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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 ?

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  • $\begingroup$ 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 $0$. $\endgroup$
    – Akshat Sharma
    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, $\endgroup$
    – Rama Ranneh
    Jun 1, 2021 at 7:01
  • $\begingroup$ @G.Smith , oh I thought $ F_x $ is just a symbol $\endgroup$
    – Rama Ranneh
    Jun 1, 2021 at 7:03

1 Answer 1

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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.

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  • $\begingroup$ 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? $\endgroup$
    – Rama Ranneh
    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

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