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My Issue

I have the mass, the horizontal acceleration and a force that acts on a body. The force forms an angle $\alpha$ of unknown degrees with the x-axis. I would like to know how I can find the angle $\alpha$ formed by the force and in which way the angle is connected with the horizontal acceleration.

What I tried

This formula calculates the horizontal acceleration given the force $F$, the mass $m$ and angle formed by the force $\alpha$:

$$a_x = \frac{F}{m} \cos(\alpha)$$

By inverting it, I can easily calculate the cosine of the angle $\alpha$:

$$cos(\alpha) = a_x\frac{m}{F}$$

However, I'm not sure about how can I revert the cosine of the angle to find its degrees.

Questions

What am I missing? In which way is the angle $\alpha$ connected with the horizontal acceleration?

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  • 3
    $\begingroup$ The reverse operation of $\cos$ is $\arccos$. Apply that to both sides and you get $\alpha$. $\endgroup$
    – DK2AX
    Commented Mar 5, 2015 at 15:57
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    $\begingroup$ This question is confusing. If the force acts completely in the x direction it makes no sense to talk about an angle other than 0 $\endgroup$
    – Sean
    Commented Mar 5, 2015 at 16:27
  • $\begingroup$ European and American punctuation on units. $\endgroup$
    – Jiminion
    Commented Mar 5, 2015 at 16:29
  • $\begingroup$ The question is not properly stated. Is the TOTAL force 10N? $\endgroup$
    – Jiminion
    Commented Mar 5, 2015 at 16:31
  • 1
    $\begingroup$ Yes, it is. Sorry for that. $\endgroup$
    – Cesare
    Commented Mar 5, 2015 at 16:45

1 Answer 1

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Assuming the total force is $10~\text{N}$, the $7~\text{kg}$ mass is accelerated by $1.0~\text{m/s^2}$, so it is experiencing $7~\text{N}$ force in x-direction. Therefore, $F_x = .7F$. The angle is roughly $45^\circ$.

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