OpenStax College Physics, Chapter 3, Problem 6 (Problems & Exercises)

This is College Physics Answers with Shaun Dychko. In problem 6, we are going to confirm that when we walk along vectors A and B in a different order compared to question (5)— in question (5), we walked first along vector A and then along vector B and ended up at some final position— now in question (6), we are going to walk first along vector B and then along vector A and see if we end up in the same final position and we can tell graphically that we are expecting that we will and we will be showing that when you have vector A plus vector B equaling some resultant that is the same as vector B plus vector A equaling the same resultant so the order of addition does not matter is what we are going to show here. We are going to add vectors A and B using components. So we have some resultant vector here and it has a component R x— the x component of the resultant— and then some component downwards R y and we will find the R x component by adding the x component of vector B to the x component of vector A and you can see that if you were to copy them and move them over... if you have vector A here and then add to that vector B... x components of course... you will end up with the x component of the resultant. So that's a graphical way of showing that and we are going to do it numerically down here and the same is true for the y component of the resultant. Now the y component of vector A is in the upward direction opposite to the y component of vector B, which is downwards but vector B has a y component downwards a bit more than vector A's y component is upwards and so we are left with a little bit downwards remaining when we add these two together. Okay! So we have to remember how to find the legs of a right triangle... this is a right triangle because it has a ninety degree angle here and this is a right triangle with a ninety degree angle there and the x component is called the opposite leg because it is opposite to the angle that we know— Θ A, which is 20 degrees— and this x component for vector B is called the adjacent because it's adjacent or beside the angle that we know—Θ B. So vector B is 40 degrees south of west so that means goes downward in this Southerly direction starting from west which is this negative x axis here. And vector A is at a direction 20 degrees west of north so 20 degrees towards the west which is left compared with north as our starting point so 20 degrees that way. Alright! So let's find the x component of the resultant by adding the x component of vectors A and B. So for vector A, this x component is pointing to the left, which is the negative direction conventionally, so we put a negative sign there to indicate that it's going to the left and then we multiply the length of the vector—the hypotenuse, in other words— by sin of Θ A because we know that the sin of an angle—Θ A, in this case— is the length of the opposite which is the x component of vector A divided by the hypotenuse which is just A and we can solve for A x by multiplying both sides by A and then switch the sides around, put A x on the left and then we are left with AsinΘ A on the right side and that's where this part comes from which is this. Okay! And the same kind of story with the x component for vector B except that we are using cosine since we have... the adjacent leg is what we are trying to find. So cosine of Θ B is the adjacent which is B x divided by the hypotenuse and then solve for B x by multiplying both sides by B. Alright! So then we plug in numbers. So we have negative 12 meters times sin of 20 degrees minus 20 meters times cosine of 40 degrees and this is negative 19.425 meters and we are keeping lots of digits here since this is an intermediate calculation— we don't round until the final answer. For the y component of the resultant, we want to combine the y component of vector A with that of vector B and that's A times cosine of Θ A minus B times sin of Θ B. So this minus appeared because we can see from the picture that the y component of vector B is downwards in the negative direction and we are using sin of Θ B because this is the opposite leg of this right triangle and we are using cosine for finding the y component of vector A because this y component is the adjacent leg of the right triangle. Okay! So we have 12 meters times cosine of 20 degrees minus 20 meters times sin of 40 and that's negative 1.5794 meters. So these are the components for vector R that we have found now and the length of R is the length of the hypotenuse of a right triangle and we can find it using the Pythagorean theorem. So we take the square of each of the legs, add them together and then take the square root of that sum. So we have the square of the x component of vector R—that's square of negative 19.425 meters— and add to that the square of the y component of the resultant— negative 1.5794 meters— and then take the square root of that sum and we have 19.489 meters is the length of the resultant. And these vertical lines here— these are the absolute value signs— with the vector, they indicate the length of the vector. And then we need to find the direction of the resultant so that's Θ R and that's this little angle in here— that's what we are finding now... also drawn here. So we will take the inverse tangent of the opposite, which is the y component divided by the adjacent which is the x component of the resultant. So it's the inverse tangent of negative 1.5794 meters divided by negative 19.425 meters and you know sometimes I wouldn't even bother putting these negatives in here because you can already tell where the position of the angle is and you know, we can just treat this as any old triangle and then indicate our direction using south of west. Okay! So we have the inverse tangent of this ratio here and we have 4.648 degrees and that is south of west as we can see from our picture up there. So the resultant is 19.5 meters, 4.65 degrees S of W—south of west.