Cort's 3/4 Physics Thread

[PB]

Hmm, forgive me if I am wrong but I think all the angle stuff are red herrings?
If the net force in this circular motion is 3125N, then it should stay as 3125N anywhere in the circle- as long as the velocity is the same and the radius is the same. Else, the cage would spiral off (too little net force towards the centre)or spiral towards the centre (too much net force).

So at the top of the circle, the net force will still be 3125N. The only things that could be changing throughout the circular motion are the constituents of the net force - weight force and tension force. We know that weight force will always stay (250kg x9.8 =2450N) - therefore it is the tension force that is changing.
At the top of the circle, you can take the weight force away from the net force (3125-2450N = 675N).  Assuming that there are two cables, 675 / 2 = 337.5N, which is kind of close to 360N

Edit: Assuming that there are two cables is the only way I can make sense of this question...

[jesterino]

yeah there are two cables. sorry my explanation of the question is very poor. the answer at the back of checkpoints is:

3125 = 2Tcos30 + mg, where T is tension i'm guessing.

So if at the top: net force = tension + mg

How did they deduce tension = 2Tcos30? I also attached the diagram of the problem below.

[RKTR]

N+mg=3125
N=3125-mg
there are two cables
divide N by 2

cos30= (N/2) / T
Tcos30=N/2
N=2Tcos30

[souka]

After getting 625N, you can create a vector sum of the two equal tensions and the normal reaction force. Thus to create a right angle triangle from the isosceles triangle you divide the normal by to hence a (adjacent) is 312.5 and angle is 30° from the vertical.  You can use cos30° = 312.5/ T
Which when solved would equal ~360.8 N

[Cort]

Reading up/doing some questions regarding torque. I know the rule is t = f*r = f*r*sin theta if the angle is involved. My problem is, for r, when do you know that you have to measure it from the centre of the object, or just use the full radius of the object? Whhich/what keywords should I be aware of that will tell me when I should be using r = half of the x length, or r = use the full length?

Currently I'm assuming that you use r= half the x length when it is used for translational equilibrium/balancing, right? This is for materials & structures.

Thanks,
Cort.

[lzxnl]

Reading up/doing some questions regarding torque. I know the rule is t = f*r = f*r*sin theta if the angle is involved. My problem is, for r, when do you know that you have to measure it from the centre of the object, or just use the full radius of the object? Whhich/what keywords should I be aware of that will tell me when I should be using r = half of the x length, or r = use the full length?

Currently I'm assuming that you use r= half the x length when it is used for translational equilibrium/balancing, right? This is for materials & structures.

Thanks,
Cort.

The radius r is measured from the axis of rotation. Identify where that is first.

[Cort]

The radius r is measured from the axis of rotation. Identify where that is first.

stupid question - axis of rotation is the midpoint right?

Ah, thanks for the heads up. The axis of rotation is where..well, where it rotates. r measures from that direction, apparently, but in different situations the set up will be different. Thanks for the thought!

[Cort]

Need help just for some understanding. Reading over Heinneman Physics 2 on structures and material:
"When cantilevers are joined at the centre of the span, there is no reduction in force. These are the same as if the beams were not connected."
 - What does it mean when there is no "reduction in force...at the centre of the span" ?

Thanks,
Cort.

Edit: For the caniliever below, why is that the force on the left goes downwards? I assumed that both the left and right pillars will have an upwards force.

[Cort]

q738 (Structures and materials)
"Two wires, A and B, are of equal length. A has twice the diameter of B. The Young's modulus of A is twice that of B. if they are subjected to the same load, and A stretches by 10mm, the wire B will stretch by x. Which of the following is closest to the value of x?"

The answers would state, that after writing out the Young's Modulus for Mat A and Mat B, you'll do Mat A/Mat B to eliminate Fa*La/Fa*Lb. I can understand the result of dividing those two together..but its understanding why they would do it so suddenly. is it meant to be some kind of general knowledge I'm supposed to know? If so, would there be another way to actually approach, and break down the ideas contained within the question?

Thanks,
Cort

[Cort]

When it says RMS voltage -- does that mean voltage = 12v?

[Thorium]

When it says RMS voltage -- does that mean voltage = 12v?

Yes in other words. Rms stands for root mean square

[Stevensmay]

When it says RMS voltage -- does that mean voltage = 12v?

RMS voltage of an AC voltage can be found by dividing it by root(2). It doesn't mean voltage is always 12V however.

[lzxnl]

RMS voltage = equivalent DC voltage that would give the same average power output. In the case of AC voltage only, it's the peak voltage / sqrt 2.

[Cort]

Thank you all. Is it required to know about rms and multiplying it and what not in the exam?

[Stevensmay]

Thank you all. Is it required to know about rms and multiplying it and what not in the exam?

From memory I believe it is.