Biophysics Problem 3

A force acts at \(\mathrm{30^\circ}\) to the horizontal. Its horizontal component is known to be \(\mathrm{300 \;N}\). What are its magnitude and its vertical components?

First make a diagram

horizontal and vertical components
 

Now write the trigonometric relation for the cosine of the \(30^\circ\) angle and solve for the magnitude of the force.

Don't forget to consider the units.

Calculation:

\(\mathrm{\cos 30 = \frac{adjacent}{hypotenuse} \\ 0.866 = \frac{200\; N}{F} \\ Therefore \; F = 200 \; N /0.866 \\ = 231\; N}\)

 Note that the answer is incomplete without units ('\(N\)' for Newtons).

Now we have to get the vertical component of \(F.\)

Use the appropriate trigonometric relation to calculate the magnitude of this component.

You could get this part by using either the sine or tangent of the angle.

\(\mathrm{\sin 30 = \frac{opposite}{hypotenuse} \\ 0.5000 = \frac{vertical \; component}{231 \; N} \\ \therefore \; vertical \; component = 0.5000 \times 231 \; N \\ = 116 \; N}\)

\(\mathrm{\tan 30 = \frac{opposite}{adjacent} \\ 0.577 = \frac{vertical \; component}{200 \; N} \\ \therefore \; vertical \; component \; = 0.577 \times 200 \; N \\ = 116 \; N}\)