Physics Chapter 2 passage question 2 and 4

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crivz17
Posts: 9
Joined: Mon Jan 07, 2019 10:35 pm

Physics Chapter 2 passage question 2 and 4

Post by crivz17 » Fri Jan 11, 2019 4:06 pm

Hello! I'm having a bit of trouble really understanding the torque problems. I understand that when we deal with equilibrium, we need to make the Torques equal to each other. I get how to find the CW torque for M2, but I don't understand why we have to find the torque at the center of the beam. Also, the explanation in the book was a bit confusing: it says that the CCW torque must be 10 = T/0.25 = 40 Nm. Is T torque or tension? and if tension, why does it have the same units as torque? Isn't it more related to force? question 4 is relatively the same question as 2 so if you could please just explain the main concepts on how to break down the question i would REALLY appreciate it! thanks so much! :D
NS_Tutor_Will
Posts: 403
Joined: Fri May 25, 2018 9:15 am

Re: Physics Chapter 2 passage question 2 and 4

Post by NS_Tutor_Will » Sun Jan 13, 2019 8:31 am

I'll try to address your questions sort of conceptually, so feel free to follow up with any specific questions/clarifications that you might have!

With torque, it's all about knowing where the point (or center) of rotation is and then considering all of the forces that would be acting on a point on the "beam" (or whatever is the thing that rotates) that isn't the center of rotation. So, on your first question, we have to take the mass of the beam into account here because its center of mass is not the center of rotation. In other words, the mass of the beam contributes a force that would cause the beam to rotate. The mass of the beam exerts a torque. We take the center of mass of the beam to be the point where that force acts and use the distance of that center of mass to the center of rotation to calculate torque. In other words, the beam's mass is like any other force acting on the beam and thus needs to be taken into account (unless the beam's center of mass is itself the center of rotation, in which case we'd be multiplying by 0 meters and the torque would be 0).

Tension forces are always in force units (Newtons usually), but the reason we have to manipulate it is because we only include forces that are perpendicular to the "beam" (or whatever is the thing that rotates) when calculating torque. Thus, we aren't interested in the entire tension force, but only the vertical component. So we have to use trig to solve for that and then plug that into our torque calculation. To move from a tension force to a torque, you have to do the same process as with any force: determine where that force acts, how far it is from the center of rotation, and then multiply the force by the distance.

Here's another recent post on this topic: http://forum-mcat.nextsteptestprep.com/ ... eb27#p4170

I hope this helps and let me know if you have any follow up questions!
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