# Thread: direction of resultant of forces (vectors & bearings)

1. ## direction of resultant of forces (vectors & bearings)

I've been given the problem: "Find the direction and magnitude of the resultant of forces of 23.75 lb acting S 56.14 degrees W and 48.61 lb acting S 33.86 degrees E.

I have found the magnitude (54.10 lb) but I don't know how to find the direction. I know the answer is S 7.82 degrees (that's what the back of the book says), but I don't know how to get it and I don't understand the textbook's explanation.

2. Originally Posted by mhunn5260
I've been given the problem: "Find the direction and magnitude of the resultant of forces of 23.75 lb acting S 56.14 degrees W and 48.61 lb acting S 33.86 degrees E.

I have found the magnitude (54.10 lb) but I don't know how to find the direction. I know the answer is S 7.82 degrees (that's what the back of the book says), but I don't know how to get it and I don't understand the textbook's explanation.

I don't know your textbook's explanation but it could be like this.

You must draw the figure so that you can see approximately the position of the R.

For the horizontal components, going West is negative and going East is positive.
For the vertical components, going North is positive and going South is negative.

Summation of horizontal components.
H = -23.75sin(56.14deg) +48.61sin(33.86deg) = 7.36178 lbs ----positive, so going East

Summation of vertical components,
V = -23.75cos(56.14deg) -48.61cos(33.86deg) = -53.59850 lbs ----negative, so going South.

Meaning, the resultant R is going S some degrees E.

This "some degrees", or theta, is arctan(H/V)
Since you can see the figure, then disregard the signs of H and V. Remember that the signs are only for directions. Since you know already that R is due SE, just use the absolute values of the H and V-----or their positive values.
So,
theta = arctan(7.36178 /53.59850) = 7.82067 degrees

Therefore, the bearing of R is S 7.82deg E --------answer.

3. Originally Posted by ticbol
I don't know your textbook's explanation but it could be like this.

You must draw the figure so that you can see approximately the position of the R.

For the horizontal components, going West is negative and going East is positive.
For the vertical components, going North is positive and going South is negative.

Summation of horizontal components.
H = -23.75sin(56.14deg) +48.61sin(33.86deg) = 7.36178 lbs ----positive, so going East

Summation of vertical components,
V = -23.75cos(56.14deg) -48.61cos(33.86deg) = -53.59850 lbs ----negative, so going South.

Meaning, the resultant R is going S some degrees E.

This "some degrees", or theta, is arctan(H/V)
Since you can see the figure, then disregard the signs of H and V. Remember that the signs are only for directions. Since you know already that R is due SE, just use the absolute values of the H and V-----or their positive values.
So,
theta = arctan(7.36178 /53.59850) = 7.82067 degrees

Therefore, the bearing of R is S 7.82deg E --------answer.
Thank you for your reply, but you are using terminology that I'm not familiar with and I don't understand your summations.

I have drawn my visual aid to show dotted north/south lines and dotted east/west lines; drew a solid line from origin S 56.14deg W and labeled it 23.75 lbs; drew a solid line from origin S 33.86deg E and labeled it 48.61 lbs (which are at right angles); and drew my hypotenuse - which I found to be 54.10 lbs (according to the answer in the back of the book, this is correct). Then, I used tangent to find both angles inside my newly formed triangle. (I'm not sure why I did that.) I don't understand how to find the direction. Your answer, S 7.82deg E, is close to the answer in the back of the book, S 7.82deg, but you've added E where the book does not. And I don't understand what it means by direction or how to get it.

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# resultant force bearings calculations

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