# eww...vectors...

• May 30th 2006, 06:58 PM
CONFUSED_ONE
eww...vectors...
here's the problem:

The magnitude and direction of two forces acting on an oject are
-70 lbs, S56[degree]E and
-50 lbs, N72[degree]E.
find the magnitude, to the nearest hundredth of a pound, and the direction angle, to the nearest tenth of a degree, of the resultant force.

i did this problem like 5 times! i keep getting different answers.
the answer in the back of the book is 108.21 lbs, S77.4[degree]E.
could someone explain it to me?

tthhhaannnk you, even for reading my post. thanks.
• May 30th 2006, 08:47 PM
Soroban
Hello, CONFUSED_ONE!

I assume you're familiar with trig forms for vectors.

Quote:

The magnitude and direction of two forces acting on an object are:
70 lbs, S 56° E and 50 lbs, N 72° E.

Find the magnitude, to the nearest hundredth of a pound,
and the direction angle, to the nearest tenth of a degree,
of the resultant force.

Answers: 108.21 lbs, S 77.4° E
Code:

              Q       |      *       |    *       |72°*       | * 18°     O * - - - - - - -       | * 34°       |56°*       |    *       |      *       |        *       |          *                   P

The object is at $O$.
Rewrite the bearings (angles) relative to the positive x-axis.

The forces are: $\overrightarrow{OP} = [70\cos(-34^o),\,70\sin(-34^o)] = [70\cos34^o,\,-70\sin34^o]$

. . . . . . . and: $\overrightarrow{OQ} = [50\cos18^o,\,50\sin18^o]$

The resultant vector is the sum of these vectors: $\overrightarrow{OR} \:= \:\overrightarrow{OP} + \overrightarrow{OQ}$

. . $\overrightarrow{OR} \;= \;[50\cos18^o + 70\cos34^o,\,50\sin18^o - 70\sin34^o]$

. . $\overrightarrow{OR}\;\approx\;[105.585,\,-23.693]$

The magnitude is: $|\overrightarrow{OR}| \;= \;\sqrt{105.585^2 + (-23.693)^2} \;\approx$ 108.21

The resultant makes an angle $\theta$ with the positive x-axis where:
. . $\tan\theta \:= \:\frac{23.693}{105.585} \;=\;0.224397405\quad\Rightarrow\quad\theta \,\approx \,12.6^o$

Therefore, the direction is: S 77.4° E