Determine the resultant force of the inclined plane

Question

So I was given this problem for my statics subject. I'm just having problems getting the x and y components of the force $120$ lb. I know that

$60$-lb Force

$$F_x = (60\mbox{ lb})\cos20^\circ$$

$$F_y = (60\mbox{ lb})\sin20^\circ$$

$80$-lb Force

$$F_x = (80\mbox{ lb})\cos95^\circ$$

$$F_y = (80\mbox{ lb})\sin95^\circ$$

but how do I solve the $120$ lb force?

My friend said that it's $5^\circ$ I've been trying to draw it several times but I still don't understand why its $5^\circ$

$120$-lb Force. But why $5^\circ$ ?

$$F_x = (120\mbox{ lb})\cos5^\circ$$

$$F_y = (120\mbox{ lb})\sin5^\circ$$

Answer 1

To confirm the results you obtained so far, from the positive $x$ axis (horizontal direction to the right), you rotate counterclockwise $20$ degrees to the direction of the $60$-pound force, then another $75$ degrees counterclockwise to reach the direction of the $80$-pound force, so the $80$-pound force is $95$ degrees counterclockwise from the positive $x$ axis.

The angle between the $80$-pound force and the $120$-pound force is $90$ degrees. One way to see this is to observe that each one is marked as being $75$ degrees counterclockwise from some other line in the figure, and the two lines those $75$-degree angles are measured from are at right angles to each other.

So from the positive $x$ direction to the direction of the $120$-pound force is a total of $(95 + 90)$ degrees, or $185$ degrees, counterclockwise from the positive $x$ axis.

Your friend may have observed that the line $185$ degrees counterclockwise from the positive $x$ axis can also be considered to be rotated $5$ degrees. But this ignores the direction of the force. You can use $120 \cos 185^\circ$ and $120 \sin 185^\circ$, or you can use $-120 \cos 5^\circ$ and $-120 \sin 5^\circ$ (either way, you get the same numbers), but $120 \cos 5^\circ$ and $120 \sin 5^\circ$ is incorrect; it is a force in exactly the opposite direction from what you want.

Answer 2

Suppose the x-axis is parallel to the ground with the positive x-direction pointing to the "right". Furthermore suppose that the positive y-axis points "up" and is perpendicular to the x-axis.

We wish to find the components of the 120 lb force (indicated in the OP's figure) with respect to the coordinate axes we just defined. We can do this if we know the angle this force makes with the positive x-axis. Lets call this angle $\theta$.

From the figure it is obvious that $\theta = 20^\circ + 90^\circ + \alpha$. Since $\alpha=75^\circ$ we have that $\theta = 185^\circ$. From this we conclude that the components are,

$$ F_x = 120 \mathrm{ lb } \cos(185^\circ), $$

$$ F_y = 120 \mathrm{ lb } \sin(185^\circ). $$

If we wanted to stick with acute angles we could notice that 185 degrees with respect to the positive x axis is the same as 5 degrees below the negative x axis in which case we would write,

$$ F_x = -120 \mathrm{ lb } \cos(5^\circ), $$

$$ F_y = -120 \mathrm{ lb } \sin(5^\circ). $$