Forces at an angle

0:00- [Instructor] Let's solve few numerical on free body diagrams and Newton's laws, here's the first one. A 10 kilogram box is being pulled by a rope that makes a 30 degree angle above the horizontal. The gravitational tension normal and frictional forces on the boxes are as given.

0:16Find the acceleration of the box. Okay, so how do we do this? Well, our first step is to draw a free body diagram of, on the object of interest. Our object of interest over here is this box because we want to calculate the acceleration of the box. So let's consider its center of mass and let's try to draw all the forces acting on it originating from this point.

0:38Okay, so what are the forces? Well, there is a gravitational force that's acting downwards, we'll call that FG and it's 100 Newtons. What else? There is a tension force due to the rope and that is at an angle of 30 degrees and that's 120 Newtons. We also have the normal force which is going to be perpendicular to the surface, to the ground.

0:57So that's over here and that's 40 Newtons. Remember that the length of the vector is proportional to the magnitudes, okay? And finally, we have the frictional force. If the block tends to accelerate to the right slide to the right, then the frictional force will oppose it, so the frictional force will be towards the left.

1:16So there will be a frictional force towards the left and that's going to be 80 Newtons. And our goal is to find the acceleration of the box. How do we do this now? Now one big problem I see is that there's a force at an angle of 30 degrees. That's a problem because our strategy is to sum up for, calculate the net force in the X direction separately, in the horizontal separately, and the net force in the vertical separately, we consider them separately and analyze them separately.

1:44But now we have a force that's neither in the horizontal nor in the vertical. It's making an angle, what do we do about that? Well, what we can do is we can write this force as a sum of this horizontal and this vertical force. So this force can be written as a sum of these two forces.

2:05So I can replace this force with these two forces. But the question is, I know the magnitude of this force that's given to me, that's tension force, 120 Newtons. What will be the magnitude of these two forces? Well, to answer that question, we need to use some trigonometry.

2:21Now remember in a right angle triangle, you can remember SOHCAHTOA, which says sine of theta is the opposite side divided by the hypotenuse. Cos of theta is the adjacent side divided by the hypotenuse and tan theta is the opposite side divided by the adjacent side, okay?

2:40Now if you have a vector along the hypotenuse, that can be written as a sum of these two vectors. So if I know what the hypotenuse value is, that is the magnitude of this vector, I can use trigonometry to figure out these two values, right? So from this, look, the opposite side becomes h times sine theta, and the adjacent side becomes h times cos theta.

3:06Again, I can use opposite side as a times tan theta, but we're not gonna use this, we're gonna use these two. So the opposite side becomes h times sine theta, and the opposite, the adjacent side becomes h times cos theta. So we now have everything needed to figure this out.

3:21So look, this is our hypotenuse, that is the tension force, 120 Newtons, okay? So from this, this now becomes the adjacent side, which is going to be hypotenuse times cos theta, so it's going to be FT cos 30 degrees. And what will this be? Well, it's going to be FT, the force of tension times sines of 30 degrees.

3:45These two vectors are called the components of this vector. And since these two vectors add up to gimme the same vector, I can now get rid of this. This is replaced by these two vectors. Let me get rid of this as well. And we now have all the vectors in the horizontal and vertical.

4:00Now just to clarify, when somebody asks you to draw a free body diagram, this is the actual free body diagram. It's just that, that sometimes we resolve forces into the horizontal and verticals, and in doing so, we draw a separate diagram. But this is the original free diagram, okay?

4:15So we will never start with the resolved components, we'll always start like this and if necessary, we'll draw separate diagrams this way. And remember, since these two components add up to give me this force, I've already accounted for that force. And so we shouldn't make the mistake of redrawing that force over here.

4:32Anyways, now we can calculate the net force in the horizontal and vertical, but our goal is to find the acceleration of the box, not the net force. But hey, we know from Newton's second law, acceleration is the net force divided by mass. So what we can do now is find the acceleration in the horizontal by using the net force in the horizontal and then find the acceleration in the vertical by using the net force in the vertical.

4:55It'll be a great idea to pause the video and see if you can continue yourself. All right, let's start with the horizontal. The acceleration in the horizontal is gonna be net force in the horizontal divide by mass, okay? So what's the net force in the horizontal?

5:08I just have to look at the horizontal. Well, I have to add up these two forces, but with sines and how do I decide which of the forces will be positive and negative? For that, I have to choose one direction positive. Let's choose right side to be positive. It's completely your choice, you're free to choose whichever direction you want to be positive, I'm just gonna choose the right side to be positive.

5:26Then this is going to be a positive number. So I'll get FT cos theta minus the force of friction. That will be a negative number. So I'll get this to be my net force in the horizontal divided by the mass. And now we can plug in, we know FT is 120, the force of friction is 80 and we know the mass is 10 kilograms.

5:49We also know cos 30, it's a known value, it's root three divided by two. If we didn't know it, you could just calculate it using a calculator or maybe we can just leave it at cos 30. Anyways, let's plug it all in so you get the acceleration in the X direction to be this much.

6:02And if we put it all in our calculators, this is what we'll end up with. So the acceleration in the X direction is 2.4 meters per second squared, but which direction is it? Well, it's positive, remember we chose right side to be positive. So this means the acceleration is to the right.

6:19If we had gotten the acceleration to be negative, that would've meant that the acceleration would be towards the left. Similarly, we can now calculate the acceleration in the Y direction. We calculate the net force in the Y direction, divide by mass. What's the net force in the Y direction?

6:33Again, let's choose upwards to be positive, completely your choice. Then I get this plus this minus this. That's the net force, okay? Divided by mass, that's 10 kilograms. So if you plug all of that in now, we get this. And if you plug this values in the calculator, you'll get the answer to be zero.

6:50This means acceleration in the Y direction is zero, there is no net force in the Y direction, it all cancels out, the acceleration is zero. That means there's only the acceleration in the X direction. This represents the total acceleration of the box. So that is our final answer and that kind of makes intuitive sense.

7:07If you were to pull a box like this, you would expect it to slide and accelerate along the horizontal. And that's what it's saying, the total acceleration is along the horizontal. Let's try another one. A flower pot of mass 25 kilograms is supported by two identical cables as shown.

7:21Each cable makes an angle of 30 degrees with the vertical. Find the tension magnitude in each wire, Assume G to be 9.8 meters per second square. So just like before, let's draw a free body diagram for our object of interest, which is the flower pot. The first force that I usually draw is the force of gravity.

7:38Then we have the tension forces acting at an angle, 30 degrees with the vertical. And again, I see force is at an angle. So my first step would be to resolve these forces into horizontal and vertical. Let's do one of them first, the other would be similar, so let's concentrate on this one.

7:53I can write this force as a sum of this vertical component and the horizontal component. And to find the values, well again, I can use my triangle, I just have to rotate it and look, this will be h sine theta, so this is going to be the force of tension times sine 30 degrees and this will be the force of tension times cos 30 degrees.

8:19So let's get rid of this vector and I can do the same thing over here. In fact, it's gonna be very similar. The horizontal component will be FT sine 30 and the vertical component will be FT cos 30. All right, so I have my free body diagram with only horizontal and vertical forces.

8:34Okay, now I look at what is asked. We are asked to find the value of the tension force. How do we do that? Well, the idea is the same, we're gonna use Newton's laws: net force equals mass times acceleration. This time, we know the acceleration of the flower pot.

8:49It's supported by two identical cables as shown, it's at rest, it's not moving at all, which means its velocity is zero all the time, its acceleration stays zero. So we know the acceleration is zero, that means the net force on that pot must be zero, which automatically means the net force in the horizontal is zero, the net force in the vertical must also be zero.

9:14So all I have to do is use the free body diagrams to set up equations for X and Y directions and set them to zero and see if I can use them to calculate the value of FT. Great idea to pause the video and see if you can try this yourself. Okay, let's do it, let's start with the forces in the X direction.

9:30Again, choose one direction to be positive. Let me choose right side to be positive. If I do that, this will be a positive number. This will be negative number, and so that'll get FT sine 30 minus FT sine 30, which is zero. So look, the net force in the X direction is independent of the values of FT.

9:45Whatever is the value of the tension force, this number will always be zero. So this is not gonna help me. So let's try the net force in the Y direction. What's net force in the Y direction? Well, we have these two forces. Let's consider upwards positive. So then these two will be positive, this will be negative.

10:01So I'll get FT cos 30 plus FT cos 30 positive values minus FG and that should also be equal to zero. This gives me 2FT cos 30 minus Fg equals zero. From this, I can calculate what the force of tension is. I can just rearrange this equation and I get force of tension as Fg divided by 2 cos 30.

10:22Force of gravity is just mg. And so now I have all the values necessary, I can just plug them in. So mass is 25 kilograms, G is 9.8 and this is two. And cos 30 is also a known value. So if I put this all in my calculator, this is what I would end up with. And depending on the question, we might be asked to round this off.

10:42Let's say we are asked to round this off two significant figures, then this would be 140 Newtons. So that is the tension force in each wire. All right, last problem. Great idea to see if you can try to do the whole thing all by yourself. All right, so we have a child that's pushing on the box and there are a bunch of forces acting on the box.

11:05So let's draw the free body diagram and draw the forces. Well, we have the force of gravity, we'll also have a normal force acting on it because of the floor pushing on it. Then there's the force that the child is putting, the child is exerting at an angle of 45 degrees downwards to the right and there will be a friction force.

11:23Now the box will end up sliding to the right, isn't it? And therefore the frictional force will be towards the left. We are also given the acceleration is one meters per second square and what direction would the acceleration be? Well, if you push the box towards the right, it will accelerate to the right.

11:38So the box is also accelerating at one meters per second square to the right. Again, our first step would be to resolve this force into horizontal and vertical, and these will be the two values. And since acceleration is given to us and we are asked to find one of the force vectors, we'll again use Newton's second law acceleration equals net force divide by mass.

11:59But the acceleration is purely in the X direction, right? It accelerates forward, if the box accelerates at one meters per second square, it's going to accelerate forward in this direction. So since acceleration is only in the X direction, I only need to look at the net force in the X direction.

12:14So let's do that. So net force in the X direction, what's that going to be? Well that's going to be, if I choose right side to be positive, this number minus this, so it's going to be Fc cos 45 degrees. Remember, Fc is the force of the child, okay? Minus the force of the friction.

12:32This is the net force in the X direction. And I know that equals mass times the acceleration in the X direction. And so from this, I can calculate what, I can rearrange to get what the force of friction is going to be. So that's gonna look like that. And now we can plug in all the values.

12:48The force of the child is 160 Newtons. Cos 45 is a known value, mass is given to be 12 kilograms, and we know the acceleration in the forward direction is one meters per second square. So if I plug this into my calculator, this is what I'll end up with. Again, we may be asked to round it off.

13:06So let's say let's round it off to three significant figures, I'll end up with 101 Newtons. So that is the frictional force acting on the box. And if you're wondering what about the vertical? Well, in the vertical direction the net force has to be zero because there is no acceleration in the vertical direction.

13:23And we know these two values, the values of these two forces, which means I can use the same thing in the vertical to figure out the magnitude of the normal force. So yeah, for extra practice, you can try figuring it out as well.