# Force vectors

1. Sep 7, 2006

### wku_tops

Please look at this pic for the problem.
I have drawn the freebody diagrams correctly.

I solved for F' to be 14.07 N, but I cant solve Fr. I think the key to the problem is the law of sines. According to autocad the angle opposite Fr is 13 degrees, but how do I solve for that myself without autocad?

Here's some work ive done:

90-20=70 degrees

F’2=(50 N)2 + (20 N)2 -2(50 N)(20 N)cos(70)
F’=47.07 N

47.07 N 50 N
----------= -------- Θ=86.46 degrees
Sin 70 sin Θ

90-86.46= 23.53 degrees

Last edited: Sep 7, 2006
2. Sep 7, 2006

### wku_tops

Basically, i just need to now how autocad computed one of the angles in the diagram to be 13 degrees.

3. Sep 8, 2006

4. Sep 8, 2006

### HallsofIvy

Staff Emeritus
F2 and F3 "added" tail to head, form a triangle having adjacent sides of lengths 20 and 50 and angle 90- 20= 70 degrees between them. By the cosine law, the length of the third side is given by c2= 202+ 502- 2(20)(50)cos(70)= 2216 so c= 47N as you got. Further the angle that vector makes with the positive x-axis is given by the sine law:
$$\frac{sin \theta}{20}= \frac{sin 70}{47}$$
so $\theta$= 24 degrees.
F1 isn't given as an angle but apparently forms a "3-4-5" right triangle. It's not difficult to calculate that the angle F1 makes with the negative x-axis is given by $tan(\theta)= \frac{3}{4}$ so $\theta$= 37 degrees.
Putting F' and F1 "tail to head" gives us a triangle with adjacent sides of lengths 30 and 47 and angle between them 37- 24= 13 degrees. Now use the cosine law again: The opposite side is given by
$$c^2= 30^2+ 47^2- 2(30)(47)cos(13)$$
You can use the sine law again to find the direction.

5. Sep 8, 2006

### andrevdh

I agree with your results for F'.

The diagram with the problem indirectly indicates the angle of F1 with the 3,4,5 triangle. This means that the angle that it makes with the x-axis can be calculated with

$$arctan(\frac{3}{4})$$

Last edited: Sep 8, 2006