Forces in 2D Legacy Problem #23 Guided Solution

In problems like this, it becomes imperative that you practice the habits of a good problem solver, that you read the problem carefully, developing a mental picture of what's going on, identify the known quantities and the unknown quantities, and plot out a strategy to get from the known to the unknown. Effective strategy centers around the use of a free body diagram. You can use it to organize the known information as well to help you plot the strategy. A free body diagram is provided here on this help page. I would use it in order to help work your way through this problem. The mass of the crate of books that's being pushed up the loading dock is 86 kg. That's the m value. Now, we're told that it's being pushed up the loading docks at a constant speed of 24 cm per second. And the important part of that phrase is not the 24 cm per second, but the constant speed. The fact that it's moving at a constant speed in the same direction means that the acceleration is 0 m per second squared, and the net force is zero. Thus, all the forces acting up on this crate of books are going to be balanced. The ones that are going parallel to the incline balance each other out, and the ones going perpendicular to the incline will balance each other out. Now, I want to begin my force analysis by taking the 86 kg, the mass of the crate of books, and multiplying it by 9.8 N per kg. That's going to give me my Fg value. Fg will be important because we're going to use it to calculate the parallel component and the perpendicular component of the forces. The equations for doing so are provided on the overview page. Now, the reason you need to know the perpendicular component of the force is because it's equal to the normal force. The normal force is important for calculating the friction force. The forces of interest here are the forces that are parallel to the plane. After all, if Ben Laborin is going to pull this crate up the loading dock at a constant speed, the force up the incline plane has got to balance the two forces that are going down and parallel to the incline plane. This F applied has got to be equal to F parallel plus F friction. So, I need to find F perpendicular first so I can get F normal and then F friction. Finding F perpendicular is a matter of taking this Mg value of 842.8 N and multiplying it by the cosine of 12 degrees. When I do that, I get 824.3828 N. Not all of those digits are significant, but I'm going to write them down and use them throughout the calculation, rounding at the end. So, now I know the normal force is 824.3828 N, and I can find the friction force if I go mu times F norm. The mu is 0.74, multiply it by F norm, and you get a value for friction of 610.04 N. That's one of the two forces that are directed parallel to the incline and down the hill. Ben's going to have to pull with enough force to balance these two. The second force is the parallel component of gravity, and it can be calculated by taking this Mg value of 842.8 N and multiplying by the sine of 12 degrees. When you do that, you get 175.228 N. Now, together, these two forces, F friction and F parallel, add up to a value of 785.27 N. If Ben is to pull this crate, or push this crate, up the rolling dock at a constant speed, he must push with that amount of force. He may need to push with a little more to get it started, but once in motion, an object will remain in motion at a constant speed, provided all these forces are balanced. So the final answer to our question, 785.27 N, can be rounded to 790 N, consistent with the idea of two significant figures.