### Video Transcript

Given that the two parallel forces π
sub one equals two π’ plus π£ and π
sub two equals negative four π’ minus two π£ are acting at π΄ negative three, negative five and π΅ five, three, respectively, determine their resultant π and find its point of action.

Now, itβs fairly straightforward to calculate the resultant of two or more forces; we simply find their vector sum. However, finding the point of action is a little trickier because it will depend on the direction in which the two forces are acting. So, letβs begin by finding the resultant; thatβs π
sub one plus π
sub two. Then, weβll sketch a diagram, and the direction in which the forces are acting will help us approximate the point of action of the resultant.

The sum of our two forces is two π’ plus π£ plus negative four π’ minus two π£. And remember, we can add a pair of vectors by simply adding their components. Two minus four is negative two. So, the π’-component is negative two. And then, we have one minus two, which is negative one. So, the π£-component is negative one. And that gives us a resultant force of negative two π’ minus π£.

So, with that in mind, how do we find the point of action? Well, letβs sketch this out first and double-check the direction in which our forces are acting. Here are our two points π΄ and π΅ are negative three, negative five and five, three, respectively. Now, weβve not actually plotted them on a coordinate plane just to keep the diagram really simple. Force π
sub one is two π’ plus π£ and that acts at point π΄, and force π
sub two is negative four π’ minus two π£. Weβre told that these are parallel forces, which is important, but we couldβve spotted this quite quickly by recognizing that force π
sub two is a multiple of force π
sub one and vice versa.

So, we know that the two forces are acting in the opposite direction to one another. Weβve calculated the vector force that describes the resultant. But we do know that if weβre dealing with the resultant of two parallel but unequal and unlike forces β in other words forces acting in opposite directions β the resultant will work in the direction of the force of greater magnitude. And in this case, the line of action of that resultant divides the distance between the lines of actions of the two forces externally from the side of the force of greater magnitude. And this is in an inverse ratio to their magnitude.

Now, the force that has the greater magnitude in this example is force π
sub two. So, point πΆ will lie on the line π΄π΅ but on the side of point π΅. We can calculate the distance between either point π΄ and πΆ or point π΅ and πΆ by looking at the ratios of these two distances. Itβs an inverse ratio to their magnitude. So, πΆπ΅ divided by πΆπ΄ is equal to the magnitude of force π
sub one divided by the magnitude of force π
sub two. And of course, we calculate the magnitude of the force by finding the square root of the sum of the squares of the components.

So, the magnitude of force π
sub one is the square root of two squared plus one squared, which is the square root of five. Similarly, the magnitude of force π
sub two is the square root of negative four squared plus negative two squared, which is equal to root 20 or two root five. So, πΆπ΅ over πΆπ΄ will be equal to the square root of five over two root five, which is equal to one-half. And this is really useful because we now know the ratio of the distances between points π΅ and πΆ and points π΄ and πΆ. Itβs one-half. In other words, the distance between π΅ and πΆ is half the distance between π΄ and πΆ. And of course, we know that πΆ lies on the line π΄π΅. So, we can, in fact, say that the vector ππ must be equal to the vector ππ.

Letβs clear a little bit of space and calculate the vector ππ. The vector ππ is the vector ππ minus the vector ππ, where π in this case is the origin. Itβs the point with coordinates zero, zero. ππ will be therefore five π’ plus three π£. And then, weβll subtract negative three π’ minus five π£, giving us vector ππ is eight π’ plus eight π£, which must also, of course, be equal to vector ππ. This means we can find the coordinates of point πΆ by moving eight units right and eight units up from the point five, three. So, point πΆ must have coordinates five plus eight, three plus eight or 13, 11.

And so, weβve completed the question. The resultant π is negative two π’ minus π£ and its point of action is 13, 11.