An object on an incline with an angle $\theta$ is being pushed at constant speed.

Constant speed implies $a = 0$. Because $F=ma$ and the object must have mass greater than 0.

  1. If I want to find the force of the push, and it is being pushed parallel to the incline then the formula to use is: $F=w\sin\theta$, where $w$ is the object's weight. Why?

  2. If it is being pushed parallel to the floor the formula is $F\cos\theta=w\sin\theta$. Why?

I tried to work them out to find the logic behind the formulas, but maybe I am making too many assumptions. So can someone explain how to arrive at these formulas?

  1. $$\sum F = w\sin\theta-F = ma = 0$$

    This is because of $x^2+y^2=r^2$, so I used that to find F.

  2. $$\begin{align} \sum F &= w\sin\theta + w\cos\theta+ F\sin\theta + F\cos\theta=0\\ \sum F &= w\sin\theta + 0+ 0 + F\cos\theta=0 \end{align}$$

    So because $a = 0$ I make my equations equal 0. Then I say that $w\cos\theta = 0$ because gravity is on the y axis, not on the x axis. Then I say $F\sin\theta = 0$ because the push is parallel to the ground.

  • $\begingroup$ Please mention what the terms mean, it is known generally, but the site demands quality. Specifically, what is w? $\endgroup$
    – Sensebe
    Commented Mar 3, 2014 at 6:43
  • $\begingroup$ Have you tried drawing a picture of the situation described? The reason for the sines and cosines are best understood geometrically. $\endgroup$ Commented Mar 3, 2014 at 6:46
  • $\begingroup$ $w$ is weight, I thought it was generally known in a physics environment. I drew it, that's mostly how I know that a bit of trigonometry was needed. I can copy paste these equations and reuse them all I want, but what is the point of that? I want to know... $\endgroup$
    – Mike John
    Commented Mar 3, 2014 at 6:52
  • $\begingroup$ @MikeJohn $F_g$ (for "gravitational force") is generally recognized as weight, $w$ is not, for some reason. Still that's no problem; you can just edit that into the question. I'll do it for you this time since I'm making other edits anyway. $\endgroup$
    – David Z
    Commented Mar 6, 2014 at 18:30

1 Answer 1


a) The reason its $F=mg\sin\theta$ is because that is how much of the weight is pointing down the slope. This makes since since when $\theta=0$ we would have to apply $0N$ so that it doesn't slide down (or remain at constant speed), and at $\theta=90$ we would have to apply $mg$ to keep it from sliding down (essentially hold the entire mass).

b) Same idea as (a) but now our applied force is parallel to plane (or $90-\theta$ from plane). The amount of the applied force that is in the direction of the plane. Hence to make the acceleration zero (in the x direction which is down the slope) you need to use newtons laws

$$\begin{align}\sum F_{x}=&0 \\\ F_{applied}\cos\theta-mg\sin\theta=&0 \\\ F_{applied}\sin\theta=&mg\cos\theta \end{align} $$

Which is what you had. Your problem is that you were taking both the x and y directions at the same time. Break them up into two different parts like we do in kinematics. I just did the $x$ direction but I am thinking of taking it down...


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