Determine the magnitude and direction of the resultant F_R= F_1 + F_2+ F_3 of the three forces by first finding the resultant F' = F_1 + F_2 and then forming F_R = F' + F_3.

Image from: Hibbeler, R. C., S. C. Fan, Kai Beng. Yap, and Peter Schiavone. Statics: Mechanics for Engineers. Singapore: Pearson, 2013.

#### Solution:

### Let us first draw the vector components excluding F_3 as follows:

### Note that we also converted the angles to degrees, thus 36.87^0 is \sin^{-1}(\frac{3}{5}).

### Now, we will calculate F' using the law of cosines. (Forgot the law of Cosines?)

### (F')^2= 20^2+30^2-2(20)(30)cos73.12^0

### F'=\sqrt{20^2+30^2-2(20)(30)cos73.12^0}

### F'=30.85N

### Now, we can use the law of Sines to figure out \theta. (Forgot the law of Sines?)

### \frac{30.85}{\sin 73.13^0}=\frac{30}{70^0-\theta}

### \theta=1.47^0

### Let us now draw the vector components with F_3, F_R and F' as follows:

### Again, we can use law of cosines to figure out F_R.

### F_R=\sqrt{30.85^2+50^2-2(30.85)(50)cos1.47^0}

### F_R=19.18N

### And now, we can use law of sines to figure out \phi.

### \frac{19.18}{\sin 1.47^0}=\frac{30.85}{\sin \phi}

### \phi =2.37^0

###### This question can be found in Engineering Mechanics: Statics (SI edition), 13th edition, chapter 2, question 2-19.