Calculating Forces: T1, T2 & EF

  • Thread starter Thread starter rafael_josem
  • Start date Start date
  • Tags Tags
    Forces
AI Thread Summary
To calculate the forces T1 and T2, the balance of vertical forces was correctly established as EF = T1 + T2 - 7.84N - 9.8N - 14.7N. However, there were errors in calculating the moments, particularly regarding the distances used for the 1.0 kg and 1.5 kg masses. The correct moment for the 1.0 kg mass should account for the full distance from point A, which is 21.1 cm, rather than just 14 cm. Additionally, there is a discrepancy in the dimensions provided, as the total distance between T1 and T2 does not match the sum of the distances for the weights. It is advisable to verify the problem's dimensions for accuracy before proceeding with the calculations.
rafael_josem
Messages
9
Reaction score
0
Hi, I need help to calculate the force of T1 and T2...
http://www.arnusa.org/Fuerza.jpg
I know that EF = 0 (the sum of all forces), so I did something like this, EF = T1 +T2 -W1(9.8) - W2(9.8) - W3(9.8) = EF = T1 + T2 - 7.84N - 9.8N - 14.7N, then I don't know what's next...
thanks...
 
Physics news on Phys.org
You've done a balance of forces.

Now do a balance of Moments.

ΣM = 0.
 
Is it correct??

I did the following...


ΣF = 0
ΣM = 0

ΣF = T_1 + T_2 -7.84N - 9.8N - 14.7N
ΣM = -55.7Ncm - 137.2Ncm -294Ncm + 61cmT_1
T_1 = 8.0N

T_2 = 7.84 + 9.8 + 14.7 - 8.0
T_2 = 24.34N

Is this correct??

Thanks...
 
Not totally correct. I'm afraid.

You had the balance of (vertical) forces correct, but got mixed up, it looks like, with taking the moments of the forces.

Moments are all taken about the same point.

You had T1*61. So you were taking moments about the left-hand edge/point. Call it point A. And call the other point, if we need it, point B.

For the 0.8 kg mass, the moment, about A, is 0.8*9.8*7.1 = 55.664 Ncm, which is what you had - correct.
But for the moment, about A, of the 1.0 kg mass, you should have had 1.0*9.8*(7.1+14) = 206.78 Ncm. The distance, from the point A, to the line of action of the 1.0 kg mass is 21.1 cm. Instead you used just 14cm, the gap between the 0.8 kg mass and the 1.0 kg mass. You had the same error for the 1.5 kg mass.

If you correct those errors you will (probably) get the right answer.

I say "probably", because there's a problem with the dimensions on your sketch (just noticed).
The distance between the two tensions, T1 and T2, is 61 cm.
But adding up the distances on the bottom gives 7.1 + 14 + 20 + 12 = 53.1 cm only. 7.9 cm seem to have gone missing !
I don't know if that's a mistake in the presentation of the problem, or if it's deliberate, and you're supposed to do something about it. e.g. add another 7.9/2 = 3.95 cm onto the dimensions at either end, giving 11.05 cm and 15.95 cm and then use these dimension to work out your moments ??
Perhaps you could ask whoever gave you this problem, if there is a mistake with the dimensions ?
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top