Two people push on a desk they are trying to move. One person pushes with 35 N at a 50^{o}
angle, while the other person pushes with 27 N at a 340^{o} angle. What is the resultant push on the desk?

Given*: push
one,*_{}

Unknown: resultant push, _{}

** **

** **

Physical (Mathematical) Principles and Ideas: rules for adding vectors, trigonometric

functions and Pythagorean theorem.

Solution: There are several ways to add vectors, which are neither co-linear (i.e., do not lie along a line) nor perpendicular. The one that can be used most easily in the greatest variety of situations is the method of components. A vector, any vector, can always be “broken” down in two (or in three dimensions—three) other vectors at right angles to each other which, when added vectorially using the Pythagorean theorem, “sum” to give the original vector. These two (or three) vectors are called components. Using this idea of components, the general outline for how we solve vector addition problems is the following:

(a) “Break” each
vector down into its components along identified axes using trigonometric
functions. That is, find the *x*
and *y* (and when necessary *z*) components of each vector in the
problem. Notice that this process
requires orienting the vectors properly on a set of coordinate axes.

(b) Add the
components for each of the axes independently, i.e., add the *x*
components to get one “*x* vector”, add the y components to get one “*y*
vector”, and add the *z* components to get one “*z* vector”. Keep in mind that these processes of
“adding” components may actually involve subtracting, since components pointing
in opposite directions would subtract.

(c) Finally, we
add the “*x* vector”, “*y* vector” and “*z* vector”, using the
Pythagorean theorem, to find the resultant (sometimes called the net) vector.

We will now carry out this program for the two vectors in the problem

(a) If
we sketch _{} with its components,
we get:

So _{} itself is the
hypotenuse of this right triangle, the x

component is the side of the triangle adjacent to
the 50^{o} angle, and the

*y* component is the side of the triangle
opposite the 50^{o} angle. That

means:

*cosine50 ^{o} = _{}_{}^{}*

* *

That is, *x*
component (35 N)(0.64) = 22.5 N and since it points to the right on our axes,
it is positive. (On our axis, right is
positive and left is negative for *x*, up is positive and down is negative
for *y*.) The *y* component
is obtained from:

* sine50 ^{o} = _{}*

* *

giving *y*
component = (35 N)(0.77) = 26.8 N

This
component points upward so it is positive also. We now have the *x* and *y* components of _{}. They are:

_{}

In a similar way we find the components of

_{}. Here we have a
preliminary step

we have to complete; we have to find the angle

q. The angle we were given was 340^{o},
which

means q =
360^{o} – 340^{o} = 20^{o}. Now we

proceed in
the same way we did with _{}

*cosine20 ^{o}
= _{} x component = (27 N)
cosine 20^{o} = 25.4 N*

* *

and this will be positive since it points to the right.

_{}

*y* component = (27
N) sin 20^{o} = 9.2 N

but this will be negative since it points down. The components of

_{} are: F(2)*x* = +25.4 N and F(2)*y* =
-9.2 N

b) In this step we add the components
independently. (Why can we do
this?) In other words, we find the *x*
component of the resultant from

F*x* = F (1) *x* + F
(2) *x* = +22.5 N + 25.4 N = 47.9 N

and the *y*
component from

F*y* = F (1) *y* + F
(2) *y* = +26.8 N – 9.2 N = 17.6 N

c) Finally, we “sum” the two components of the resultant using the Pythagorean theorem, and then determine the angle for the resultant. First, we get the magnitude of the resultant

_{}

Now the angle can be found using either the sine, cosine, or tangent.

_{}

_{}

_{}

Note that
this means 20.17^{o} above the *x* axis (i.e., in the first
quadrant) since F*x* and F*y* are both positive. (What quadrant would the angle be in if F*x*
was negative and F*y* positive?
What about if F*x* and F*y* were both negative?)

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