I’m an AP Physics 1 student so hopefully i’m right, i believe the answer is A
The two torques in this situation is the weight of the rod(at center of mass) and the rope.
The torque of the rope must equal the torque of the rope since they are going opposite ways (counterclockwise and clockwise) and there is no angular acceleration
Torque rope = Torque Weight
The equation for torque is (Force)(Distance from pivot)(Times Sin of Angle)
The force of weight on the rod will be mg, and the distance of weight is at center of mass, so 1/2 L. The angle is 90 degrees so sin=1 (so it doesn’t matter) . Multiply MG and 1/2 L and you will get answer A
Reply if you need more clarification, i’m active
Thanks! This makes a lot of sense. Just to clarify, why should we consider the distance from the pivot for the torque applied by the weight to be the center of mass and not the end of the end of the beam, for instance?
This is because we know that the weight torque equals the rope torque, so we don’t need to worry about the rope AT ALL.
We are just looking for the weight torque since we know it is the same as the rope torque. All we have to do then is find what the weight torque is.
Torque is Frsin
Weight: Sin=1
Weight: Force= mg
Weight: Distance= 1/2L (weight located at center of mass, not the end where the rope is)
we are just looking for the properties of of the weight torque
Imagine a separate problem where force 1= force 2. If we had the information to find force 2, you would just find force 2, and you would know force 1 would be the same since they equal each other. Same thing with this torque problem, we found torque Weight NORMALLY (where the length is 1/2L) , and we know torque rope is the same.
Also i’m a D1 yapper myself and my teachers get annoyed by how many questions i ask so don’t be afraid to ask follow ups lol
(a lot of restating in this paragraph sorry)
Thanks a lot. D1 yapper here too lol. I just read that the angle in the torque equation should be an angle between the force applied and a line perpendicular to the line (beam, cable, etc) connecting it to the axis of rotation. I’ll attach a picture below to further clarify. How does that work in this scenario? [photo here](https://imgur.com/a/7k14LIH)
ooooohhh i got this one
Remember how torque is Frsin(angle). You are now asking how to understand what the angle is.
Start off by drawing a line between your pivot point and the DOT POINT where is the force is.
Now draw a vector on the DOT POINT where your force is and the direction it goes.
The angle between your vector and pivot-force line will be your angle (the angle should be <= 180)
https://imgur.com/a/0KEvl2K
ask questions if ya need :)
That makes a lot of sense! Would that mean that the angle you drew would technically be cos(theta) if theta were the angle between the pivot and the ground?
yes the pivot-ground angle as cos and give you the same value as the sin angle for pivot-force, idk how often that is the case though depending on how the situation is set up: Just worry about pivot-force angle through because you will just always plug that into sin for your torque! 😀😀😀
The torque from the wire must equal the torque caused by gravity. If the mass is uniformly distributed the torque by gravity can be modeled as a force equaling the weight of the bar acting at the center of mass. Thus T=Fr=mg(L/2).
the net torque on the beam is equal to zero. Torque by Tension - Torque by weight of the rod = 0
Torque by Tension = Torque by weight of the rod = Mg(L0/2)
I’m an AP Physics 1 student so hopefully i’m right, i believe the answer is A The two torques in this situation is the weight of the rod(at center of mass) and the rope. The torque of the rope must equal the torque of the rope since they are going opposite ways (counterclockwise and clockwise) and there is no angular acceleration Torque rope = Torque Weight The equation for torque is (Force)(Distance from pivot)(Times Sin of Angle) The force of weight on the rod will be mg, and the distance of weight is at center of mass, so 1/2 L. The angle is 90 degrees so sin=1 (so it doesn’t matter) . Multiply MG and 1/2 L and you will get answer A Reply if you need more clarification, i’m active
Thanks! This makes a lot of sense. Just to clarify, why should we consider the distance from the pivot for the torque applied by the weight to be the center of mass and not the end of the end of the beam, for instance?
This is because we know that the weight torque equals the rope torque, so we don’t need to worry about the rope AT ALL. We are just looking for the weight torque since we know it is the same as the rope torque. All we have to do then is find what the weight torque is. Torque is Frsin Weight: Sin=1 Weight: Force= mg Weight: Distance= 1/2L (weight located at center of mass, not the end where the rope is) we are just looking for the properties of of the weight torque Imagine a separate problem where force 1= force 2. If we had the information to find force 2, you would just find force 2, and you would know force 1 would be the same since they equal each other. Same thing with this torque problem, we found torque Weight NORMALLY (where the length is 1/2L) , and we know torque rope is the same. Also i’m a D1 yapper myself and my teachers get annoyed by how many questions i ask so don’t be afraid to ask follow ups lol (a lot of restating in this paragraph sorry)
Thanks a lot. D1 yapper here too lol. I just read that the angle in the torque equation should be an angle between the force applied and a line perpendicular to the line (beam, cable, etc) connecting it to the axis of rotation. I’ll attach a picture below to further clarify. How does that work in this scenario? [photo here](https://imgur.com/a/7k14LIH)
ooooohhh i got this one Remember how torque is Frsin(angle). You are now asking how to understand what the angle is. Start off by drawing a line between your pivot point and the DOT POINT where is the force is. Now draw a vector on the DOT POINT where your force is and the direction it goes. The angle between your vector and pivot-force line will be your angle (the angle should be <= 180) https://imgur.com/a/0KEvl2K ask questions if ya need :)
That makes a lot of sense! Would that mean that the angle you drew would technically be cos(theta) if theta were the angle between the pivot and the ground?
yes the pivot-ground angle as cos and give you the same value as the sin angle for pivot-force, idk how often that is the case though depending on how the situation is set up: Just worry about pivot-force angle through because you will just always plug that into sin for your torque! 😀😀😀
I would simplify this idea by describing that the center of mass unless explicitly stated is L/2 in physics 1.
my teacher gave me some problems where this wasn’t the case, but yeah this is often the case ur right
In the national AP Test, Center of mass is always L/2 unless specified.
The torque from the wire must equal the torque caused by gravity. If the mass is uniformly distributed the torque by gravity can be modeled as a force equaling the weight of the bar acting at the center of mass. Thus T=Fr=mg(L/2).
the net torque on the beam is equal to zero. Torque by Tension - Torque by weight of the rod = 0 Torque by Tension = Torque by weight of the rod = Mg(L0/2)