What am I missing?

Is the usual ramp analysis wrong, in that mgxsin A or 2gx(1/2) or g can't the force needed to move the object up the 2 meter ramp? Should that force be (1+square root of 3)g instead?

What am I missing? Thank you so very much for helping out!

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    $\begingroup$ What is the work necessary to move an object horizontally a distance $d$? $\endgroup$ – BowlOfRed Mar 7 '17 at 4:21
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    $\begingroup$ I think I got it. Work is NOT necessary to move the object horizontally assuming constant velocity and absence of friction. This also assumes that negligible acceleration and deceleration to get the object to start moving and stopping over the horizontal distance. Therefore, there is no horizontal work component, right? $\endgroup$ – h92475 Mar 7 '17 at 5:28
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When an object mass $m$ is moved up a slope whose angle with the horizontal is $\alpha$ it is subjected to three forces.

  • The weight of the object $mg$ downwards.
  • The normal reaction due to the slope which is at right angles to the slope.
  • A force up the slope $mg\sin \alpha$.

Let the vertical height the object is raised be $h$ which means that the force up the slope moves a distance $\dfrac{h}{\sin \alpha}$.

No work is done by the normal reaction as it is at right angles to the direction of motion of the object.

The work done by the force up the slope is $mg \sin \alpha \, \dfrac{h}{\sin \alpha}=mgh $ which is independent of the angle of the slope.


Now look at the motion in a different way including the horizontal displacement of the object as you have.
What you have omitted is the fact that the normal reaction will now do some work as now vertical and horizontal components of the normal reaction are displaced vertically and horizontally.
Using this method you find that the net upward vertical force on the object is $mg$ and the net horizontal force is $0$.
This means that the total work done on the object is $mgh$ as found using the other method.

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The work is the dot product, not just a simple multiplication. If the same magnitude of force is used in both the cases, but with directions different (one upward and one in direction of ramp), the work will be different. This difference will manifest as the greater-work-done case having more final acceleration of the object, assuming initial velocity is same. Work-energy theorem.

For the work done to be same in both cases, you will find that the force required in one case (ramp) will be higher, as only the vertical component of the force will contribute to the work done (hence the dot product) against gravity. The rest of the force is used to maintain the trajectory in the ramp (constraint, due to the normal force that the ramp provides).

I have assumed there is no friction, as implied from your question. And this analysis should ideally be done without assuming any acceleration, i.e., the force required to exactly counter the gravitational force. That will give the same results with almost similar arguments.

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