Three horizontal forces act on a box (mass = 8 kg) sitting on a smooth surface. F1 is 30 N acting at 53° counterclockwise from the +x axis, F2 is 13 N acting at 67.4° clockwise from the +x axis, and F3 is 20 N directly along the -x axis. What are the magnitude and direction of the box's acceleration?

1.5 m/s 2 , 76° ccw from +x-axis

Three horizontal forces act on a box (mass = 8 kg) sitting on a smooth surface. F 1 is 30 N acting at 53° counterclockwise from the +x axis, F 2 is 13 N acting at 67.4° clockwise from the +x axis, and F 3 is 20 N directly along the -x axis. What are the magnitude and direction of the box's acceleration?

Hey, guys. So we've got these 3 horizontal forces that are all pulling on this box here. What we wanna do in this problem is we wanna figure out the magnitude and it and the direction of the box's acceleration. So let's get to it. Basically, what we wanna do is we wanna figure out the magnitude of a and we also wanna figure out theta a. That's gonna be the direction of this two dimensional vector. Right? We've got these two dimensional forces that are gonna produce a two dimensional acceleration and that's what we wanna find. Alright. So there's actually 2 different ways that we could solve for the acceleration, the magnitude. Remember that you have the magnitude of any vector. It's just the, Pythagorean theorem of a squared or a x squared plus a y squared, right, of its components squared. However, the other way that you could figure out the acceleration is by using f equals m a. So I'm gonna scoot this down for just a second. Remember that the acceleration is also just equal to the net force divided by the mass. If you can figure out the magnitude of the net force and you divide it by the mass, that's going to be the magnitude of the acceleration. In 2 dimensions, one dimension, it doesn't matter wherever it is. Right? So basically, what we can do here is instead of trying to calculate a x and a y separately, it actually makes a little bit more sense for us to figure out the magnitude of the net force, because we've got these three forces. They're all pulling in different directions. And what we wanna do is we wanna figure out where is the magnitude of that net force going to be. So one way we can do this is that the magnitude of this net force here, actually gonna write this as sigma f just so you know it's the net force, is really just if we add together f 1 plus f 2 plus f 3 altogether, but they point in different directions. We're actually gonna have to do this in the x and the y axis. We're gonna have to do this by components. Right? So, basically, what happens is, we're gonna have to go ahead and draw the free body diagram which is already done for us. Then we can we're gonna have to decompose all of our 2 dimensional forces so we can figure out what this net force is. And to do that, we're gonna need the x and y table. So I've got my x, my y table like this. And so let's see. That looks about right. And then I've got my f one, my f two, and then I've gotta add another one, which is f 3. And then if I add all those things together, their x and y components, that's gonna give me the sum or the net force of all these things added together. Right? So I'm gonna have to do that. So I'm gonna have to decompose all of these two dimensional vectors or forces. So I've got this f one here that we'll start off with. This has a component in the x axis f one x, component in the y axis f one y, and I can calculate this by just using my normal sine and cosine, right? So f one x goes with cosine. So really this f one, remember, we're given in the problem, this f one is 30 newtons, this f two is 13, and this f 3 is 20. So to find out the x component, I'm just gonna use 30 times the cosine of the angle, which is 53, and that gives me 18. Then I do the same thing, 30 times the sine of 53, that's gonna give me the y components, That's gonna be 24. Now we just look at the directions. Right? So we have this up, which I'm gonna choose as positive, and the right as as positive as well. And so basically anything in this quadrant here is gonna be positive. So these both get written as positives here. This is gonna be 18 and 24. Let's do the same exact thing for f 2 now. So this f 2 x is gonna be this, 13 times the angle, which is 67.4. This is gonna be my component f2x. This is gonna be 13 times the cosine 67.4 and you're gonna get 5. And then when you do this for the y, f two y's, you're gonna get 13 times the sine of 67.4. Notice how we didn't put a negative sign here even though the angle was below the x axis, and that's because we worry about the directions later when we're looking at the components, and their directions. Right? So this is gonna be 12. However, because this f two x points to the right, it's positive, but this f two y points downwards. So what happens is it picks up a negative sign. So I've got 5 and then negative 12. And then f 3 is actually pretty straightforward because it's 20 purely along the negative x axis. What that means is that the x component is not 20 but negative 20, right, because this is our negative x axis and then our y component is 0. So when you when you go and add up all these things together, you're gonna get 185 and negative 20, that gives you 3, and then this one gives you 12. So basically, these are your x and y components of the net force. So basically, what we can do now is the magnitude of this net force here, the magnitude of our net force is really just gonna be the pythagorean theorem of fx squared plus fy squared, or really just the sums. Right? Basically those two numbers that I just got there. So basically when we do this, we're gonna get 3 squared plus 12 squared, and you're gonna get 12.4. That is the magnitude of this resultant vector or this net force. So this is 12.4 newtons. Alright. So you'll notice that 12.4 isn't actually part of our answer choice, and that's because we're not done yet. Remember that this net force here, actually, in order to get the acceleration, we're gonna have to divide our net force by the mass. Alright? So, basically, what happens is we're gonna get a net force of 12.4, and then you divide it by the 8 kilograms of the mass. And this is where you get 1.5, and this is gonna be meters per second squared. So this is one of your answer choices right here. 1.5. So that eliminates c and d. Now to figure out which one, we have to figure out actually the direction. This is theta. And to do this, what we're gonna do is we're gonna do tangent inverse. Now normally, what we would do is we would do a y over a x. But remember, we didn't calculate the acceleration. But remember that this two dimensional force, right, once we figure it out with the components, it actually points somewhere in this direction here. This force is also gonna be the same direction as the acceleration. Right? The forces components and the accelerations components are gonna have the same ratio. So what that means is that another way to calculate this theta a is by using the tangent inverse of not a y over a x. We can also use fy over fx. It's gonna give you the exact same answer. So basically we're gonna get our theta a is the tangent inverse and we have 12 over 3 and when you do this you're gonna get 76 degrees. So 76 degrees means we're looking at answer choice b, and a is not the correct one. Alright? So that's it for this one, guys. Let me know if you have any

Three horizontal forces act on a box (mass = 8 kg) sitting on a smooth surface. F 1 is 30 N acting at 53° counterclockwise from the +x axis, F 2 is 13 N acting at 67.4° clockwise from the +x axis, and F 3 is 20 N directly along the -x axis. What are the magnitude and direction of the box's acceleration?

1.5 m/s 2 , 76° ccw from +x-axis

1.5 m/s 2 , 14° ccw from +x-axis

6.9 m/s 2 , 1.1° cw from -x-axis

6.9 m/s 2 , 0.46° cw from -x-axis

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6. Intro to Forces (Dynamics)

Forces in 2D

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6. Intro to Forces (Dynamics)

Forces in 2D

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Multiple Choice

Three horizontal forces act on a box (mass = 8 kg) sitting on a smooth surface. F₁ is 30 N acting at 53° counterclockwise from the +x axis, F₂ is 13 N acting at 67.4° clockwise from the +x axis, and F₃ is 20 N directly along the -x axis. What are the magnitude and direction of the box's acceleration?

1.5 m/s², 14° ccw from +x-axis

1.5 m/s², 76° ccw from +x-axis

6.9 m/s², 1.1° cw from -x-axis

6.9 m/s², 0.46° cw from -x-axis

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Verified step by step guidance

Step 1: Break down each force into its x and y components. For F1, use trigonometry: F1x = F1 * cos(53°) and F1y = F1 * sin(53°). For F2, F2x = F2 * cos(67.4°) and F2y = F2 * sin(67.4°). F3 is along the -x axis, so F3x = -20 N and F3y = 0 N.

Step 2: Calculate the net force in the x-direction by summing up all the x-components: F_net_x = F1x + F2x + F3x.

Step 3: Calculate the net force in the y-direction by summing up all the y-components: F_net_y = F1y + F2y + F3y.

Step 4: Determine the magnitude of the net force using the Pythagorean theorem: F_net = sqrt((F_net_x)^2 + (F_net_y)^2).

Step 5: Calculate the acceleration of the box using Newton's second law, a = F_net / m, where m is the mass of the box. Then, find the direction of the acceleration using the arctangent function: θ = arctan(F_net_y / F_net_x).