Deriving Equation: g = aave(M + m / m)

  • Thread starter Thread starter vrobins1
  • Start date Start date
  • Tags Tags
    deriving
AI Thread Summary
The discussion revolves around deriving the equation g = aave(M + m / m) using two pairs of equations involving tension and kinetic friction. Participants are advised to eliminate the unknown tension from the first pair of equations and the kinetic friction from the second pair. A key suggestion is to isolate the tension in both equations to facilitate comparison. The confusion primarily stems from understanding the process of derivation and how to manipulate the equations correctly. Ultimately, clarity is achieved through guidance on equating the derived expressions.
vrobins1
Messages
21
Reaction score
0
I am supposed to derive this equation: g=aave(M + m / m)

It gives me two pairs of equations:
T1-Fr = Ma1
mg-T1= ma1

and

T2 + Fr = Ma2
mg - T2 = ma2It also says, "hint: for the first pair of equations, eliminate the unknown tension. Do the same for the second pair of equations. In this new pair of equations, eliminate the unknown force of kinetic friction Fr.

So I tried just canceling out the T in the first set of equations, leaving me with
-Fr = Ma1 and mg = ma1, but that doesn't seem right. I am mostly just confused on how to "derive an equation". I'm just looking for a little direction--do I set them equal to each other? I'm just confused on what it's asking, I've never done a problem like this before and have been trying to understand it forever! Any help would be greatly appreciated. Thanks!
 
Physics news on Phys.org
vrobins1 said:
I am supposed to derive this equation: g=aave(M + m / m)

It gives me two pairs of equations:
T1-Fr = Ma1
mg-T1= ma1

and

T2 + Fr = Ma2
mg - T2 = ma2


It also says, "hint: for the first pair of equations, eliminate the unknown tension. Do the same for the second pair of equations. In this new pair of equations, eliminate the unknown force of kinetic friction Fr.

So I tried just canceling out the T in the first set of equations, leaving me with
-Fr = Ma1 and mg = ma1, but that doesn't seem right. I am mostly just confused on how to "derive an equation". I'm just looking for a little direction--do I set them equal to each other? I'm just confused on what it's asking, I've never done a problem like this before and have been trying to understand it forever! Any help would be greatly appreciated. Thanks!

When it says to cancel the T's out, it doesn't mean to literally pretend they never existed. To eliminate the T's from the first two equations, re-write them so that the T's are isolated in both equations.

Now it should be obvious that if A=C and B=C, then A=B. Use the exact same idea with the equations, and see if you can work from there.
 
Thanks that helped, I think I got it now!
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging 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