In FIGURE EX12.19, for what value of Xaxle will the two forces pr...

Question

In FIGURE EX12.19, for what value of Xaxle will the two forces provide 1.8 Nm of torque about the axle?

Accepted Answer

Hey everyone. Welcome back in this problem. A rectangular plate rotates about an axis passing through point A with coordinates X A Y A as shown in the figure two forces acting on the plate produced a net torque of magnitude 2.3 Newton meters. And we're asked to calculate X A for giving me a diagram of this situation. We have this rectangular plate that is 12 centimeters tall and eight centimeters wide. And we have this line passing through our point X A Y A which goes from the top left corner to the bottom right corner, we have force one of 30 newtons acting to the left from the bottom left corner and force two of magnitude 13 newtons acting to the right from the top right corner. And we have four answer choices all in centimeters. Option A 3.3, option B 4.4, option C 5.1 and option D eight. Now we wanna calculate X ay. The information we're given in this problem is about a net torque. So let's think about this network. OK. Now, if we think about force one or force two, OK? If we apply those forces. Both of those are gonna cause a clockwise rotation. Now, we're interested only in the net to so what we care about is the magnitude and because these are both causing the same direction, we can just write them as positive. So we have that the torque to, is going to be equal to the torque from force one. What the torque from? For soup? I recall that the torque can be written as R multiplied by F multiplied by sine theta. OK. So we have R one, F one science data one plus R two, F two sign beta two. OK? Where R one and R two are the distance or that vector that goes from the force to the point of interest. Now our point of interest, we're gonna be rotating about point A and that's how we're gonna introduce that X A and Y A. So let me draw on this diagram here in red. You can imagine, I'm gonna draw this line connecting the force F one. Let me draw a bit straighter F one to this point A and that's gonna be R one and I'm gonna do the same for force two. That's gonna be R two. Now, with these lines we've drawn and we have R one and R two. We want to think about how we can write this R one and R two. And we want to think about how we can write sign of the angle because we don't know what these angles are. So let's go ahead and write down these triangles and try to break it up for ourselves. So if we think about the triangle in the bottom left, OK. We have our distance R one on the hypos, we have some wide distance and we have some distance. Now, with X distance, this is X A A starts on the far left and then this line goes up to this point A and so the bottom of our triangle is going to be X A. OK? And the same for the right hand side, that's gonna be Y A. All right. So now we have a relationship between this R one value the X A we're interested in and Y A as well. Now when we think about angles and torques, it's gonna be the angle between the force F and vector R. So our angle theta one is actually gonna be this outer angle. Uh which means that the angle in our triangle is going to be pi minus theta one. OK? Because together they make an entire straight line 180 degrees or pi Yeah, it can be really tempting to write a cosign because we have X A on our adjacent side, we could also think about writing the angle on the top right of our triangle in terms of the other two angles to use that with a sign to calculate X A. The problem is that, that is going to be difficult to simplify. So what we're gonna do is we're gonna calculate Y A. First, we're gonna use that to calculate X A. And so we're gonna write all of these angles in terms of Y A. So we can write sign of pi minus theta one is equal to Y A divided by R one. Now recall sign of pi minus theta one is just equal to sign of pi uh theta one, right? So that's why we use sine. That's why we use this particular angle because we can simplify this. So that sine of theta one is equal to Y A divided by R one. All right. And we're gonna use that in our equation. I'm just gonna put a blue box around it. So we don't lose track of it. We're gonna substitute that into our torque equation. Now, we're gonna look at the other triangle we have as well. The other triangle we're gonna draw, we have R two and the hypotenuse, we're gonna have our two point down to the left and then we're gonna complete the triangle up into the right on the left side of our triangle. Why is this bottom portion? We know that 12 centimeters is the total height. So the distance of this triangle, it is gonna be 12 centimeters minus Y A. And we're talking about this triangle here similar on the top side here, right? We know that the left hand side up to our dotted line is going to be X A. We know the entire length is eight centimeters. And so to the right of the dotted line is gonna be that eight centimeters minus X A and we can write our angle here as well. We know that the, or the angle between F two and R two is going to be theta two. So the angle inside of our triangle is gonna be pi minus theta two. We have sine of pi minus the two gonna be the opposite side 12 centimeters minus Y A divided by the hypotenuse of R two. And just like for the first triangle, we know that we can write sine of pi minus beta two as just sign of theta two. We're gonna do that. We have signed a beta two is equal to centimeters minus Y A divided by R two. So we've written sign of our angles out in terms of Y A. OK. We're gonna have to calculate Y A first before we can calculate X A. And now we're gonna sub these into our torque equation. So we have that the torque is going to be equal to R one multiplied by F one multiplied by sine of theta one which we found to be Y A divided by R one. What are two, F two multiplied by 0. m. OK. Converting centimeters to meters, we divide by 100. So we can rate this as 0.12 m minus Y A divided by art. All right. So the units or sorry, not the units but these values of R one and R two, we're gonna divide it. Now, we have one in the new murder and one in the denominator. So now what we have is we have our torque equation. We've gotten rid of these R one and R two values. We've written sine of theta one in terms of Y A. The only other thing in our equation is F one and F two, which we know the values of. So we can now use this equation to calculate Y A. Let's go ahead and substitute in all the values we know we're told that the torque is equal to 2.3 newton meters. This is gonna be equal to F one which is 30 newtons multiplied by Y A, both F two which is 13 newtons multiplied by 0.12 m minus Y A. OK? And now on the right hand side, you can see why we've converted this to meters. When we multiply, we're gonna end up with new meters, which is what we want so that we can simplify with the left hand side that already has those units. OK. So let's go ahead and do that simplification. Now, 2.3 newton meters is equal to newtons multiplied by Y A, distributing the 13 newtons to both terms inside of our brackets. We get plus 1.56 Newton meters minus 13 newtons multiplied by Y A. We can move our 1.56 Newton meters to the left hand side by subtracting. We get 0.74 Newton meters. We have 30 newtons, Y A minus 13 newtons Y A which gives us 17 newtons Y A. And finally to solve for Y A, we divide by 17 newtons and we get a value of 0. 2941 m. And you can see I've left a lot of digits here. Always check with your professor or your textbook on what the convention is that you're using. I've left lots of digits to avoid any round off error in the future. Now the answer choices are in centimeters. We wanna be working in centimeters. So let's go ahead and convert this back to centimeters. Now that we're done with our torque equation, this is gonna be 4. centimeters. And this is our value of Y A. And again, this is not the value we're looking for. Remember, we're looking to calculate X A but because of the trig functions we had and the equation we had, we needed to find Y A first. Now that we have Y A, we want to think about how we can relate it to X A. So let's go back to our diagram and see how we can do that. From our diagram, we were given this dotted line. What we can see is that our point A lies on this dotted line because we have the left corner at the origin, we can write the equation of this line and we know that the point is going to lie. OK. So that's how we're gonna use our point Y A to calculate X A. We're gonna write the equation of that dotted line. So let's recall the equation of a line. We're gonna write as Y is equal to M X plus B. OK. B is going to be equal to our Y intercept. OK. Going back to our diagram. Where does this line cross the Y axis? Well, it crosses at 12 centimeters. So B is gonna be 12 centimeters and then we can calculate our slope based off the rise and run using the height and width of this rectangle. Hey. All right. So our Y intercept is going to be wow, centimeters and M represents the slope. And again, we can calculate this using rise divided by a run. The rise is gonna be the height of that rectangle but the line is going down to the right. OK. So right is actually gonna be negative. We're going from 12 centimeters down to zero centimeters. We get a rise of negative 12 centimeters a run of eight centimeters as we can simplify our slope to negative three half. Hey, we're getting really close. We now have the equation of our line Y is equal to negative three halves multiplied by X plus 12 centimeters. We have a Y value. If we substitute in our Y value, we're gonna get the corresponding X value. So if we substitute Y A, we'll get X A. OK. So we're gonna substitute and let's call the equation of our line equation star. And we're gonna substitute Y A equals 4.3529 or one centimeters into that equation star, we get 4. centimeters is equal to negative three has X oh last 12 centimeters. We're gonna subtract 12 centimeters from both sides. Negative 7. 059 centimeters is gonna be equal to negative three halves X and finally dividing by negative three halfs X. And let me write that this is X A to make it very clear. So dividing by negative three halves to isolate X A, you get that X A is equal to 5.98. OK? And that is that final answer we were looking for, we found the X coordinate of our point A if we go up to our answer choices, OK? And we round to one decimal place, we can see that our answer corresponds with option C 5.1 C meters. Thanks everyone for watching. I hope this video helped see you in the next one.

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In FIGURE EX12.19, for what value of Xaxle will the two forces provide 1.8 Nm of torque about the axle?

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