Review packet
Solutions to example problem and Equilibrant force problem
1. Find the magnitude and direction of the resultant force acting on an object if a 10 N force
acts north, a 15.0 N force acts west, and a 15 N force acts 60° north of east (measure up
from east).
- Alright, so we are doing some vector addition. Check out the steps that are in your notes.
We are going to follow those procedure steps, and analyze where the resultant (Net) force
will end up.
Step 1: resolve the vectors into perpendicular components
-
A quick analysis of the compass points on an x-y plane will help us with the angles here
- The 1st force acts North, which
would fall along the positive y axis,
which is at 90 degrees, which is the
angle that we will use for this force.
- The 2nd force acts west, which
would fall along the negative x axis,
which is at 180 degrees, which is the
angle that we will use for this force.
- The 3rd force acts at 60
degrees north of east, so the angle
that we will use to analyze this
motion is given, we’re good.
Ok, we are now ready to break them down. Here’s how it works out:
x – component
F1x = (10 N) x (cos 90 ) = 0 N
F2x = (15 N) x (cos 180) = -15 N
F3x = (15 N) x (cos 60) = 7.5 N
y – component
F1y = (10 N) x (sin 90) = 10 N
F2y = (15 N) x (sin 180) = 0 N
F3y = (15 N) x (sin 60) = 13 N
Step 2: Add up the x – components, then the y – components. This will give us the legs of the
right triangle that will give us our resultant force. We will call the sum of the x – components
FRX and the sum of the y- components FRY.
x – component
F1x = (10 N) x (cos 90 ) =
F2x = (15 N) x (cos 180) =
F3x = (15 N) x (cos 60) =
FRX
0N
-15 N
+7.5 N
y – component
F1y = (10 N) x (sin 90) =
F2y = (15 N) x (sin 180) =
F3y = (15 N) x (sin 60) =
10 N
0N
+13 N
FRY
= 23 N
= -7.5 N
Step 3: Determine the magnitude of the resultant. Again the parts that we just found are the legs
of a right triangle, the resultant will be the hypotenuse. We apply the Pythagorean theorem to do
this. (to find the mag, use Pythag).
FR2 = FRX2 + FRY2
FR2 = (-7.5 N) 2 + (23 N)2
FR2 = 585.25 N
FR = 24.2 N
Step 4: OK, so far so good. Keep up the good work, we are almost there. We will use the
tangent function in order to determine the angle at which our resultant force acts.
tan  = y – component
x – component
 = tan-1(y-component)
(x-component)
 = tan-1 23 N
-7.5 N
 = -72 degrees
Let’s take a look at where this is on our x-y plane to help us figure out the angle stuff.
We will use a quick head to tail
method to determine the location of
the resultant.
- The FRX = -7.5 N, which
puts the vector along the negative x
axis
- The FRY = 23 N, which
puts the vector along the positive y
axis
- Remember that these are the
legs of a right triangle, whose
hypotenuse is the resultant vector
-
The angle that we just calculated was the angle between the negative x axis (west) and
the resultant vector. Therefore we will say that the angle is 72 degrees North of West.
The answer to the question is that the resultant vector quantity is
24.2 N, 72 North of East.
Two forces are exerted on an object. A 36 N force acts at 225° and a 48 N force acts at 315°.
What are the magnitude and direction of the equilibrant force?
Step1: resolve the vectors into perpendicular components
Ok, we are now ready to break them down. Here’s how it works out:
x – component
F1x = (36 N) x (cos 225) = -25.5 N
F2x = (48 N) x (cos 315) = 33.9 N
y – component
F1y = (36N) x (sin 225) = -25.5 N
F2y = (48 N) x (sin 315) = -33.9 N
Step 2: Add up the x – components, then the y – components. This will give us the legs of the
right triangle that will give us our resultant force. We will call the sum of the x – components
FRX and the sum of the y- components FRY.
x – component
F1x = (36 N) x (cos 225) = -25.5 N
F2x = (48 N) x (cos 315) = 33.9 N
FRX
y – component
F1y = (36N) x (sin 225) = -25.5 N
F2y = (48 N) x (sin 315) = -33.9
= 8.4 N
FRY = -59.4 N
Step 3: Determine the magnitude of the resultant. Again the parts that we just found are the legs
of a right triangle, the resultant will be the hypotenuse. We apply the Pythagorean theorem to do
this. (to find the mag, use Pythag).
FR2 = FRX2 + FRY2
FR2 = (8.4 N) 2 + (-59.4 N)2
FR = 60 N
Step 4: OK, so far so good. Keep up the good work, we are almost there. We will use the
tangent function in order to determine the angle at which our resultant force acts.
 = tan-1(y-component)
(x-component)
 = tan-1 -59.4 N
8.4 N
 = -82 degrees
Let’s take a look at where this is
on our x-y plane to help us figure
out the angle stuff.
We will use a quick head to tail
method to determine the location
of the resultant.
- The FRX = 8.4 N,
which puts the vector along the
positive x axis
- The FRY = -59.4 N,
which puts the vector along the
negative y axis
- Remember that these are
the legs of a right triangle, whose
hypotenuse is the resultant vector
-
The angle that we just calculated was the angle between the positive x axis and the
resultant vector. Therefore we will say that the angle is 82 degrees below 360. We don’t
have the compass points to base this on, so we have to figure out the angle below 360.
To do this we subtract the 82 degrees from 360, which gives an angle of 278 degrees.
-
The resultant vector must be 60. N, 278 degrees.
-
The question asked to determine the equilibrant force, which is equal in magnitude but
opposite in direction to the resultant. Therefore we need to determine the force that acts
opposite of 278 degrees. Adding 180 will put us over 360 degrees, so we will have to
subtract the 180, which gives us an angle of 98 degrees.
-
The equilibrant force is 60. N, 908 degrees.