(II) Two large snowcats are towing a housing unit north, as sh...

Question

(II) Two large snowcats are towing a housing unit north, as shown in Fig. 4–42. The sum of the forces F_A and F_B exerted on the unit by the horizontal cables is north, parallel to the line L, and F_A = 4200 N. Determine F_B and the magnitude of F→_A + F→_B .

Accepted Answer

Hello, fellow physicists today, we're gonna solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Two trucks are pulling a large storage container north as shown in the figure below the sum of the forces vector F one and vector F two that are exerted on the container by the horizontal steel cables in the north direction are also parallel to line L note that vector F one equals 4400 newtons, calculate vector F two and the magnitude of vector F one plus vector F two. Awesome. So that's our end goal is we're asked to find the value of vector F two and we're trying to find the magnitude of F one or vector F one plus vector F two. And that's the final answer that we're ultimately trying to solve for. We're trying to figure out what the value of F two is the vector F two to be precise and the magnitude value. Awesome. We're also given a figure to help us visualize this problem. So we have our two trucks, which we have like a gray backside and then it's like a yellow green front for the trucks. We have two trucks and they're pulling a large storage container north which the storage container is. This steel colored boxes, gray rectangular boxing. And as we could see from our figure that's provided to us, it says that this is a top view. So if let's just say we are looking down on it with like a drone. So the drones up in the air looking down at our two trucks, pulling this large storage container. So we have a vector fa going diagonal from our container to our truck on the left. And then we have a vector FB going from the container to our second truck. And these are denoted with our green arrows for both vector fa and vector FB. And then the angle between vector fa in our L direction. So and this is the steel cables in the north direction are also parallel to line L. So we have our line L. So between vector fa and line L is a 46 degree angle and from our line L to vector FB, it's 30 degrees. Awesome. So we're also given some multiple choice values. Let's read them off to see what our final answer pair might be. Let's note that all of our units are in units of newtons and our magnitude. So both the F two value is in units of newtons and our magnitude is in units of newtons and as the magnitude of total force is blank. So instead of saying that every single time, I'm just gonna say what the total, the magnitude of total force is. So A is 6330 and 8538 B is 433,538 C is 344,358. And finally D is 8358 and 6330. OK. So first off, let us create a detailed diagram of this problem to help us better visualize the forces at work. So I've gone ahead and found a fancy diagram for us to work with here. So we have in our figure that's provided to us, we called it vector FA. But on the far left, let's call this vector F one which is equal to 4400 newtons. And they have a line that denotes F one X. And then going to the far right, we have our vector F two with our line denoting F two X and then in the middle going up is we have our line are parallel to line L. So we have our line L in the middle and we have our 46 degrees and our 30 degrees respectfully for the angle. And this blue rectangle denotes our container in this case. OK. So now we need to write down all of our given information. So let's start off with the Y direction. So why direction, the right direction is the forward direction? So I'm just gonna say equals forward. So we don't forget is the forward direction and the X direction is to the right. OK. And the storage container will move forward in a straight line which implies that there is no acceleration in the X direction. Let us also note that vector one or F one. So vector F one is equal to 4400 newtons. And that is applied at the 46 degree angle towards the forward direction. And we have vector F two which is equal to, we do not know this value because this is the force exerted by the second truck. And this is the unknown value that we are solving for. So now to solve for this problem, we must recall and use Newton's second law which states that the net force in any direction where there is no acceleration will be zero. Therefore, we could write the following equation for the X direction. We can write that the sum of vector F one X plus vector F two X is equal to zero, given that vector F one is at a 46 degree angle. And the X and the X component can be found by recalling and using the following equation vector F one X equals F one multiplied by sine of 46 degrees. So we're using sign because as you can see where we have our angles for 46 and 30 we could see that it forms two triangles. So we're gonna be using geometric identities to help us solve for this problem. So now if we assume that vector F two is applied at a 30 degree angle towards the forward direction, its X component will be written as vector FX one equals vector F one multiplied by sine of 46 degrees. OK. As shown in our diagram, vector F two is applied at an angle of 30 degrees in the forward direction. And its X component can be written as vector F two, X equals vector F two multiplied by sine of 30 degrees. So now we can add these X components together and set them equal to zero to help us solve for F two. So with that in mind, let's start setting them equal to each other. So negative vector F one multiplied by sine of 46 degrees plus vector F two multiplied by sine of 30 degrees is equal to zero. So when we rearrange this equation to isolate and sulfur F two or vector F two, we will find that vector F two is equal to vector F one multiplied by sine of 46 degrees all divided by sign of 30 degrees which now we can plug in all of our known variables to sol for the numerical value of vector F two. So vector F two is equal to 4400 newtons multiplied by sign of 46 degrees. And we have to divide this value by sign of 30 degrees. Awesome. So when we plug that into our calculator, we will find that vector F two is equal to 6000 three hun sorry. So 6330 newtons when we round to the nearest hall number. And this is our first answer that we're trying to solve for. Now, we have to solve for the magnitude of the total force. So now to find the total force in the Y direction, where at vector Fy is the sum of the Y components of vector F one and vector F two, which we can write as the sum of. So some of vector F one Y plus vector F two Y is equal to vector F one multiplied by cosine of 46 degrees plus vector F two multiplied by cosine of 30 degrees. So now at this stage, we can substitute in all of our known values and solve for the magnitude, the final answer for the magnitude. So let's do that. Shall we? So vector fy equals 4400 newtons multiplied by cosine of 46 degrees was 6330 newtons multiplied by cosine of 30 degrees, which is equal to when we plug that into our calculator. 8538 newtons will be round to the nearest hole number. And this is our final answer for the magnitude hooray we did it. Therefore, the force vector F two exerted by the second truck is 6330 newtons. And the total force in the Y direction or the forward direction is 8538 newtons. And these answers correspond to multiple choice answer letter A vector F two exerts a 6330 newtons and the magnitude of total force is 8538 newtons. Thank you so much for watching. Hopefully, that helped and I can't wait to see you in the next video. Thank you so much for watching. Bye.

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(II) Two large snowcats are towing a housing unit north, as shown in Fig. 4–42. The sum of the forces F_A and F_B exerted on the unit by the horizontal cables is north, parallel to the line L, and F_A = 4200 N. Determine F_B and the magnitude of F→_A + F→_B . <IMAGE>

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