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I'm fairly new to physics (high school) but really enjoy. I recently came across a problem of my own that I tried to solve: what is the best angle to pull an object at? I derived $F=F\cos \theta-\mu mg-F\sin\theta$ after deciding mass would be a constant and canceling out all the masses leaving me with just accelerations in stead of forces. But I came across a very peculiar graph that showed acceleration shooting off to infinity at $2\tan^{-1}(u)$, but if this was true it would that mean it would take no force to accelerate something at that angle. This obviously can't be right so I was looking for some explanation.

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  • $\begingroup$ A little diagram would replace a 1000 words here. $\endgroup$ – Gert Nov 25 '15 at 22:51
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You arrived at almost correct answer. I get net horizontal accelerating force as: $$ F_x(\theta) = F_{pull}(\cos\theta+\mu\sin\theta) - \mu m g $$

You are just missing the $\mu$ before the sin and it also has a different sign. Double check your diagram, especially the orientation of forces. But nevertheless you still seem to arrive at almost correct outcome.

Let's just say we want find $\theta$, so that $F_x$ is maximised (because maximal horizontal force will yield maximal horizontal acceleration, which is what we want to achieve). So then let's take a derivative with respect to theta: $$ \frac{dF_x}{d\theta} = F_{pull}(-\sin\theta+\mu\cos\theta) $$

Placing this derivative equal to zero gives: $$ \mu\cos\theta = \sin\theta \\ \\ \theta = \arctan\mu $$

Which actually makes perfect sense! For $\mu = 0$ you get $\theta = 0$, so with frictionless surface the obvious best direction of pull would be parallel to the surface. With increasing $\mu$ the optimal angle increases somewhat, because it's suddenly optimal to sacrifice some of the x component of pulling force, to lift the thing a little and thus reduce drag. But if you look at properties of arctan function, optimal theta will never exceed $90^o$, and that also makes perfect sense. You will be pulling more and more upwards, but you will always retain some horizontal/forward component.

Of course, this solution is only valid when the vertical component of the pulling force is smaller than the gravitational force on the object. Otherwise you will lift it from the surface completely.

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  • $\begingroup$ This makes sense since tan($)=u $\endgroup$ – JD Brown Nov 26 '15 at 0:02
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Drawing a free body diagram, the following equations can be found:

$F_y = F_\text{pull}\sin(\theta) + F_\text N - mg$

$F_x = F_\text{pull}\cos(\theta) - F_\text R$

Stipulations on variables:

  • $F_\text N$ (normal force) must be positive; the ground cannot exert an attractive force.
  • $F_\text R$ (frictional force) must be less than or equal to $\mu$ (coefficient of kinetic/static friction depending on what you're examining) times the normal force.
  • $F_y$ should probably be zero; otherwise you'd be lifting up the object or crushing it into the ground!
  • If you're looking for constant velocity, $F_x$ is also zero.

Solve the equations... and you'll get something.

Now, this only holds true when $F_\text N$, $F_\text R$ match the above equations. So you may be looking at a case where there is negative normal force, or higher than possible frictional force. Verify that your solution obeys constraints that aren't applied through the equations alone.

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