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.
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
Steve's diagram explanation is false:
SinR/212 = Sin118.79/125 results in...
Sin^-1(212Sin118.79/125) = ERROR
To find the second force and the direction of the resultant force, we can use vector addition. The resultant force is the vector sum of the individual forces.
First, let's break down the given forces into their components. Since the first force is pulling due west, its x-component is -125 N, and its y-component is 0 N since there is no force in the north-south direction.
The second force is given in both magnitude and direction. In order to find its components, we can use trigonometry. The magnitude of the second force is 28.79 N, and it is pulling in a direction of N 28.79 W. This means that the angle between the second force and the positive x-axis is 28.79 degrees in the northwest direction (west and slightly north).
Using trigonometry, we can find the x-component and y-component of the second force:
x-component = magnitude * cos(angle)
= 28.79 N * cos(28.79°)
≈ 24.89 N
y-component = magnitude * sin(angle)
= 28.79 N * sin(28.79°)
≈ 13.18 N
Next, we can find the components of the resultant force. Since the resultant force is given as 212 N, we can assume it has components in the x and y directions.
Let the x-component of the resultant force be R_x and the y-component be R_y.
Now, we can set up two equations based on the vector addition:
R_x - 125 N + 24.89 N = 0 N (sum of x-components equal to zero)
R_y + 0 N + 13.18 N = 0 N (sum of y-components equal to zero)
Solving these equations, we get:
R_x = 100.11 N
R_y = -13.18 N
Finally, we can find the magnitude and direction of the resultant force:
Resultant magnitude (R) = sqrt(R_x^2 + R_y^2)
≈ sqrt((100.11 N)^2 + (-13.18 N)^2)
≈ 101.07 N
Resultant direction = arctan(R_y / R_x)
= arctan((-13.18 N) / 100.11 N)
≈ -7.43°
Therefore, the magnitude of the resultant force is approximately 101.07 N, and its direction is approximately 7.43 degrees south of west.