Full transcript
0:02good morning three forces act on an
0:04object Force One acts Southward and has
0:07a magnitude of 15 Newtons Force two acts
0:11Westward and has a magnitude of 29
0:13Newtons and force three has a magnitude
0:16of 31 Newtons and acts at an angle of
0:1935° north of East what is the net force
0:22that acts on the
0:27object so we have our three4 forces
0:30Force 1 15 Newtons South Force 2 which
0:32is 29 Newtons West and Force 3 which is
0:3531 Newtons at an angle of 35° north of
0:38East we should of course draw a vector
0:46diagram we have our Vector diagram with
0:49Force 2 Force One and force three and
0:51we're solving for the net force we're
0:53going to add all those forces together
0:56before we're going to do so or before we
0:58can we need to break Force three into
1:00its components because it is neither in
1:01the X direction or the y direction you
1:03can see Force 2 and Force One are
1:05already in the either the X Force two or
1:08the y direction Force One so again we're
1:10going to break Force three into its
1:15components the components of force three
1:17are Force three in the y direction Force
1:193 in the X direction we can use Theta
1:23and to do
1:26so s of theta equal opposite over
1:29hypotenuse
1:30opposite Theta is Force 3 in the y
1:32direction the hypotenuse is Force 3 now
1:34we can multiply by Force 3 to solve for
1:36the Force 3 in the y
1:40direction therefore Force 3 in the y
1:42direction equals force 3 * the S of
1:44theta and we can substitute in
1:49values so Force 3 in the y direction
1:51equal 31 * the S of 35° or 17.78 1
1:56Newtons we can do this we can do the
1:58same thing with the cosine of theta
2:05the cosine of theta equals adjacent over
2:07hypotenuse or the Force 3 in the X
2:09Direction divided Force 3 multiplying
2:11through by Force 3 gives us Force 3 in
2:13the X directional Force 3 * the cosine
2:15of TH or 31 * the cosine of 35° which is
2:2025394
2:25Newtons then of course we need to redraw
2:27the free body diagram we still have
2:29Force 2 and force one but instead of
2:31having Force three we have Force 3 in
2:33the y direction and Force 3 in the X
2:35Direction now considering we're trying
2:38to solve for the net force we should sum
2:40the forces let's start uh in both
2:42directions and let's start with summing
2:44the forces in the y
2:47direction the net force in the y
2:49direction is equal to force 3 in the y
2:52direction minus Force 1 Force 3 in the y
2:54direction is positive because it's up
2:56Force One is negative because it's down
2:59we can now subtit itute in
3:02values so we get 17.78 1 - 15 therefore
3:06the net force in the y direction is 2.78
3:091 Newtons now we can sum the forces in
3:12the X
3:16Direction the net force in the X
3:18direction is going to be equal to the
3:20force 3 in the X Direction minus Force 2
3:22Force 3 in the X direction is positive
3:24because it's to the right Force two is
3:26negative because it's to the left so
3:2825394 - 29 so the net force in the X
3:32direction is -
3:343.66 NT so notice the net force in the y
3:38direction is positive and the net force
3:40in the X direction is negative so the
3:41net force in the y direction is up the
3:44net force in the X direction is to the
3:46left so now we should draw a vector
3:50diagram with the net force in the X
3:52Direction and the net force in the y
3:54direction which are the components of
3:56the net force
4:03so the net force in the X direction is
4:05to the left and the net force in the y
4:07direction is up therefore the net force
4:09is going to be the hypotenuse of this
4:11triangle I do want to point out that I
4:13actually had to draw this bigger than to
4:16scale because these numbers are very
4:18small it actually would have been too
4:19small for you to see on the board would
4:21have been about 1/3 of that size uh so
4:25now uh in order to find the net force we
4:27use the Pythagorean theorem
4:35so A2 + b s = c^2 therefore the net
4:38force in the X Direction squar plus the
4:40net force in the y direction squar is
4:41equal to the net force squar and we can
4:43take the square root of the whole
4:46equation to get the net force is equal
4:49to the square root of the net force in
4:50the X Direction squar plus the net this
4:53net force in the y direction squared and
4:55we can plug in numbers
5:01the square root of the quantity -
5:043.66 squared make sure you put the
5:07negative inside the parentheses there
5:09plus the
5:112781 squared so the square root of that
5:14whole thing gives you the net force of
5:164.55 457 or with two sigfigs 4.6 Newtons
5:21and because force is a vector we need
5:23the direction so we we can now use to to
5:27figure out the direction
5:34the tangent of theta equals opposite
5:36over adjacent opposite our Theta is the
5:39net force in the y direction adjacent
5:41Theta is the net force in the X
5:43Direction taking the inverse tangent of
5:45the whole equation gives us Thea is
5:46equal to the inverse tangent of the net
5:48force in the y direction divided by the
5:49net force in the X Direction therefore
5:52the inverse tangent of 2781 / -
5:563.66 which works out to be 37.640311
5:59or with two sigfigs
6:0438° which means the net force acting on
6:07this object is 4.6 Newtons at an angle
6:11of 38° north of West that is north from
6:17West moving the Northerly direction from
6:20the West direction thank you very much
6:23for learning with me today I enjoyed
6:25learning with you