The cm of a 95,000-kg train locomotive starts across a 260-m-l...

Question

The cm of a 95,000-kg train locomotive starts across a 260-m-long bridge at time t = 0. The bridge is a uniform beam of mass 23,000 kg and the train travels at a constant 80.0 km/h. What are the magnitudes of the vertical forces, F_A(t) and F_B(t), on the two end supports, written as a function of time during the train’s passage?

Accepted Answer

Hello, fellow physicists today, Ernest solved 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. A 33,000 kg dump truck carrying gravel is passing across a an 88.0 m flyover span from north to south. The flyover span can be considered a uniform structure of mass. 200,000 kg, the center of mass of the truck is moving at a constant speed of 65.0 kilometers per hour, determine the magnitude of the vertical force FSFT acting on the south end of the span during the travel of the truck as a function of time assume at T equals zero. The center of mass of the truck is at the north end on which the force FN of T acts. So that's our goal. Our angle is ultimately we're trying to determine what the magnitude of the vertical force is. So we're trying to find the equation for FS of T and that will be our final answer that we're ultimately trying to solve for awesome. So considering that looking at all of our multiple choice answers, we have that Fsft which is our vertical force equation is equal to some fun some equation. So Fsft is equal to something. So let's read them off to see what our final equation for Fsft will be. Or our function I should say. And for a, it is 1.61 multiplied by 10 to the power of five newtons plus zero. TB is 6.64 multiplied by 10 to the power of four newtons per second T plus 2.00 multiplied by 10 to the power of five newtons. C is 6.64 multiplied by 10 to the power of five newtons per second T plus 9.81 multiplied by 10 to the power of six newtons. And finally D is 6.64 multiplied by 10 to the power of four newtons per second T plus 9.81 multiplied by 10 to the power of five newtons. Awesome. So first off, right off the bat, let us draw a diagram to help us better visualize this problem. So we can get an idea of what exactly we're dealing with. So in our little diagram here, imagine that this blue rectangle is denoting our flyover span from north to south. And on the far left of our span here, on the far left of our blue rectangle, our skinny blue rectangle, we have our vector F one and then on the far right, we have our vector F two and quickly here. Now let us con let us note that M subscript T, so MT is our mass of our dump truck and M subscript S is the mass of our flyover span. So the mass of the span, so let us note that MT multiplied by vector G and MS multiplied by vector G is the weight force of the dump truck when it's MT multiplied by G and then the weight force of the flyover span is MS multiplied by vector G. Awesome. And let us not from our distance between vector F one from the far left to the far right. Vector F two is some length L and in this case, L will be 88.0 m because that is the length of our flyover span. And then from our distance from where vector F one is to the wait force of the dump truck is some distance X indicated by arrows. So from the far left to where the wait force is, and note that the wait force for the dump truck and the wait force for the flyover span are both pointing in the downward direction. OK. So now that we have an idea of what's going on in this problem, we need to note that since the span has uniform mass distribution, its center of gravity will be found at the middle. Thus, the weight MS multiplied by G will act at the middle as seen in our free body diagram that we just drew together and discussed together. So moving right along, we also need to note that we will need to assume that the dump truck will travel from the left, which is the north direction to the right, which is the south direction. And at T equals zero seconds, the truck will be located at the left end. Our next step in solving this problem is we need to consider that the equilibrium conditions for the torque are about the left or north end and that the equilibrium condition for the vertical forces to find the expression of the forces. So now we need to consider the equilibrium condition for the vertical forces specifically. So in order to do that, we first need to figure out what the speed of the truck of the dump truck is in meters per second. But let's take a moment here that let's call you the speed of our dump truck. And let's note that the speed of our dump truck is 65 0.0 kilometers per hour. But we need to convert this to meters per second in order to help us solve for our problems that have the proper units at the very end when we're solving. So using dimensional analysis, we will take 65.0 kilometers per hour and we will as we should be able to call recall, hopefully, we need to note that in one hour, there is 3600 seconds. And then we need to note that there's 1000 m in one kilometer. Awesome. And as you can see, hopefully you should be able to note that the units per kilometer will kilometers will cancel out and our units for hours will cancel out, leaving us with meters per second. And when we type that into our calculator, we will find that the final answer is equal to 18.05556 m per second. Awesome. So with that in mind, we can move right along here. So therefore, thus, the position of the center mass of the dump truck from the left end is given by X is equal to U multiplied by T or T is the time and you once again is the speed of our dump truck. So now we need to consider the equilibrium condition for the torque about the left north end by taking the clockwise moments about the left end as positive and the counterclockwise moments as negative. Thus, we could write that. So I'm gonna put a little arrow here that the sum of the torque which you're gonna use Tau to denote torque that the sum of torque is equal to mt multiplied by G multiplied by X plus MS multiplied by G multiplied by one half multiplied by L minus FS multiplied by capital L once again, which is all equal to zero. We also need to note at this point that sense the force FN, which once again FN is our normal force is at the origin point and it corresponds to the torque for it will be zero. Thus, we could write that FS multiplied by L is equal to MT multiplied by G multiplied by X plus MS multiplied by G multiplied by one half multiplied by L. So now at this point, we need to rearrange this equation to isolate and solve for FS because FS of T is what we're ultimately trying to solve for. So, FS is equal to, when we use a little bit of algebra to rearrange everything, we will get that MT multiplied by G multiplied by X divided by L plus MS multiplied by G multiplied by one half. Awesome. So now we need to substitute in X equals U multiplied by T. So when we do that, we will get that FS is equal to MT multiplied by G multiplied by U all divided by capital L multiplied by T plus one half multiplied by MS multiplied by G. So now at this point, we need to plug in all of our known variables and sulfur, the numerical value of FS. So when we do that, we will find that FS is equal to. And let us note that the mass of our dump truck is 33,000 kg. Awesome. And then we need to multiply it by the acceleration due to gravity which is 9.81 m per second squared. Awesome. And then we need to multiply it by our speed of our dump truck, which we found to be 18.05556 m per second. Awesome. And then we need to divide all of this by the length of our flyover span which is 88.0 m. And then we need to multiply our entire division by tea And then we need to add one half. So we need to take one half and then we need to multiply it by and don't forget that MS our mass of our flyover span is 200 1000 kilograms. Awesome. And then we need to multiply this by the acceleration due to gravity again, which is 9.81 m per second squared. Awesome. And that got a little jumbled up there, didn't it? There we go second squared. Perfect. So when we plug that into our calculator, we will find that FS of T is equal to. So when we write our final answer in scientific notation and round 23 significant figures, we will get 6.64 multiplied by 10 to the power of four and its units will be newtons per second. And this will be multiplied by T, then we will add 9.81 multiplied by 10 to the power of five newtons. Awesome. And that's it we've solved for this problem. Hooray, we did it So looking at our multiple choice answers are correct answer pass to be the letter D which states that Fsft is equal to 6.64 multiplied by 10 to the power of four newtons per second T plus 9.81 multiplied by 10 to the power of five newtons. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Newton's Second Law of Motion

Static and Dynamic Equilibrium

Center of Mass and Load Distribution

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