Forces - calculating the net force

[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 :)