This problem, surprisingly, is very similar to example #2. It’s a force problem, and you’re given a force vector which you need to break into components. The major difference in this problem is that we’re given actual numeric values instead of variables. You will find out that all of these problems are really just the same person with different faces, as we press through the next example. Notice that the steps don’t really change, and that the answer is very easily acquired by simply following the steps.
Step 1: Identify your coordinate axis and draw it on the paper. Because the force that we’re looking for is horizontal component of the pulling force, we want to align our x-axis horizontally. If you’re having trouble determining how to align your axis, this will come easier as you do more problems. Running down false alleys and seeing how the problem may not work out nicely will help give you a better sense of how to orient your coordinate axis in the future. The very nice and important thing to remember is that no matter which axis you choose the final answer has to be the same. It will not change based on your selection in this step. However, the math to achieve it may be messier or cleaner, which will improve based on your expertise.
Step 2: Identify your vector and draw it on the paper. In the problem the rope is being pulled upwards with respect to the x-axis, so we can guestimate the direction that it’s pointed like so.
Step 3: Identify your angle and draw it on the paper. This is given to us in the problem, so the only thing left to do at this point is draw it where it goes.
Step 4: Use sine & cosine relationships for component definition. In this scenario, because of the position of the angle, our x-component will be the adjacent side, and the vector our hypotenuse. This means we’ll be able to use cosine to calculate the horizontal component of the force. This will be the amount that directly opposes the frictional force, when you drag the box across the floor. This is why this force is of interest to us. We’ll also just calculate the y-component for fun. Remember, when you calculate the sin/cos(35), remember that this is 35 DEGREES, or else you’ll have to convert to radians!
Step 5: Box your answer, you’re done! So, a box being pulled 35 degrees w/r/t the horizontal will pull with 8.91 Newtons of force in the x-direction and 5.74 Newtons in the y-direction. So, are you getting the feel for it yet? Do you see how these problems are all basically asking the same question even though the scenarios are all different? If not, go back and refresh the last few examples to see if you see the similarities now.
