if I want to push the bowling ball in a direction which is not parallel to the lane, then I need to make a Projection on the axes (the x axis is parallel with the width of the lane and the y axis is parallel with the length of the lane) ... using the Newton's laws of motion $\sum \vec F = m \vec a$ ,when I projected the force(which is the friction force) and the acceleration I got this:

$$\ -F_x \cos(\theta) = m\cdot a_x\cdot \cos(\theta)\quad \text{ and }\ -F_y\cdot \sin(\theta) = m\cdot a_y\cdot \sin(\theta)$$

by canceling out what is similar it gave me the same relation if the ball was moving forward with no angle

$$\ -F_x = m\cdot a_x \quad \text{and} \quad \ -F_y = m\cdot a_y$$

why did I get that ?

  • $\begingroup$ Where did those negative signs come from? $\endgroup$
    – G. Smith
    Jun 1, 2021 at 6:10
  • $\begingroup$ $F_x=F\cos\theta$ is already the $x$-component, etc. It doesn’t need to be multiplied by another $\cos\theta$. $\endgroup$
    – G. Smith
    Jun 1, 2021 at 6:13
  • $\begingroup$ Your last two equations don't say the ball is moving forward without any angle, that would only be true if $a_x$ was $0$. $\endgroup$ Jun 1, 2021 at 6:17
  • $\begingroup$ @G.Smith I put the negative sign because the force is the friction force and when I projected it, it was in the opposite direction comparing to the axes direction, $\endgroup$ Jun 1, 2021 at 7:01
  • $\begingroup$ @G.Smith , oh I thought $ F_x $ is just a symbol $\endgroup$ Jun 1, 2021 at 7:03

1 Answer 1


You have the $\cos(\theta)$ and $\sin(\theta)$ doubled.

Start by simply ackowledging that you need the force component along the x-axis without plugging anything in yet:

$$F_x = ma_x.$$

Next, with $a_x$ being the wanted unknown and $F_x$ being a needed unknown, find an expression for $F_x$. Such an expression might be set up trigonometrically as $\cos(\theta)=\frac{F_x}{F}$:

$$F\cos(\theta) = ma_x.$$

Note how this includes $F$ and not $F_x$. And we here only have one cosine term. From this you can easily solve for $a_x$.

Same approach along the y-axis and then you find $a_y$, finalising you acceleration vector.

  • $\begingroup$ I think I understood, so if I want to find the a and I have the value of F I use $Fcos(θ)=ma_x$ and the same for y axis, and if I want to find the F and I have the value of a I use $F_x=macos(θ)$ and the same for y axis, is that right? $\endgroup$ Jun 1, 2021 at 7:17
  • $\begingroup$ @RamaRanneh Yes, that is correct. Whether to use sine or cosine depends on the exact setup. You ought to draw out an imagined right-angled triangle every time to be sure which to use. When you know that, then you can set up the equations as you've outlined in the comment. $\endgroup$
    – Steeven
    Jun 1, 2021 at 7:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.