Forces in 2D Legacy Problem #9 Guided Solution

A good problem solver is going to read a problem carefully and develop a mental picture of what's going on. You're going to identify the known and unknown values and plot out a strategy as to how to get from the known values to the unknown value. Here the act of actually visualizing the situation is centered around drawing the forces which are acting upon a crate of books. There is one force stated here, 692 Newtons at an angle of 36 degrees, and I'm going to use the organizational structure provided on this help page to organize this information. So that is the F applied, 692 Newtons, and the theta or the angle it makes with the horizontal is 36 degrees. The mass of the books is 112 kilograms, and the coefficient of friction, mu, between the crates and the vinyl floor is 0.548. Now what I wish to do is determine the acceleration. That's my major unknown. There's no way to determine the acceleration until I first determine the net force. So I want to know what all these forces add up to. Now the immediate problem you encounter is that the 692 Newtons at 36 degrees is neither a horizontal nor a vertical force. So it doesn't add very well to any of the friction forces or gravity and normal forces. So what we must do is to break it up or resolve it into its components, into its effect in the horizontal and the vertical direction. So you'll notice there's a little force triangle here, and we know the 692 Newtons is the hypotenuse of that force triangle, and if we wish to determine the Fx value, we'll have to find the side adjacent to 36 degrees on the force triangle. So we use the cosine function to do that. We go 692 multiplied by the cosine of 36 degrees. That gives us the Fx value, and it comes out to be 559.8398 Newtons. Now to determine the y value, the y component of this force, what we need to do is use the sine function to find the side opposite the 36 degrees. So we go 692 Newtons multiplied by the sine of 36, and we determine the Fy value to be 406.7474 Newtons. And I hope you've written these force values down in the blanks that are provided. Once you've resolved the 692 Newtons into its two components, you no longer have to bother with the 692 Newtons. Erase it from the page or from your mind, or somehow ignore it, so that now you recognize what you have is a situation with five individual forces. Two of them are going left and right, and three of them are going up and down. Now we want to determine what the net force and then the acceleration is. To determine the net force, we need to determine the friction force going to the left. The only way to determine that is to know the F-norm, and to use F-frict equal mu times F-norm. The only way to determine F-norm is to recognize that all the vertical forces sum up to zero. So that would mean that the F-grab is going to be equal to the Fy value we just calculated plus the F-norm. The F-grab can be calculated as mg, 112 kilograms times 9.8 Newtons per kilogram. That comes out to be 1097.60 Newtons, and that's equal to the Fy value plus the F-normal value. So the F-normal value is going to be 1097.60 Newtons minus the 406.7474 Newtons. That comes out to be 690.8526 Newtons. I'm using an exorbitant number and significant digits here. I'll do my rounding at the end. Now that I have the F-normal value at 690.8526 Newtons, I can go F-frict equal mu times F-norm. 0.548 times 690.8526 gives me an F-friction value of 378.5872 Newtons. Now I need to determine the net force, the sum of all the forces. The verticals have balanced since we're not accelerating up and down. That usually is the case when you're pulling a crate across the floor. What we're doing is accelerating to the right, and so we can determine the net force by first taking the Fx value and then subtracting from it the F-frict value. The 559.8398 Newtons minus the 378.5872 Newtons is equal to 181.2525 Newtons. Now if I take that 181.2525 Newtons and divide it by the mass, that gives me the acceleration. So when you divide F-net by 112 kilograms, you get 1.6183 meters per second per second. And rounding that to three significant figures would give you a value of 1.62 meters per second per second.