force of 125 N pulls due west on a body, and a second force pulls N 28.79 W. The resulta force is 212 N. Find the second force and the direction the resultant.
3 answers
Who is "jonnathan"?
suppose we say that vectors
u = force of 125 N pulls due west
v = force of magnitude v pulls N 28.79 W
r = resultant force is 212 N at angle b
expressing those in x-y coordinates,
u+v = r
<-125,0> + <-v sin 28.79,v cos28.79> = <212 cosb, 212 sinb>
<-125,0> + <-0.482v,0.876v> = <212 cosb, 212 sinb>
That means that
-125 - 0.482v = 212 cosb
0.876v = 212 sinb
solving, we get
v = 121.281
b = 30
so,
the 2nd force is 121.281N
the resultant acts at W30N
You can also solve by drawing the diagram. If we label the sides u,v,r and the angles U,V,R, then we have
u = 125
r = 212
U = 118.79°
sinR/r = sinU/u
Now, having U and R, V is easy, since they sum to 180°. Then
v^2 = u^2+r^2 - 2ur cosV
u = force of 125 N pulls due west
v = force of magnitude v pulls N 28.79 W
r = resultant force is 212 N at angle b
expressing those in x-y coordinates,
u+v = r
<-125,0> + <-v sin 28.79,v cos28.79> = <212 cosb, 212 sinb>
<-125,0> + <-0.482v,0.876v> = <212 cosb, 212 sinb>
That means that
-125 - 0.482v = 212 cosb
0.876v = 212 sinb
solving, we get
v = 121.281
b = 30
so,
the 2nd force is 121.281N
the resultant acts at W30N
You can also solve by drawing the diagram. If we label the sides u,v,r and the angles U,V,R, then we have
u = 125
r = 212
U = 118.79°
sinR/r = sinU/u
Now, having U and R, V is easy, since they sum to 180°. Then
v^2 = u^2+r^2 - 2ur cosV
Steve's diagram explanation is false:
SinR/212 = Sin118.79/125 results in...
Sin^-1(212Sin118.79/125) = ERROR
SinR/212 = Sin118.79/125 results in...
Sin^-1(212Sin118.79/125) = ERROR
6 answers