Is this the proper method to find net force?

[Kmcquiggan]

Homework Statement

Calculate the net force acting on each object indicated in the following diagrams

Relevant Equations

sine law=sinA/a=sinB/b=sinC/c
c^2=a^2+b^2-2abcosC°1/2

c^2=a^2+b^2-2abcosC°1/2
= 32^2 =38^2 - 2(32)(38)cos35(0.5)
= 2468-995.904
= √1472.096
= 38.36 N

tanΘ= 32/38 tan= 0.842 = 40°
This is my attempt, I am not sure that I am understanding these formula right

 

[Doc Al]

c^2=a^2+b^2-2abcosC°1/2
= 32^2 =38^2 - 2(32)(38)cos35(0.5)

Looks like you're trying to use the law of cosines. (Correct the formula -- where did you get that 1/2?) To do that, you'll need to define your triangle and the angle between the sides of that triangle. (It's not 35 degrees.)

You can also use the method of components to find the net force.

 

[Kmcquiggan]

I miss understood the formula that i was reading so I realize it doesn't need to be in there. As for the angle would I be adding the 35 degrees to the 90 angle or would I be using the 55 degrees? I am still struggling on which angle to be usuing

 

[Doc Al]

Why don't you draw the triangle that would represent $\vec{a} + \vec{b}$.

 

[Kmcquiggan]

I did and I am thinking that it needs to be the angle of 125 degrees with the other angles being 20 and 35 degrees. I could be doing this right or completely wrong, but I am doing this as correspondence and I have no professor to ask questions too so if i am way off can you explain where I am going wrong?

 

[Doc Al]

Please post a diagram of that triangle.

 

[Kmcquiggan]

 

[Doc Al]

That is not the correct triangle. Do this: Keep the 32 N force where it is. Move the other force so that its tail touches the head of the other. (You can slide the force vector around as long as you don't change its direction or magnitude.) That way, you'll have a triangle representing the sum of those forces. Try it!

 

[Kmcquiggan]

ok thank you