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I have a problem that involves pushing an object with a:

  • Mass of 80 kg
  • Up a sloped plane (angle of incline is the unknown)
  • With an applied force of 700N
  • At an acceleration of 4.88 m/s^2
  • With a coefficient of friction 0.07 acting against it

My goal is to find the angle that allows for that acceleration, using the 700N of force. I first flipped the axis to make the x-axis go up the inclined plane. So I can find the angle using the sum of the x-forces.

The sum of the x-forces (or m*a) = Applied force - (Weight force y component * coefficient of friction) - (Weight force of the x component).

I rewrote this as m*a = Applied - (coefficient of friction * w * cos theta) - (w * sin theta).

After trying for solve for the angle, I got stuck at (Applied - m*a) / w = (coefficient of friction * cos theta) + sin theta.

Is there a better way to solve this, without ending up with two different trig functions?

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closed as off-topic by John Rennie, Qmechanic Sep 27 '16 at 23:04

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Construct an auxiliary right-angle triangle with angle $\alpha$, opposite side $\mu$ and adjacent side $1$. The hypotenuse is hence $\sqrt{1+\mu^2}$, and therefore

$$\mu \cos\theta+\sin\theta=\sqrt{1+\mu^2}\sin\alpha\cos\theta+\sqrt{1+\mu^2} \cos\alpha\sin\theta$$ $$=\sqrt{1+\mu^2}\sin(\theta+\alpha)$$ where $$\alpha=\tan^{-1}\mu$$

Can you take it from here?

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