Video: Resultant Motion and Force

Two ropes tied around a heavy stone are pulled with the magnitudes and at the angles shown in the diagram. The 90 N force acts at an angle 30° above the horizontal. What is the magnitude of the net horizontal force acting on the stone due to the pull from the ropes? Answer to the nearest newton. What is the magnitude of the net vertical force acting on the stone due to the pull from the ropes? Answer to the nearest newton.

08:12

Video Transcript

Two ropes tied around a heavy stone are pulled with the magnitudes and at the angles shown in the diagram. The 90-newton force acts at an angle 30 degrees above the horizontal. What is the magnitude of the net horizontal force acting on the stone due to the pull from the ropes? Answer to the nearest newton.

Okay, so, in this diagram, we can see that there are two forces, this 50-newton force and this 90-newton force, acting on a stone. Now we can assume that this stone is located here at the origin of the two axes. And as well as this, we’ve been told that the 90-newton force is acting at an angle of 30 degrees above the horizontal. And the diagram tells us that this angle here is 50 degrees. In other words then, the 50-newton force is acting at an angle of 50 plus 30. So, that’s 80 degrees above the horizontal.

Now because both forces are acting on the stone at the same time, the resultant force on the stone is going to be the vector sum of these two forces. And to add forces vectorially, we can think about this 50-newton force being moved into this position here. And then, the resultant force on the stone will look something like this. We can call that resultant force 𝑅.

Now the first part of the question is actually asking us to find the magnitude, or the size, of the net horizontal force acting on the stone. And so, what we’re being asked to find is the horizontal component of the net force 𝑅. However, because forces add vectorially, we don’t actually need to work out what 𝑅 is in the first place. We could just find the horizontal component of the 90-newton force and the horizontal component of the 50-newton force and add them together.

So, let’s first start by thinking about the horizontal component of the 90-newton force. Now, we know that the angle that this 90-newton force makes with the horizontal is 30 degrees. And then, at this point, we can draw a right-angled triangle showing the horizontal and vertical components of this 90-newton force. So, this orange arrow’s representing the horizontal component. Let’s call it 𝐻 subscript 90. And this vertical blue arrow’s representing the vertical component. Let’s call it 𝑉 subscript 90.

Now at this point, we can recall that the horizontal and vertical components are at 90 degrees to each other. Therefore, what we’ve got here is a right-angled triangle. And hence, we can solve for the length of this vector, which represents the magnitude of the horizontal component of this 90-newton force, by using trigonometry.

We use the useful memory trick SOH CAH TOA to tell us which trigonometric function we’re going to need to use in this situation. SOH tells us that the sine function needs to be used when we’re looking at the side of the triangle that’s opposite to the angle that we’re considering, in this case 30 degrees, as well as, the hypotenuse of the triangle. CAH tells us that cosine needs to be used when we’re thinking about the adjacent side to the angle and hypotenuse of the triangle. And TOA tells us that tan, or tangent, needs to be used when considering the sides of the triangle that are opposite to the angle and adjacent to the angle.

Now in this case, what we’re trying to find is 𝐻 90. That’s the horizontal component of the 90-newton force. And so, this side of the triangle is adjacent to this 30-degree angle. And as well as this, the quantity that we do know is the 90-newton force. That happens to be the hypotenuse. So, we’re working with adjacent and hypotenuse, adjacent, hypotenuse. Therefore, we need to use cosine.

And hence, we can say that the cosine of the angle that we’re considering, which is 30 degrees, is equal to the length of the side that’s adjacent to it, which is 𝐻 90, divided by the length of the hypotenuse, which we know to be 90 newtons.

And then at this point, we can rearrange this equation to solve for 𝐻 90. We do this by multiplying both sides by 90 newtons. This way, 90 newtons on the right-hand side will cancel, as we can see here. And so, what we’re left with is that 90 newtons multiplied by the cosine of 30 degrees is equal to 𝐻 90. That’s the horizontal component of the 90-newton force.

Evaluating the left-hand side, we find that 𝐻 90 is equal to 77.94228 dot dot dot newtons. So, we’ve just calculated the horizontal component of the 90-newton force.

Now whilst we’re here, we may as well calculate the vertical component. This will become useful in a later part of the question. So, let’s use SOH CAH TOA once again to work out what the vertical component of the 90-newton force is. Once again, the angle that we know is 30 degrees. And this time, we’re trying to calculate the vertical component. So, that side of the triangle is the side opposite to the angle that we know. And as well as this, once again, we know the magnitude of the hypotenuse. It’s 90 newtons. And so, we’re looking at the side opposite, as well as the hypotenuse, opposite, hypotenuse. We’re going to need to use sine.

So, we can say that the sine of 30 degrees is equal to the magnitude of the opposite side, which happens to be 𝑉 90, divided by the magnitude of the hypotenuse, which is 90 newtons. Rearranging this equation, once again, as we did earlier, we see that 90 newtons multiplied by sine of 30 degrees is equal to the vertical component. And then evaluating the left-hand side, we find that 45 newtons is the magnitude of the vertical component. And so, at this point, we found both the horizontal and the vertical components of the 90-newton force.

Let’s do the same thing for the 50-newton force. So, here is our 50-newton force. And, yet again, we can split it into vertical and horizontal components. We can see that this is the horizontal component of the 50-newton force, which we will call 𝐻 subscript 50. And this is the vertical component, 𝑉 subscript 50. And as usual, the angle between the horizontal and vertical components is 90 degrees. But what is this angle here?

Well, as we said earlier, that angle is the angle between the 50-newton force and the horizontal. In other words, it’s this 50 degrees plus this 30 degrees. And hence, this angle here is 80 degrees. Now based on all this information, we can use a similar method to before to work out the horizontal and vertical components of this 50-newton force.

Starting with the horizontal component, we’re looking at the adjacent side to the angle that we know, that’s 80 degrees, and the hypotenuse of the triangle, which is the 50-newton force itself. So, once again, we use cosine, because we’re talking about the adjacent and hypotenuse. This time, we say that the cosine of 80 degrees is equal to the adjacent side, that’s 𝐻 subscript 50, divided by the 50-newton force. And then, when we rearrange and simplify as before, we find that the horizontal component is 8.6824 dot dot dot newtons.

And then working at the vertical component, once again, we use sine. Because this side is now opposite to the angle that we know. And we’re also talking about the hypotenuse, the 50-newton force. And so, we can say that the sine of 80 degrees is equal to the vertical component 𝑉 50 divided by the 50-newton force. Hence, we’ve found that the vertical component is 49.24038 dot dot dot newtons.

Now the first part of this question, as we said earlier, is asking us to find the magnitude, or size, of the net horizontal force acting on the stone. So, the magnitude of the net horizontal force is simply going to be the sum of the horizontal components of the two forces that are acting on the stone. And this is because forces add like vectors. So, the horizontal component of the 90-newton force is going to add to the horizontal component of the 50-newton force, which we can imagine as being placed at the end of this vector.

And so, the net horizontal force is going to be this whole vector here. So, we can say that the net horizontal force, which we’ll call 𝐻 subscript 90 plus 50, is equal to the horizontal component of the 90-newton force plus the horizontal component of the 50-newton force. And then, when we add these two together we find that this value becomes 86.62468 dot dot dot newtons.

But we need to give our answer rounded to the nearest newton. And so, this is the value that we need to round. And we look at the next position, the first decimal place, to tell us what happens to this value. Now the value in the first decimal places is a six. Six is larger than five. Therefore, this value is going to round up. And hence, we’ve found our answer to this part of the question. The magnitude of the net horizontal force acting on the stone due to the pull from the ropes is 87 newtons to the nearest newton. So, let’s look at the next part of the question.

What is the magnitude of the net vertical force acting on the stone due to the pull from the ropes? Answer to the nearest newton.

So, this question is very similar to before. Except, we’re now finding the net vertical force, in other words, the vertical component of the net, or resultant force. And as we did with the horizontal components, we can add the vertical component of the two forces acting on the stone to find the net vertical force acting on the stone.

In other words then, we can see that the vertical component of the net force, which we we will call 𝑉 subscript 90 plus 50, is equal to the vertical component of the 90-newton force plus the vertical component of the 50-newton force. And then, when we evaluate the right-hand side of this equation, we find that it becomes 94.24038 newtons. Or, rounded to the nearest newton, our answer becomes 94 newtons.

Nagwa uses cookies to ensure you get the best experience on our website. Learn more about our Privacy Policy.