Video: Calculating the Scalar Product of Two Vectors Shown on a Grid

An object moves along a straight line. As the object moves, it is acted upon by a constant force. The diagram shows the displacement of the object 𝐝 and the force acting on it, 𝐅. Each of the grid squares in the diagram has a side length of 1. How much work is done by the force?

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Video Transcript

An object moves along a straight line. As the object moves, it is acted upon by a constant force. The diagram shows the displacement of the object 𝐝 and the force acting on it 𝐅. Each of the grid squares in the diagram has a side length of one. How much work is done by the force?

In this question, we’re presented with a diagram showing two vectors, 𝐝 and 𝐅. We are told in the question that 𝐝 is the displacement of an object and 𝐅 is the force acting on that object. We are also told that this force 𝐅 is constant, which means that it always has the same value, and we don’t need to worry about it changing with time.

We are asked to find how much work the force does on the object. So let’s begin by recalling the definition of the work done by a force. The work done on an object by a force, written here as π‘Š, is equal to the scalar product of the force 𝐅 and the displacement 𝐝 of the object. Work done is a kind of energy, so it has units of energy. Specifically, if the force is measured in units of newtons and the displacement is measured in meters, then the work done, the scalar product of the force and the displacement, will have units of newton meters, which are equal to joules.

The question is asking us to find the work done by the force. And we’ve just said that this is equal to the scalar product of the force and the displacement. So let’s recall a definition of the scalar product of two vectors. We’ll consider two general vectors that we’ll label 𝐀 and 𝐁. If we suppose that both vectors lie in the π‘₯𝑦-plane, then we can write them in component form as an π‘₯-component labeled with a subscript π‘₯ multiplied by 𝐒 hat plus a 𝑦-component labeled with a subscript 𝑦 multiplied by 𝐣 hat.

Remember that 𝐒 hat is the unit vector in the π‘₯-direction and 𝐣 hat is the unit vector in the 𝑦-direction. The scalar product 𝐀 dot 𝐁 is then given by the π‘₯-component of 𝐀 multiplied by the π‘₯-component of 𝐁 plus the 𝑦-component of 𝐀 multiplied by the 𝑦-component of 𝐁. In other words, the scalar product of two vectors is given by the product of the vectors’ π‘₯-components plus the product of their 𝑦-components.

This general expression for the scalar product of two vectors tells us that if we want to find the scalar product 𝐅 dot 𝐝, then we need to work out the π‘₯- and 𝑦-components of our vectors 𝐅 and 𝐝. These vectors are given to us in the form of arrows drawn on a diagram. And we’re told in the question that the grid squares in this diagram each have a side length of one. What this means is that a length of one square in the diagram represents one unit of either displacement or force.

Assuming that both our force and displacement are measured in SI units, this means that, for our vector 𝐝, one square in the diagram represents one meter. And for 𝐅, one square represents one newton. We’ll define the horizontal direction in our diagram to be the π‘₯-direction and the vertical direction to be the 𝑦-direction. We see that our vector 𝐝 points along our horizontal grid line, while our vector 𝐅 points along our vertical grid line. This means that, for our vector 𝐝, the 𝑦-component is zero, while for our vector 𝐅, the π‘₯-component is zero.

Writing out two vectors in component form, we then have that 𝐝 is equal to some π‘₯-component multiplied by 𝐒 hat plus zero meters multiplied by 𝐣 hat and that 𝐅 is equal to zero newtons multiplied by 𝐒 hat plus some 𝑦-component multiplied by 𝐣 hat. To find these missing components, we need to start at the tail of each vector and count the number of squares until we reach the tip.

Let’s start with vector 𝐝. We begin at the tail of the vector and we count one, two, three, four, five, six, seven, eight squares in the negative π‘₯-direction until we reached the tip. Since for our vector 𝐝 one square is equal to one meter, this tells us that the π‘₯-component of 𝐝 is equal to negative eight meters.

Now let’s do the same thing with vector 𝐅. Again, we start at the tail of the vector, and we count one, two, three squares in the negative 𝑦-direction until we reach the tip. Since for 𝐅 one square is equal to one newton, this tells us that the 𝑦-component of 𝐅 is equal to negative three newtons.

So we now have expressions for both of our vectors 𝐝 and 𝐅, written in component form. We have that 𝐝 is equal to negative eight meters 𝐒 hat plus zero meters 𝐣 hat and that 𝐅 is equal to zero newtons 𝐒 hat plus negative three newtons 𝐣 hat. Now that we have our force and displacement vectors in component form, we are ready to calculate the scalar product 𝐅 dot 𝐝.

From our general expression for the scalar product of two vectors, we see that the first term is given by the product of the π‘₯-components of the vectors. So in the case of the scalar product 𝐅 dot 𝐝, that’s the π‘₯-component of 𝐅, which is zero newtons, multiplied by the π‘₯-component of 𝐝, which is negative eight meters. We then add a second term equal to the product of the 𝑦-components of the vectors. In our case, that’s the 𝑦-component of 𝐅, which is negative three newtons, multiplied by the 𝑦-component of 𝐝, which is zero meters.

Now we just need to evaluate this expression here. The first term is zero newtons multiplied by negative eight meters, which gives us zero newton meters. Then, the second term is negative three newtons multiplied by zero meters, which also gives us zero newton meters. Then we have zero newton meters plus zero newton meters. And when we add zero to zero, we get a result of zero so that our scalar product 𝐅 dot 𝐝 is equal to zero newton meters.

Now, we already said that the units newton meters are equal to units of joules and that the scalar product 𝐅 dot 𝐝 gives us the work done by the force. So if we replace our units of newton meters by units of joules and replace our scalar product 𝐅 dot 𝐝 by the work done, then we have that the work done by the force is equal to zero joules. So our answer to the question is zero joules.

But it turns out there was another way that we could’ve got to this answer that might provide us with a little more physical insight. The scalar product of two vectors 𝐀 and 𝐁 can also be written as the magnitude of 𝐀 multiplied by the magnitude of 𝐁 multiplied by the cosine of the angle between 𝐀 and 𝐁, which here we’ve labeled πœƒ. If we look at the function cos πœƒ, then we see that for a value of πœƒ equals 90 degrees, cos πœƒ equals zero. From our expression for the scalar product of two vectors, we see that this cos πœƒ term means that whenever πœƒ equals 90 degrees, or in other words whenever we have perpendicular vectors, then the scalar product of these two vectors will be zero.

In our calculation of the scalar of product 𝐅 dot 𝐝, then as soon as we’d got our two vectors in component form, we could see that the 𝑦-component of our displacement vector 𝐝 was zero, while the π‘₯-component of our force vector 𝐅 was zero. This meant that we knew our vector 𝐝 lay purely along the π‘₯-direction, whilst our vector 𝐅 lay purely in the 𝑦-direction. So we could see that our vectors 𝐝 and 𝐅 were perpendicular to each other.

We could also have seen this by measuring the angle between the two vectors on the diagram. And once we knew that the angle between our two vectors 𝐅 and 𝐝 was equal to 90 degrees, then using this expression here, we could’ve said straight away that their scalar product would be zero.

Since the work done on an object by a force is equal to the scalar product of the force and the displacement, we could then have concluded immediately that the work done by the force would be zero joules. So why does this method have the potential to provide more insight? Well, we’ve said that whenever we have two perpendicular vectors, we know that their scalar product is zero. And we know that the scalar product of the force and displacement gives us the work done on the object by the force. So this means that we can generalize from our result to say that any force that is perpendicular to an object’s displacement does no work on the object.

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