# Inclined Plane: Example

by Jared Rovny
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00:01 Now that we have an overview of how a slope problem works, let's take a look at an example.

00:07 We've a block sliding down a smooth inclined plane, meaning that we're not talking about friction yet, and we have an angle given to us at 30 degrees that that plane makes with the horizontal.

00:16 The question is, how long would it take this block to slide 5 meters down the slope? So you can see what's happening in this problem.

00:22 We have both Newton's laws and the equations of motion coming into play together where we have to worry about gravity, the normal force, and this object sliding down the slope.

00:31 So before I jump into how to do this problem, you should pause and give it a shot and use what we've just introduced, the normal force and the gravitational force as well as how to break the gravitational force up into different vector components, and see if you can find this time, the time it takes for the block to slide 5 meters down the slope.

00:49 If you've done this problem, here's a good overview of how it should have looked.

00:57 We've given it a mass of 10 kg, and we're wondering how long it takes to slide down the slope.

01:01 So we're looking for a time and we know that the forces are as we've drawn here from the analysis we've already done.

01:08 We have a normal force, which is the one pointing out of the slope.

01:10 We also have the gravitational force directly downwards, which we've already split up in two, Fg cosine of theta and the Fg times the sine of theta.

01:18 And again, we also know that the angle of the slope is 30 degrees.

01:21 The question, one more time, is that the object will move 5 meters and we wanna know how long it will take with the forces that we've shown for it to move those 5 meters.

01:31 Again, the first thing we're gonna do in any problem like this is create a coordinate system, so we have our plus x direction and our plus y direction.

01:39 And again, this is different from any other problems where we just consider plus x to be always horizontal and plus y to be only vertical.

01:46 We align them along the slope because that makes our problem much easier to find motion along the slope itself.

01:51 So here's one way you might wanna solve this problem.

01:54 Since we're trying to figure out how long it will take a block to slide 5 meters, you know that this is an equations of motion type problem.

02:02 So we can look at our equations of motion and say that the motion or the position of the object will be whatever its original position was plus its original velocity in the x direction times time plus 1/2 times the acceleration times the time it took the object to slide squared.

02:20 So there are some things here we don't know.

02:22 For example, we don't know what this acceleration is so that's something we need to find.

02:25 This initial velocity in the x direction is something that we know because its initial velocity is zero, if the object started from rest, so we can get rid of that.

02:36 And then, we can say that this initial position of the object is zero.

02:40 We can say it's starting at the zero point and it's going to slide 5 meters down to its final position, so we'll also call that zero.

02:49 But again, the reason that this is a force problem is that we need to find this acceleration and figure out what that is.

02:55 So we're gonna do that using Newton's second law.

02:57 So Newton's law says that F equals ma and as we said what we're gonna do is treat this in the two directions independently, so let's say that the forces in the x direction is equal to the mass, that's 10 kg, times the acceleration in the x direction.

03:13 So we're just considering anything that's horizontal, so none of these vertical forces are something that we're going to consider right now, just this horizontal force along the x direction.

03:22 And again, what I'm calling horizontal is up and down the slope, in other words, our x direction. So, we can plug this in immediately.

03:29 We can see that the forces in the x direction here are just the Fg sine of theta and I don't see any other forces so we have on the left-hand side here, Fg times the sine of theta is equal to the mass of the object times its acceleration.

03:42 So not too hard to solve this.

03:45 The acceleration of the object is equal to Fg over the mass times the sine of theta.

03:52 And so we've already figured out the acceleration just from applying Newton's second law.

03:57 So now we have two components to this problem that we've done.

04:00 First, we have the equations of motion up top and then we did Newton's second law on the bottom and as we've discussed, the thing that's always going to link these two aspects of a problem will be the acceleration of your object.

04:19 Rewriting the equations of motion using the acceleration we just found, we have that the final position is 1/2 this acceleration, which is Fg over m times the sine of theta times the time it took the object to slide squared.

04:35 Now in this problem, we're not looking for the position, we're looking for the time, so we will have to rearrange this, so let's do that.

04:41 We have t squared equals, we have 2 times the mass, times the position, having multiplied both sides by the 2 and the m.

04:49 We're dividing by the force of gravity and the sine of theta, and this will be equal to the time squared.

04:57 So we still have to take the square root at the end of this problem, so don't forget to do that.

05:00 It's a very common mistake.

05:01 So let's plug in the values and see what we get.

05:03 We have 2 times the mass times the position divided by the force of gravity.

05:10 And as we've already discussed, this force of gravity, Fg, is always equal to the mass of the object times gravity.

05:16 So this is equal to m times g times the sine of the angle.

05:21 Something really important here is that the mass is cancelled, so the answer you get for this problem for how long it takes this object to slide actually doesn't depend on the mass at all.

05:30 So plugging in the values that we have remaining we have 2, we have the position or the distance which is 5 meters, we have g which again for the reasons that I've discussed already we're gonna approximate as 10 meters per second squared, so we have g is 10, and then we have the sine of theta which is the sine of 30 degrees and the sine of 30 degrees from the table that we showed before is equal to 1/2, so that's a good thing to remember.

05:56 So now we're ready to solve.

06:00 We can put this 2 up top since the denominator over denominator, it goes to the numerator.

06:06 So 2 times 2 is 4, 4 times 5 is 20, divided by 10 and we have 2.

06:12 Now, one more time don't forget your square root.

06:14 Very often people will stop their problem right here and say 2 is my answer, it was 2 seconds, but don't forget we have a square root to take because t squared equals 2. So now that we have 2, we have t equals the square root of 2, which it turns out is approximately 1.4 seconds that this object took to slide 5 meters down the slope.

06:33 So, this is a good example problem of a slope and how to deal with many forces that we've put in different directions.

The lecture Inclined Plane: Example by Jared Rovny is from the course Force.

1. mgCos θ.
2. mg Sin θ.
3. mg Tan θ
4. mg.
5. F=ma.
1. gSin θ.
2. g Cos θ.
3. mgSin θ.
4. g.
5. mg.
1. 980N.
2. 9.8N.
3. 1131N.
4. 11.3N.
5. 100N.

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