Determining Resultant of Concurrent Forces: Law of Cosines

[RingWraith2086]

A 60-N force is acting at 30° east of north and a second 60-N force acting in the direction 60° east of north are concurrent forces. Determine the resultant of the forces.

Using the Law of Cosines:
c²=a²+b²-2ab(CosC)
c²=13435.38
c=115.91 N for the magnitude of the resultant

Then, for the direction, I have no idea what to do.

Can someone check what I have done so far and explain how to get the direction? Thanks...

 

[cookiemonster]

I've always liked to separate 2d vectors into 1d components, which then add nicely and algebraically. But to keep in line with your method, the next step would be to use the law of sines.

$$\frac{\sin A}{a} = \frac{\sin B}{b}$$
where A is the angle opposite side a and B is the angle opposite side b.

But if you do it this way, be sure to draw a picture! You'll need to add something to the angle you get to get it to work out properly.

cookiemonster

 

[M98Ranger]

Your calculation for the magnitude of the resultant force seems correct. To determine the direction of the resultant force, you can use the Law of Sines. According to this law, the sine of an angle in a triangle is proportional to the length of the opposite side. In this case, the resultant force is the opposite side to the angle we are trying to find. So, we can set up the following equation:

SinA/a = SinC/c

Where A is the angle we are trying to find, a is the length of the side opposite to angle A (which is 60 N in this case), C is the angle between the two given forces (60°) and c is the magnitude of the resultant force (115.91 N).

Substituting the values, we get:

SinA/60 = Sin60/115.91

Cross multiplying and solving for SinA, we get:

SinA = (60 x Sin60)/115.91
SinA = 0.51
A = Sin^-1(0.51)
A = 30.96°

Therefore, the direction of the resultant force is 30.96° east of north.

To verify this result, you can also use the Law of Cosines again, but this time to find the angle between the resultant force and one of the given forces.

c² = a² + b² - 2ab(CosC)
Where c is the resultant force, a is one of the given forces (60 N) and b is the other given force (60 N).

Substituting the values, we get:

115.91² = 60² + 60² - 2(60)(60)(CosA)
13435.38 = 7200 - 7200(CosA)
CosA = (7200-13435.38)/(-7200)
CosA = -0.25
A = Cos^-1(-0.25)
A = 104.48°

Since we know that the angle between the resultant force and one of the given forces is 60°, we can subtract 60° from 104.48° to get the direction of the resultant force, which is 44.48° east of north. This is close to our previous result of 30.96°, which confirms the accuracy of our calculation.