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AP Physics 1 - Forces on an Incline — Transcript

by Mr. Marek Physics Funhouse · 2,083 words · 302 segments · language en · Watch on YouTube

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0:02hey there y'all in this video we're
0:04going to study the forces on an object
0:06which is on an inclined surface so the
0:10question for today is how do we analyze
0:12the forces of an object if the surface
0:15that it is on is tilted so examples of
0:18this might be something like a car on a
0:21steep driveway like it goes up a hill or
0:23something like that or moving something
0:25up or down a ramp um for example you may
0:29have sometime in your life pushed
0:31something up a ramp so if you're moving
0:33and you got a real heavy box you may
0:35have to push it up a ramp into the
0:37moving truck or whatever and so the
0:40general picture would look something
0:42like this here's our surface and then
0:45horizontal is the dash line and so the
0:48surface makes an angle which we'll call
0:50Theta relative to the
0:53horizontal and there's our box so if we
0:56consider the forces acting on this box
0:58some of them are going to be the same
1:01like for instance gravity always pulls
1:03us
1:05downwards but some of them are going to
1:07be a little bit different remember that
1:09the normal force exerted by the surface
1:12is always perpendicular to the surface
1:15that's what normal means perpendicular
1:17to a
1:18surface uh if you push something up a
1:21ramp it'd be pretty pretty silly not to
1:23push it straight up the ramp so the push
1:26force in this case the way I've drawn it
1:29I've drawn it par parallel to that
1:31surface and then friction would also be
1:35parallel to the
1:37surface so what you might kind of note
1:39right away is that all those forces are
1:43either parallel or perpendicular to the
1:45surface with the exception of
1:49one the exception is the weight the
1:52gravitational force which is always
1:55downwards is not parallel or
1:57perpendicular to the ramp that's going
2:00to be an important thing to realize
2:01moving
2:07forwards so just moving my um free body
2:10diagram that I drew over here the next
2:13step in analyzing these forces or
2:15finding the acceleration of this box
2:18would be to resolve all the forces into
2:20X or Y
2:22components and then write net force
2:24equations for each
2:27Direction now here we have two options
2:31we didn't learn any differently we'd
2:33probably choose the harder option which
2:35is to leave X and Y um to the right and
2:39up and then resolve the frictional force
2:43the push and the normal force into
2:45components that are parallel and
2:47perpendicular to that coordinate system
2:50well that's going to be really hard
2:52because you're going to have to do three
2:53Vector diagrams and then you're going to
2:56have an acceleration in both directions
2:59because this thing would accelerate up
3:00into the right or down to the left and
3:03so option one is not a very good option
3:07so there's option
3:09two if we turn the coordinate system our
3:12X and Y AIS to match the
3:16surface in other words make X parallel
3:19to the surface and Y perpendicular to
3:22the surface like that then the only
3:26force that we would have to resolve
3:28would be the force of gravity
3:30so the trick is instead of making our
3:33coordinate system up and to the right
3:38we're match the incline so option two is
3:43going to be our preferred option
3:49here so how do we resolve the
3:51gravitational
3:53force we kind of consider just the force
3:56of gravity like that
4:00we can figure out the angles that would
4:03be in our Vector diagram so this angle
4:07here Theta 2 would be 90° minus the
4:12angle with horizontal the gray angle I'm
4:14going to leave that as
4:16Theta if I kind of extend that line a
4:19little bit and then draw a line that's
4:22perpendicular to the um surface remember
4:26that's what we want our components to be
4:28as perpendicular parallel to that
4:31surface that line makes a right angle
4:34with the
4:36surface meaning that the angle in
4:38between the force of gravity and the
4:41dashed red line that I drew has to be
4:44Theta since Theta 2 and Theta have to
4:47add up to
4:5290 and so when we resolve the force of
4:55gravity we're going to set up our Vector
4:58diagram such that that angle Theta is
5:01part of our Vector diagram so kind of
5:05drawing a little bit bigger down here
5:07force of gravity still goes
5:11downward the Y component of gravity
5:14would be the part that's perpendicular
5:16to the diagram so that would look like
5:19that and then there's my X component so
5:22you can see it's still makes a triangle
5:25which our Vector diagram should the
5:28right angle of our triangle oops oops
5:30get the right pin is right
5:34there and then that angle right there at
5:37the top is the same as the angle that
5:40the ramp makes with horizontal so these
5:43angles are the same and that's important
5:46to
5:47remember kind of taking you one step
5:50further if we can't see that the um long
5:54side of the triangle would be this side
5:58right here making that the the
6:01hypotenuse this side is the adjacent
6:04side this side is the opposite side so
6:08even though our triangle isn't turned um
6:11to where the y and x axis are parallel
6:14perpendicular like the lines on our
6:17paper um the hypotenuse is still the
6:20long side of the triangle and so we need
6:22to be able to kind of turn that in our
6:25minds a little
6:27bit so if I wanted to know know what the
6:30Y component is I could use the cosine
6:33function if I wanted to know what the X
6:35component was I could use the S
6:42function what might make it slightly
6:44easier to visualize is if you took the
6:47diagram that we drew the one that looks
6:50like that we could take that whole thing
6:54and rotate it so that our X and Y AIS
6:58are pointing up into the right like
7:00we're accustomed to them being and so
7:02when we do that now we can kind of see
7:05how the frictional force normal force
7:09and the push are parallel and
7:12perpendicular to our coordinate
7:15system the only one that's not is this
7:19one and so again we did our Vector
7:22diagram we got a component that did
7:25something like this and a component
7:29did something like
7:34that so let's look at a simple
7:39example suppose we have a 2 kilogram box
7:42which is at rest on a surface and then
7:44let's suppose a surface is inclined at a
7:4737° angle from the horizontal and we
7:50want to know what are all the forces
7:52acting on this thing so the first step
7:54would be to draw a free body
7:56diagram so there is my
8:00surface there's force of gravity always
8:02goes
8:03downward normal force is perpendicular
8:06so it go up to the right or excuse me up
8:08to the left like that and then friction
8:12would be holding this thing up the ramp
8:14it's not allowing it to slide down so
8:16friction must go up the incline and
8:19friction is always parallel to the
8:21surface so there's my free body
8:24diagram the force of gravity that should
8:26be easy just multiply by G mass by G
8:30that is and so we would get something
8:33like 20 newtons and then don't forget to
8:36indicate the direction the direction is
8:39down the next step would be to write a
8:42net force equation in each Direction net
8:44force in the y direction so remember Y
8:47is that way perpendicular to the
8:51surface would be the normal force minus
8:54the Y component of the
8:56weight and in the X Direction
9:00remember that ways
9:01X we've got friction minus the X
9:04component of
9:06gravity since this thing's at rest I
9:09know that both of those net forces are
9:12zero and so I know that the normal force
9:15and the Y component of gravity have to
9:17be balanced and I know that the
9:20frictional force and the X component of
9:22gravity have to be
9:25balanced and so resolving the weight
9:27Vector y component comp and X
9:31component 37° angle goes at the
9:36top and so I can use the S function to
9:39find
9:41fgy so fgy would be FG sin
9:4637° and so 20 newtons times the S of 37
9:50should be something like
9:5312 and so the normal force will be equal
9:56to 12 Newtons and then don't forget to
9:59indicate the direction so the normal
10:01force is perpendicular to the surface um
10:05and that's kind of all we really need
10:07for the direction on something like
10:09that so take a minute and see if you can
10:12pause this and figure out what the um
10:15force of friction
10:23is so when I did it I got the force of
10:28friction to to be about 16 Newtons up
10:32the incline so again don't forget to
10:34include the
10:38direction let's look at another
10:41example so kind of a similar setup got a
10:4450 kogam box which is being pulled up a
10:47ramp by a Cable cable is kind of like a
10:49rope and we know the ramp makes a 37°
10:52angle with the horizontal that gives us
10:54nice um s and cosine values and that the
10:58tension in the cable is 600 Newtons we
11:01also know that the force of friction on
11:03the box is 200 Newtons so we're going to
11:06kind of do this um you do part of it
11:09check yourself and then um I'll give you
11:12something else to do with it but the
11:14first thing I want you to do is just
11:15press pause and see if you can draw a
11:17free body diagram showing the forces on
11:20this thing press pause see if you can do
11:22that
11:31so my free body diagram would look
11:33something like this gravity goes down
11:36normal force is perpendicular tension is
11:39up the
11:40incline and friction in this case would
11:43be down the incline so since the box is
11:45sliding up friction would oppose its
11:47relative motion and so it would go
11:50down so next do this see if you can
11:53resolve the force of gravity again press
11:56pause see if you can do that
12:07so the triangle will look something like
12:11this that angle right there is
12:1537° doing M * G gives you 500 Newtons
12:19for the force of
12:20gravity and you might recognize that
12:23that was a 345 right triangle or you can
12:26use your sign and cosine either way you
12:28get FG y to be 400 newtons and fgx to be
12:32300
12:35Newtons so do this now see if you can
12:38take the Free by diagram and the vector
12:41diagram that we did and see if you can
12:43use that to find the acceleration of the
12:46Box again press pause see if you can do
12:48it on your by yourself and then press
12:51play
12:58again hopefully you could write a net
13:00force equation in the X
13:04Direction and so the net force equation
13:06will look like tension minus friction
13:09minus fgx so tension goes up the incline
13:13friction and fgx goes down the incline
13:16so plugging in the values give you
13:19something like 100 newtons that would be
13:21the net force on this
13:25object and then we know Newton Second
13:27Law acceleration is equal to net force
13:30over
13:32Mass not sure why there's an equal sign
13:36there and so plugging in your net force
13:39of 100 Newtons and then the mass is 50
13:43kg would make the acceleration 2 m/s
13:47squared and again don't forget about the
13:51direction so the accelerations up the
13:54incline because the net
13:56force was up the incline
14:01so let's kind of summarize the things we
14:03need to remember if you have an inclin
14:06surface then tilt your X and Y AIS so
14:09that they are parallel and perpendicular
14:11to the surface so X is parallel Y is
14:16perpendicular resolve your weight into
14:19the components perpendicular and
14:21parallel so that they're X and Y and
14:23then we need to remember that when we
14:25draw the triangle for FG like that our
14:28vector diagram that top angle is the
14:31same as the angle that the surface makes
14:34with horizontal if you can remember
14:36those three things and do everything
14:40else we've learned about forces so far
14:43then you shouldn't have any trouble
14:45understanding and analyzing a surf an
14:48object which is on a surface that is
14:51incline so until next time tataa

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