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Send uniform calculated

if you haven't done so yet please pause the video and try to answer the question on your own before listening on our first step will be to draw the strut and then show the forces that are acting on it so we'll go through each of these forces slowly but surely at the top of the strut here we have the tension that is pulling the strut backwards to the left we have the weight of the block that's hanging and pulling down on this end of the strut we have the weight of the strut itself notice we used a lowercase M to indicate the mass of the strut and then an uppercase M to indicate the mass of the block that's hanging from the strut and then down here we have a couple of forces that might not be obvious we have a force F X pointing to the right and a force F Y pointing up the reason that we have a force pointing to the right is because the tension is pulling on the strut to the left and in order for the strut to remain in equilibrium we're going to have to have a force that's pointing to the right to sort of cancel out that leftward tension and that's why there is this force that we have labeled F X pointing to the right and then this F Y comes about because as the strut pushes into the ground the ground in response pushes up on the strut it's basically a normal force and so with these forces labeled and knowing that the strut is in equilibrium we can set the sum of the torques equal to zero and the reason that we know it's equal to zero is because the strut again is in equilibrium so that means that the sum of the torques is zero and also the sum of the forces will be zero as we will see later now with any torque problem what we want to do is choose a pivot point on our object our object is the strut so we want to be selecting a pivot that is located somewhere along the strut and it's usually advantageous to select a point where the greatest number of unknown forces is passing through and so since we don't know the force F X or the force F Y we're going to select our pivot at this point right here and the reason for doing that is that the torques produced by these two unknown forces will actually be zero any force that's passing through the rotational Point will produce zero torque so we're basically going to ignore these two forces in the torque equation now over here on the side we want to recall that torque is equal to a force times a distance times the sine of an angle and furthermore that any torque that tends to rotate an object in a clockwise fashion is considered negative torque and torque that goes in an anti-clockwise direction is considered positive torque so for example let's look at this force that's marked T we're going to apply the torque equation so we'll take the force of T well multiply it by the distance to this pivot point right here now the distance from where T is acting to the pivot point is simply the length of the strut so we can plug in L for the length and then we have the sine of an angle now the angle is going to actually be the angle between the force and the strut and that's going to be this angle right here back in this diagram that would have been this angle here and actually since the original diagram already included a theta why don't we step back and use alpha to represent that angle and so we're gonna have these sine of alpha here now we should note that the tension is pulling to the left on the strut and that's causing the strut to try to rotate in a counterclockwise fashion which is in this direction and so that torque will indeed be positive we can next move along to the torque produced by the force labeled mg so we're going to take that force of Mg we're going to multiply it by the distance from where that force is acting to the pivot which again is L and then we're gonna multiply by the sine of an angle now again that angle is going to be between that force and the strut itself which is actually this angle right here back over in the original diagram that angle would be right here why don't we mark that angle beta and so we'll have the sine of beta right there now hopefully we can see that mg as it pulls downward on the strut is tending to try to cause the strut to rotate in this direction and as we noted before that's a negative torque so we're gonna stick a negative sign in front of that torque finally we move over to the force labeled lowercase mg so there's the force the distance to the pivot is only half of the length so we'll have L divided by two from a little bit of geometry we can see that the angle between mg and the strut is the same angle that we marked beta so we'll include the sine of beta in this torque term and this force is causing are tending to cause the strut to rotate in a counterclockwise direction so it's going to have a negative torque value and that's it remember these two forces down here produce no torque because they pass through the pivot point we can set the sum of the torques equal to zero and we're going to solve this equation for the tension T but perhaps before we do that we can note that we can divide each term of this equation by L so these ELLs are actually going to cancel out and now to solve for T we'll add this term as well as this term over to the right-hand side and then we'll divide both sides of the equation by the sine of alpha now we're almost ready to plug in the known values because the masses were given as far as the angle beta and alpha are concerned we can go back to the original diagram this angle we know was 30 degrees theta was 45 degrees for beta if we look carefully we can see that we have a nice right triangle right there so if this angle is 45 degrees in this angle would be 90 degrees then beta is also 45 degrees and then for alpha there's a couple of ways of finding it but one way would be to consider this right triangle right here might be helpful to redraw it and so this angle was 30 this is 90 this angle right here would have to be 60 total but remember that 60 degree angle was made up of both beta and alpha we just found beta to be 45 that means alpha would have to be 15 degrees in order for the entire angle to add up to 60 so alpha is 15 with all these known values we can plug in and solve for the tension and when we crunch that down in our calculators we should get a value of approximately six point six three times 10 to the power of 3 Newtons so this would be the correct answer to Part A which is the tension located in that Kable now for Part B of the question we're going to note that the sum of the forces in the x-direction is equal to zero and we will also note that there are two forces acting in the X direction there is the force that's marked FX that's pointing in the positive x direction and then there's going to be a force that points in the negative x direction and it turns out to be the component of tension in the X direction so to see that we can redraw the tension over here we know that there would be a Y component pointing straight down and then the X component would be pointing to the left now we recall that this angle right here we look at the original diagram is this entire angle right here and that entire angle we determined earlier was 60 degrees the X component is opposite from that 60-degree angle so we're going to be using the sine so we would have the tension multiplied by the sine of 60 degrees and then we can solve this equation for FX by adding the T sine 60 over to the right hand side and since we already determined the tension earlier we can plug it in and when we crunch this down we should get roughly five point seven four times ten to the power of 3 Newtons so this is the correct answer to Part B and finally for Part C we can note that the sum of the forces in the Y direction is also equal to zero we have the force marked FY which is what we're looking for because it's pointing straight up we'll keep it positive we have the two mg forces that are pointing downward and will therefore be negative and then we have the Y component of the tension now the y component of the tension is adjacent to the 60 degree angle so we're going to have the tension multiplied by the cosine of 60 degrees and this force is also negative because that Y component is pointing straight down we can set this equal to zero and solve for FY by adding all these three terms over to the right side and then we'll just plug in the known values for the masses G and tension and when we compute this we should get roughly five point nine six times ten to the power of three Newtons and so this is the correct answer to Part C