Forces - calculating the net force

[The Vinh]

@The Vinh
i am aware that using Coordinates system axis is easier, however the lesson that i am doing requires that i learn their method, and it involves the cosine law etc..
they don't provide any other ways to solve and they are expecting me to solve it using their method.

Oh, ok, so if you must use it then magnitude and angle are easy to find but the real problem is the direction of net force.

 

[The Vinh]

Oh and you have a really " cool " teacher

 

[haruspex]

what do you mean?

You quote the sine formula, but you did not say which angle in the picture is the A in the formula.
It needs to be opposite the side you divide the sine by.
Which angle do you need to find to answer the question?

 

[alexandria]

@haruspex

This was the previous diagram i had made, it forms a triangle instead of a parallelogram. I labelled angle A .
SinA / 38.0 N = Sin55° /32.9 N

 

[haruspex]

SinA / 38.0 N = Sin55° /32.9 N

Yes, that looks good.

 

[alexandria]

Ok so here is my final answer38.

c = (a^2 + b^2 – 2abcosC) ½

c = ([32.0 N]^2 + [38.0 N]^2 - 2[32.0 N] x [38.0 N]cos55°) ½

c = 32.9 NSinA/a = SinB/b = SinC/c

SinA / 38.0 N = Sin55° /32.9 N

A = 72 degrees (approx.)

net force = 32.9 N (North 72 degrees East)

does this look right?

 

[haruspex]

(North 72 degrees East)

Draw that direction on your diagram. Does it look right?

 

[alexandria]

is the correct direction (East 72 degrees North)

 

[haruspex]

is the correct direction (East 72 degrees North)

I didn't say your answer was wrong, just asked if it looked reasonable. You should always check that where possible.

 

[alexandria]

oh, ok, so i based the direction of this diagram.

the force is pulling the object 72 degrees east of north

 

[haruspex]

oh, ok, so i based the direction of this diagram.
View attachment 97361
the force is pulling the object 72 degrees east of north

Ok!

 

[alexandria]

thank you so much for your help, i really appreciate it :)