A 4.0-m-long, 500 kg steel beam extends horizontally from the ...

Question

A 4.0-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70 kg construction worker stands at the far end of the beam. What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?

Accepted Answer

Hey, everyone in this problem, a 55 kg athlete preparing to dive stands at the end of a homogeneous horizontal diving board of mass, 38 kg. The diving board is six m long and the other end is anchored through a screw to the diving tower. Ok. So we're gonna come back to what this question is asking us to find. But first, we're gonna draw a little diagram of what the question has told us so far. So we have our horizontal diving board and we have our athlete standing on it. We know that the mass of the athlete which we'll call ma is equal to kg and the mass of the diving board which we'll call MD is equal to 38 kg. Now, this diving board has a length of six m and at the opposite end of where we have the athlete, we have a screw that anchors this diving board to the diving tower. So like something like this. Now this question is asking us to calculate the magnitude of the torque exerted by the athlete and the board about the screw. Ok? We're given four answer choices all in meters. Option A 444. Option B 912 option C 2119 and option D 4356. So we're asked to find this torque from the athlete and the board. Now, the athlete standing on the end of the board and you can imagine that they're pushing down on this board. So they're gonna exert a torque. We can imagine that this torque is gonna cause the diving board to rotate clockwise. OK. So that's gonna actually end up being a negative torque, right? Because it's going clockwise. Now, the torque from the board, we're told that this is a homogeneous board. What that means is that the weight of the board is gonna act at the center of mass and the center of mass is gonna be directly in the middle of the board. And we're gonna draw that in be blue and we're gonna have the torque T D of the diving board. OK, acting directly in that center. And that is gonna be at a distance of three m from the screw. OK? Because again, the diving board is six m long, it's homogeneous. So the center of mass is directly in the middle. OK. Halfway to six m is gonna be three m. So these are our two torques, the torque of the diving board just like the torque of the athlete it's pulling down. OK. It's gonna cause clockwise rotation So we expect that it's gonna be negative. So let's start by just calculating each torque individually and we're gonna take the sum and look at the magnitude of the torque exerted by both of them together. Recall it, the torque is equal to F multiplied by R multiplied by sine data. A where F is the force acting R is that displacement from the pivot and then sine data is the angle between the force and that displacement vector R. Now, for the athlete, the force acting is gonna be the force due to gravity acting on the athlete. Now this is gonna be negative acting downwards. We have negative M G for that force of gravity and in this case, ma multiplied by G. Um then we're gonna multiply by R A that distance or that displacement vector to our athlete sign of data. Now we actually know all of these values. So the hard part's done, we just need to substitute in the known values. So we get negative, the mass of the athlete is 55 kg multiplied by the acceleration due to gravity which is 9. m per second squared multiplied by the distance from where this torque acts to our pivot. And the athlete is at one end, the pivot is at the very other end, the length of the diving board is, is six m and so the distance between them is going to be six m and then we have sign of the angle between that force and this displacement factor. OK. So our force is acting straight down vertically, our displacement vector because this is a horizontal diving board is acting perfectly horizontal. And so these two are perpendicular to each other, that means they make a 90 degree angle. So we get sine of 90 degrees now sin of 90 degrees is just one. So what we really have here is negative 55 kg multiplied by 9.81 m per second squared, multiplied by six m. For units we get kilogram meter per second squared from the first two terms. OK. We're called it, that's equivalent to a Newton. And then we multiply by a meter. So we have Newton meter, which is the units that we have in our final answer. Um So that's what we want. That's great. And when we work this out, we get negative 3237. Newton meters. OK? So that's the torque generated from the athlete. Now, let's do the same with the torque generated from the diving board. OK? Or the weight of that diving board. So again, we're gonna have the force multiplied by the displacement, oops multiplied by sign of the angle between them. This is gonna be a negative torque because that force is gonna cause a clockwise rotation. And again, it's a force due to gravity. OK? So we get negative MD, the mass of the diving board multiplied by G multiplied by D D that displacement vector or the actually let's call it R D displacement vector or that distance from the pivot to where this force acts multiplied by sine of theta the angle between the two. Again, we have all these values, we get negative 38 kg multiplied by 9.81 m per second squared. Now, the weight of the diving board we drew in our diagram. OK. Remember that this is homogeneous. That means that the center of mass is directly in the middle at three m from the end. And that is where that weight is gonna act. OK. So in this case that our value is gonna be three m and sine is gonna be a sign of 90 degrees again because once again, the force is acting straight down vertically, the displacement vector because we have a horizontal diving board is acting perfectly horizontal. They make a 90 degree angle, they are perpendicular. Now we have the same units as before. We're gonna end up with Newton meters. Sine of 90 degrees is equal to one. And when we work this one out, we get negative 1118. Newton meters. OK? So we figured out the torque from the athlete, we figured out the torque from the diving board, what we want to calculate is the magnitude of that total torque. OK. So we wanna find the magnitude of Tau, which is gonna be the magnitude of Tau A plus Tau D and define the magnitude. We're just taking the absolute value and we just want that the magnitude we just want um yes, the magnitude without the direction now to a be found to be negative at 3237. Newton meters minus 1118.34 new meters. For Tau. Really, we're gonna subtract the two, we're gonna take the absolute value and we get that the magnitude of the torque exerted by both of these is 4355. Newton meters. And that is the value that we were looking for. Now, if we go up and compare this to our answer choices, OK? Our answer choices are rounded to the nearest Newton meter. Ok? That nearest whole value. And so our answer is going to correspond with answer choice. D 4356 Newton meters. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Torque

Center of Mass

Equilibrium

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