Video Transcript
The forces π
one equal to ππ’ plus three π£, π
two equal to negative three π’ plus ππ£, and π
three equal to ππ’ minus three π£ acted on a body for three seconds. If their combined impulse on the body was π equal to three π’ minus six π£, find the values of π and π.
In this question, weβre dealing with a body or object that has three forces acting on it: π
one, π
two, and π
three. Now weβve drawn a diagram of these forces acting on the body, but this diagram canβt be completely accurate. This is because weβre missing some information about each of these forces. The π₯-component of π
one, the π¦-component of π
two, and the π₯-component of π
three are all unknown. In the question, these values are represented by the quantities π and π, and itβs our job to find the values of π and π.
The question gives us some information about the amount of time for which these forces act on the body and the impulse that is exerted on the body during this time. To answer this question, we need to use an equation that links together the force acting on an object, the amount of time for which the force acts, and the impulse thatβs produced. This equation is πΌ equals πΉ times Ξπ‘. The impulse πΌ produced by a force is equal to that force πΉ multiplied by the amount of time Ξπ‘ for which it acts. Because impulse and force are actually both vector quantities, we can draw a half arrow over the top of the π and the π
to signify this.
Now that we have this equation, we might notice that it deals with a single force, whereas in the question we have three forces. This equation is true for each of the individual forces. For example, the impulse due to π
one, which we can call π one, is equal to π
one times Ξπ‘. But notice that in this question, weβre talking about the combined impulse. This is the impulse thatβs applied to the body due to the combined effects of all three forces. In other words, itβs the impulse due to the net force.
So, in order to answer this question, we can start by finding an expression for the net force in terms of the components of π
one, π
two, and π
three that weβve been given in the question. We can then multiply this by Ξπ‘, thatβs three seconds, in order to come up with an expression for the combined impulse in terms of π and π. We can then compare this expression to the actual impulse given in the question in order to find the values of π and π.
So, letβs start by finding an expression for the net force. This is equal to the sum of π
one, π
two, and π
three. Weβre told that π
one is equal to ππ’ plus three π£. Then, weβre adding on π
two, which is equal to negative three π’ plus ππ£. And then we add on π
three, which is equal to ππ’ minus three π£. And here where weβre adding negative three π’, we can just write minus three π’. Next, we just add up the π’- and π£-terms separately. Looking at the π’-terms, we have ππ’ minus three π’ plus ππ’. So, in total, thatβs π minus three plus π times π’. And this simplifies to two π minus three. Next, we can look at the π£-terms. We have three π£ plus ππ£ minus three π£. So, in total, thatβs three plus π minus three times π£. Three minus three is, of course, zero, which just leaves us with π times π£. So, our net force π
net is equal to two π minus three π’ plus ππ£.
Next, we can obtain an expression for the combined impulse by multiplying π
net by Ξπ‘. So, weβre multiplying our expression for π
net, thatβs two π minus three times π’ plus ππ£, by the amount of time for which these forces act on the body. Weβre told in the question that this is three seconds. So, we just multiply all of this stuff by three. Here, weβre multiplying a vector by a scalar quantity. So, we multiply each term inside the parentheses. In other words, we multiply each vector component by three. First, we have three times two π minus three π’, and then we have three times ππ£. Next, we can simplify this term by multiplying across the parentheses again. So, we have three times two π, which is six π, and three times negative three, which is negative nine. This is then multiplied by π’, giving us six π minus nine times π’ plus three ππ£.
So, weβve now come up with an expression for the combined impulse of the three forces given in the question. This expression is in terms of the unknown constants π and π. But the question tells us that this impulse vector is equal to three π’ minus six π£. Since both of these expressions describe the same impulse vector, we can say that theyβre equal to each other. And this equation enables us to find the values of π and π. All we need to do is equate the π’-components and the π£-components, separately.
Letβs deal with the π’-components first. We can start by canceling the factor of π’ on each side, leaving us with six π minus nine equals three. We can then add nine to both sides, which gives us six π equals 12. And finally, dividing both sides by six gives us π equals two. So, this is one half of our answer. To find the other half, we equate the π¦-components. Once again, we can start by canceling a factor of π£ on both sides to leave us with three π equals negative six. Then, dividing both sides by three tells us that π is equal to negative two. So, this is our final answer.
By finding an expression for the net force that acts on the body, we were able to come up with our own expression for the impulse exerted in terms of π and π. We could then equate this to the actual impulse given in the question to determine the values of π and π. So, when the forces described in the question act on the body for three seconds and produce an impulse of three π’ minus six π£, we can work out that π equals two and π equals negative two.