(II) Determine the net torque on the 2.0-m-long uniform beam s...

Question

(II) Determine the net torque on the 2.0-m-long uniform beam shown in Fig. 10–56. All forces are shown. Calculate about

(b) point P at one end.

Accepted Answer

Hey, everyone in this problem, we're asked to find the total torque they exerted on this 3.6 m long uniform beam that's depicted in the provided diagram. We're gonna look at that diagram in just a minute and we're asked to compute the torque with respect to the point Q. So if we take a look at this diagram, we have our beam, OK. We're told that it's 3.6 m long. We have point Q on the left hand side and we have a force acting at point Q pointing down into the left making an angle of 49 degrees with the horizontal that is a force of 58 newtons. And we have another force acting at point P which is 2.7 m from that left end of our beam. It's also pointing down into the left. It has a magnitude of 25 newtons and is acting at 51 degrees to the horizontal. And then at the opposite end of the beam, we have point R, we have a force acting there acting up into the right 63 newtons and it makes an angle of 36 degrees with the horizontal. Now we have four answer choices in this problem. All in Newton meters. Option A 1.9 multiplied by 10 to the exponent two. Option B 1.8 multiplied by 10 to the exponent two. And option C 1.4 multiplied by 10 to the exponent two. And finally option D 81. Now what we're gonna do, we wanna calculate, calculate this torque. So we're gonna think about the total torque. OK. So the total torque ta and that torque is gonna be made up of different parts. OK? We're gonna have torque that's produced from all of these different forces. So we're gonna have the torque produced by force one. We'll call that Tau one and that, that force is going to be the force on the left hand side. No, what we can think of. OK? Is that at point Q, if we're rotating about point Q, the force is applied at the exact same point. OK. That's actually not gonna produce a torque. OK. So ta one is actually just gonna be zero. So we don't have to worry about that force. That is at the same point that we are looking at the torque with respect to. Now, we can move to the next force we'll call it F two. That's our 25 Newton force that's kind of in the middle of this beam. Now, if we imagine this torque, OK, we're rotating about point Q, we can imagine pushing on our beam, like this force is acting that's gonna cause clockwise rotation of our beam. And when we talk about torque, we call that clockwise is the negative direction. So this is going to be a negative torque. So we have minus T two and then for force three, which will be the one at the far end. Hey, this is acting up into the right. So we can imagine pushing up our beam that's gonna cause counter clockwise rotation. So that's a positive torque. So we're adding to three. OK. So we're already done with the first torque we have no torque from that force that's acting at the point that we're looking at. And we're gonna do the other two and recall that the torque is given by the force multiplied by the distance R to the rotational axis multiplied by sine of the angle beta. OK. So we can write that out for our equation. So we're gonna have negative F two multiplied by R two, multiplied by sine of the two plus F three multiplied by R three multiplied by sign of theta three. So these forces we know were given in our diagram. OK. R is gonna be the distance from the point that we're interested in. So point Q where we're rotating about two, the application point of the force on RB. OK. So we can think about R two is gonna be the distance from Q two point P which is 2.7 m. We're given that in the diagram and then R three is just gonna be the total distance of this beam. What we're told is 3.6. So we have our forces, we have these distances. The last thing we need is the angle theta. OK. So we need the angle theta for both forces. And we have to be a little bit careful when we look for angles. When we're dealing with torque. They remember that the torque equation comes from a cross product. We're taking the cross product of this force vector and our um position vector or that distance vector. OK. So when we take that cross product and we think about the angle between the two vectors, we take the angle when the two vectors are tail to tail. OK? All right. So if we think about R OK, the R vector, it's pointing from Q down the beam towards where these forces are applied. Hm Now, if we think of force F two, OK, if we were to translate it, so that our two vectors were tail to tail and we looked at the angle between the two vectors, it would actually be the angle here on the right hand side. So it's not gonna be 51 degrees, it's actually gonna be 180 degrees minus 51 degrees. OK. So whenever you're doing these torque problems, always be careful that you're taking that angle when the vectors are tail to tail. Now, for F three F three is OK, if we put these vectors tail to tail that distance from the horizontal, that angle from the horizontal as it's labeled in the diagram is the same distance. OK. So that one, we're gonna use 36 degrees. All right. So let's substitute in all of these values that we know and we'll solve this. So we get negative 25 noons multiplied by 2.7 m multiplied by sign of 180 degrees minus 51 degrees. What? 20 well, not 2563 Newton. Sorry, we're on to force three. So 63 noons multiplied by 3.6 m multiplied by sign of 36 degrees. Ok. When we work all of this out, what we get is that, that total torque tow is gonna be equal to approximately 80.8523 Newton meters. All right. So that's the total to, with respect to point Q. If we round up the two significant digits, like our answer choices, we can see that this corresponds with answer choice D 81 Newton meters. Thanks everyone for watching. I hope this video helped see you in the next one.

Struggling with Physics?

(II) Determine the net torque on the 2.0-m-long uniform beam shown in Fig. 10–56. All forces are shown. Calculate about
(b) point P at one end.
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Struggling with Physics?

(II) Determine the net torque on the 2.0-m-long uniform beam shown in Fig. 10–56. All forces are shown. Calculate about(b) point P at one end.<IMAGE>

Watch next

Net Torque & Sign of Torque practice set

(II) Determine the net torque on the 2.0-m-long uniform beam shown in Fig. 10–56. All forces are shown. Calculate about
(b) point P at one end.
<IMAGE>