F1x = 6.90 cos55.0°
F1y = 6.90 sin55.0°
F2x = 0
F2y = 5.00
So, if F = F1+F2 then
Fx = F1x + F2x
Fy = F1y + F2y
|F| = √(Fx^2 + Fy^2)
at an angle θ such that
tanθ = Fy/Fx
This should all look familiar ...
A force
F1
of magnitude 6.90 units acts on an object at the origin in a direction 𝜃 = 55.0° above the positive x-axis. (See the figure below.) A second force
F2
of magnitude 5.00 units acts on the object in the direction of the positive y-axis. Find graphically the magnitude and direction of the resultant force
F1 + F2.
magnitude _________ units
direction _________ ° counterclockwise from the +x-axis
can someone explain it step by step like I was in the first grade?
2 answers
F1,
x component = 6.9 cos 55 = 3.96
y component = 6.9 sin 55 = 5.65
F2
x component = 5.0 cos 90 = 0
y component = 5.0 sin 90 = 5.00
Resultant = F
F
x component = 3.96 + 0 = 3.96
y component = 5.65 + 5.00 = 10.65
magnitude = |F| = sqrt (10.65^2+5^2) = 11.77
tan theta = 10.65/3.96 = 2.69
theta = 69.6 degrees above x axis
x component = 6.9 cos 55 = 3.96
y component = 6.9 sin 55 = 5.65
F2
x component = 5.0 cos 90 = 0
y component = 5.0 sin 90 = 5.00
Resultant = F
F
x component = 3.96 + 0 = 3.96
y component = 5.65 + 5.00 = 10.65
magnitude = |F| = sqrt (10.65^2+5^2) = 11.77
tan theta = 10.65/3.96 = 2.69
theta = 69.6 degrees above x axis