If we have an object on a frictionless inclined bench, why does the minimum force applied to keep it steady have to be parallel to the incline (or perpendicular to the reaction force)?

Object on an inclined bench

In other words, why does the angle θ in the image have to be 0 in order for the force to be minimum?

Consider we know the weight of the object and the reaction of the incline.

  • 2
    $\begingroup$ Otherwise part of the force is used to lift the object or force it into the bench ... $\endgroup$ – Solar Mike Sep 27 '17 at 11:40
  • $\begingroup$ By that way, we let the reaction force deal with the component of the weight that is perpendicular to the incline then. It makes sense. Thank you! $\endgroup$ – Elruz Rahimli Sep 27 '17 at 11:49
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    $\begingroup$ Exam question part B: Is this still true if the surface is not frictionless? $\endgroup$ – agentp Sep 27 '17 at 20:32

The minimum force is in the direction the object would move without that force. Clearly the object would move along the inclined plain, down and right in your diagram.

Put another way, its the component of the force along the direction of travel that matters. More mathematically, it's the dot product of the force with the acceleration unit vector, with negative values contributing to cancelling the acceleration, and positive values enhancing it. The dot product is -1 when the force is exactly opposite the acceleration direction. A sideways force would do nothing, and has a dot product of 0. A force pushing down the incline has a dot product of 1, and just like the math says, would make the object go down the incline faster.

  • $\begingroup$ Great answer, I went the calc way thinking about this as a limit equation, As theta approaches zero, the force components in the direction of action (i.e. up the slope) maximizes and the normal force becomes zero. $\endgroup$ – Diesel Sep 27 '17 at 16:23

The force you're applying would have an effect that counteracts the body's tendency to move. That effect will be calculated by [applied force] $\cos \theta$.

If $\theta$ is not zero, $\cos{\theta}$ would be less than 1, so you would have to apply a larger force to produce the same result. For example, 10 degrees for $\theta$ would equate to a force parallel to the plane of 0.9848 times the applied force. You would need a larger applied force (approximately 1.015 times) to produce the same effect.

When $\theta = 0$, $\cos{\theta} = 1$.

  • $\begingroup$ Thanks, Jem. I took a bit of a shortcut there to avoid messing with character maps and ASCII. $\endgroup$ – ItWasLikeThatWhenIGotHere Sep 28 '17 at 7:18
  • $\begingroup$ Ah. I see how it's done. Got the one in the first paragraph, too. $\endgroup$ – ItWasLikeThatWhenIGotHere Sep 28 '17 at 7:20

Since I posed the question in comment, here is how to treat the friction case, and of course at the end it reduces to the expected friction free result.

Assume the applied force has components $f_x$ parallel (up) the plane and a normal component $f_y$ downward into the plane:

The normal force between block and plane is:

$$f_n = W\cos(t) + f_y$$

The force $f_x$ required to hold against sliding, assuming static coulomb friction with coefficient $u$ is

$$f_x = W\sin(t) - u f_n = W\sin(t) - u (W\cos(t) + f_y)$$

The magnitude of the applied force is then

$$f_{mag} = \sqrt{f_x^2+f_y^2}$$

without showing all the messy steps, setting $d f_{mag} / d f_y == 0$ we can solve for $f_y$:

$$f_y = u W\sin(t)\cdot\dfrac{1 - u /\tan(t)}{1 + u^2}$$


$$f_x = W\sin(t) ( 1 - u\cdot\dfrac{1/\tan(t) + u}{1 + u^2})$$

For the friction free case ($u=0$) this results in the expected $f_x=W\sin(t),f_y=0$ (ie parallel to the plane) but with friction angling the force toward the plane reduces the required force magnitude.

  • $\begingroup$ Is it possible to prove this using differential calculus? $\endgroup$ – Jem Eripol Sep 30 '17 at 0:12
  • $\begingroup$ @JemEripol Not sure what you mean, I think that's what I did. Note for completeness we should check that the extrema I found is a minimum. $\endgroup$ – agentp Oct 1 '17 at 16:15
  • $\begingroup$ Oh, I did not see the differential part. Thank you. $\endgroup$ – Jem Eripol Oct 2 '17 at 0:07

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