# Finding the Magnitude and Direction of Force one with the Resultant Given

Determine the magnitude and direction theata of F1 so that the resultant force is directed vertically upward and has a magnitude of 800N.

I have difficulty of where to start the problem? Do I sum up the forces of the x axis with unknowns in it and forces of the why axis with the unknown forces...where should I start....and how do I start solving the problem?

Please someone help me with this....

Thanks,

Nacha

 ebaines Posts: 10,055, Reputation: 5539 Expert #2 Jul 7, 2008, 02:38 PM
Start by breaking the problem in two - consider the components of the forces in the horizontal direction seprarately from the components of the forces in the vertical direction. You want to have $Sigma F_x = 0$ and $Sigma F_y = 800N$. The 600N force that goes up to the left has a horizontal component of 600N * (-4/5) (note that its sign is negative as it goes to the left), and a vertical component of 600N * 3/5. The 400N force going up to the right has a horizontal component of 400N * cos(30), and a vertical componenet of 400* sin(30). The unknown force $F_1$ has a horizontal component of $F_1 * cos( heta)$ and a vertical component of $F_1 sin( heta)$. So summing all the horizontal forces and setting that equal to 0 yields:

$
Sigma F_x = 600(-4/5) + F_1 cos( heta) + 400N * cos(30) = 0
$

From this you can determine the value for $F_1 cos( heta )$.

Then sum the forces in the vertical direction:

$
Sigma F_y = 600*3/5 +F_1 sin( heta ) + 400*sin(30) = 800
$

This gives you $F_1 sin( heta )$. The value of $F_1$ can then be found from:

$
F_1 = sqrt { (F_1 cos ( heta ))^2 + (F_1 sin ( heta ))^2 }
$

And the direction of $F_1$ is:

$
heta = Atan ( frac {F_1 sin ( heta ) } {F_1 cos( heta ) }
$
 Nacha Posts: 4, Reputation: 1 New Member #3 Jul 7, 2008, 03:44 PM
Hi,

Thanks a lot for helping me solve this problem, really it was a great help.

It's good to know that people like you are there for help.

Thanks,

Nacha
 bayley86 Posts: 52, Reputation: 10 New Member #4 Jan 4, 2010, 09:04 AM
A force of 2n acts upon a point 'A' at an angle of 55 degres. From the same point 'A' another force of 5n at an angle of -20 degres is applied. Calculate the resultant force and its direction using trigonometric functions

If anyone can help me with this it would be much appreciated

Thanks

Andy
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #5 Jan 5, 2010, 07:26 PM
Bayley86, you should have started another thread.

Your problem involves 2 forces. So, make a sketch. From that, use the parallelogram law of forces and you'll have a rough idea of how and where the resultant is. From that, you'll be able to find everything.

You'll need the use of the cosine rule to find the magnitude of the resultant. Then, some simple trigonometric ratios to find the angle at which it occurs.
 bayley86 Posts: 52, Reputation: 10 New Member #6 Jan 6, 2010, 07:10 AM
Thanks for your reply I have started a couple of threds now with different questions I didn't no how to use this website at first.

Please could you do an example using different values so I can understand it better would help me a lot.

Andy
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #7 Jan 6, 2010, 08:59 AM
Hmm, I'll take the question that Nacha asked. I'll take the why axis as the line of reference for the measurements of angle. And, I'll take the angle from the why axis to the 600 N force to be 45 degrees for ease of work.

So, we need to find the resultant of the 600 N (45 degrees from why axis) and 400 N (-60 degrees from why axis) forces.

Using parallelogram law of forces, you'll get a figure similar to that:

Well, you will need to find the 'length' (magnitude) of the resultant (double arrowhead). You can find this by using the cosine rule and taking the two sides of the triangle with sides 400 N, 600 N and resultant.

The angle between the 400 N and 600 N is found to be 75 degrees ([45 + 30] degrees, or [180 - (60+45)] degrees)

Now, the cosine rule:

$A^2 = B^2 + C^2 - 2BC\ cos \hat{a}$

A = ?
B = 600
C = 400
$\hat{a}$ = 75

Solving, you'll get A = 629.1 N

Now, to find the angle, I'll use the sine rule.

$\frac{sin\hat{a}}{A} = \frac{sin\hat{b}}{B}$

Here, angle a is 75
A is 629.1
B is 600
Angle b = ?

From the sine rule, you get angle b = 67.1 degrees.

But you know that the angle between the why axis and the 400 N force is 60 degrees and the angle in the triangle is more than that. So, the angle from the why axis is [67.1-60] = 7.1 degrees.

Hence, resultant force = 629.1 N, 7.1 degrees (anticlockwise) from why axis.
 bayley86 Posts: 52, Reputation: 10 New Member #8 Jan 7, 2010, 09:19 AM
I couldn't get the same answers as you on the example but I tried it for the question and I got
75 degres for the angle between 55 and -20
2^2+5^2-2(2x5)cos75=23.824

Last time I used the cosine rule I had to cos-1 the answer but I get a math error
 Unknown008 Posts: 8,147, Reputation: 3745 Uber Member #9 Jan 7, 2010, 09:33 AM
Well, using the cosine rule to find the magnitude of the resultant;

$A = \sqrt{B^2 + C^2 - 2BC\cos \hat{a}} = \sqrt{2^2 + 5^2 - 2(2)(5)\cos (55+20)} = 4.88 N$

Do not forget that the cosine rule in its original form gives the square of the side you're looking for.
 bayley86 Posts: 52, Reputation: 10 New Member #10 Jan 7, 2010, 09:57 AM
Thanks

Is the angle

Sin75/4.88=0.198 degres

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