A uniform 300-N trapdoor in a floor is hinged at one side. Find the net upward force needed to begin to open it and the total force exerted on the door by the hinges (a) if the upward force is applied at the center and (b) if the upward force is applied at the center of the edge opposite the hinges.

Video transcript

Hey everyone in this problem, we have a 400 newton hatch. Okay. Located at the upper end of a stairway were asked to determine the total upward force. A person standing on the staircase must apply to initiate the door opening. Okay? And the net force applied on the hatch by the hinges were told to take the upward force to be applied midway of the edge opposite to the hinges. Alright, so let's draw a little diagram here. Suppose we have the hinges here. Okay, we're gonna draw the hatch just as a straight line. So this is our hatch. Now what forces do we have? Well the hatch has weight. Okay, so in its center of match, which we're going to assume is right in the middle, we're gonna have the weight from the hatch, w now the person is going to have an upward force applied midway of the edge opposite to the hinges. Okay? So we're opposite to the hinges were on this edge, we have an upward force applied and we're going to call this F You okay? F and then the upward force you and we also have a force applied on the hatch by the hinges. Okay? So we're going to assume that this force it's going to be upwards, but we may be wrong, okay? And you'll know when you calculate it based on the sign that you're getting whether it's positive or negative, whether you've drawn it in the right direction, Okay? We're going to assume that it's up and we'll see afterwards if that's true. Alright, so we've drawn our diagram. Okay, we're gonna pick our coordinate system. So we have up into the right as positive. We have our X and Y directions. And now what we have to think about is equilibrium condition. Okay? We have known an external forces acting so we know that we're going to have the sum of the forces equal zero. Okay? And in this case we only have forces in the Y direction. Ok. So we're only gonna have some of the forces and we can say in the Y direction equal zero. Okay? And we have no net external torque. Right? And so what this means is that we're gonna have the some of the torques equal to zero as well. All right. So if we start with the sum of the forces. Okay, The sum of the forces equals zero. What forces do we have? Okay, well, we've written that we have in the positive Y direction, the force of the hinges. We also have the force. The upward force that a person applies the positive Y. Direction and in the negative Y direction. The weight. W We know w We're told in the problem the weight of that hatch, but we don't know F. H and F. You. Okay, So we have two unknowns here. We can't solve. So let's switch over to the some of the torques see what that gets us. Some of the torques are going to be equal to zero. We need to choose a pivot. Okay, let's go ahead and take the pivot to be the hinge. Okay, that's gonna eliminate this F. H. From our equation because the force right here at the hinge is not going to contribute to the torque. Okay, So let's take pivot at the Hinch. Okay, No, what will our torque be then? Okay, well we're gonna have the torque from the upward force and you'll see this is pointing up, this is going to cause counterclockwise rotation. So this is a positive torque. Okay? And then we have the torque produced by the weight. W this is pointing down. This will produce a clockwise torque and so that's going to be negative. Okay, so we have negative the torque due to the weight. Okay, And that's gonna be equal to zero. All right now, we're gonna say that the length of our hatch is l The length is elf. And the torque from this force, this upward force is going to be the distance from our pivot which is going to be l Okay, times a force which is F. U. Times sine of the angle. Okay, And we can see that between this hatch in our force we have an angle of 90 degrees. So we have L. F. You sign 90. Okay. And for our torque related to the weight. Okay, well this is going to be an L over two. It's at our center of mass, so it's gonna be halfway through the hatch. Okay? The force acting is W And again, this is also acting at 90 degrees. Those are going to be equal to zero. And what we see now is we can solve for F. You you can divide by L. We get F. U. Sine of 90 degrees. They will move this term to the right hand side. This is going to be equal to W over two. Sign of 90°. We divide by sign 90 on both sides. Let me just get that f. You is equal to the W over two. And we know the weight is 400 newtons. We're told in the problem. So, we got 400 newtons divided by two, which is gonna give us 200 newtons. Okay? All right, So 200 Nunes is our upward force. Okay? Now, if you're in a test situation, when you figure out one answer, okay, go ahead and look at the possible answer choices. Okay? And we'll see that A D. And F. I'll have an upward force of 200 newtons. Okay, So we can narrow it down to those three options right now, and let's try to figure out now the force due to those hinges. Alright, so we have fhk we're gonna switch back over to this equation because now we know f You okay, it's going to be 200 newtons in? Our weight Is 400 Newtons. This is all equal to zero. Yes, we have FH -200 newtons is equal to zero, which tells us that the force of the hatch is equal to newtons as well. Okay, Alright. So we used our equilibrium conditions that some of the forces was zero and some of the torques was zero. We found that the total upward force of a person standing on the staircase must apply to initiate the door opening was 200 newtons, and that the net force applied on the hatch by the Hinges was also 200 newtons. And so we have answer choice f. That's it for this one. Thanks everyone for watching. See you in the next video.