Find Force to Keep M3 in Equilibrium: K&K Problem

  • Thread starter Thread starter WiFO215
  • Start date Start date
  • Tags Tags
    Review
AI Thread Summary
To keep mass M3 in equilibrium, the force F applied on M1 must counterbalance the gravitational force acting on M3. The initial approach incorrectly applies fictitious forces and does not consider the overall system's acceleration. Instead, the entire setup should be treated as a single mass to determine the correct horizontal acceleration. The correct relationship leads to the conclusion that F can be expressed as F = (M3.g.M1)/M2, but the reasoning needs to be adjusted for accuracy. Clarifying the approach will help identify the mistake in the initial calculations.
WiFO215
Messages
416
Reaction score
1

Homework Statement


I got this problem from K&K. A pedagogical machine is illustrated in the sketch above. Refer link : http://www.slideshare.net/brigittperalta/sol-maquina-pedagogica-1546585 . What force F must be applied on M1 to keep M3 from moving up or down. All surfaces frictionless.


Homework Equations





The Attempt at a Solution


If I go and sit on the M1 block, I can mark a fictitious force on M2 to the left with magnitude (M2.F)/M1. On M2 there is also tension from the string (T) to the right. If M3 is to remain in equilibrium, M3.g = T and T = (M2.F)/M1 to prevent block 2 from moving which will cause block 3 to move.

Therefore, I get my answer to be F = (M3.g.M1)/M2. I am making a mistake here but cannot see where. Can someone please tell me where?
 
Physics news on Phys.org
Hi anirudh215,

anirudh215 said:

Homework Statement


I got this problem from K&K. A pedagogical machine is illustrated in the sketch above. Refer link : http://www.slideshare.net/brigittperalta/sol-maquina-pedagogica-1546585 . What force F must be applied on M1 to keep M3 from moving up or down. All surfaces frictionless.


Homework Equations





The Attempt at a Solution


If I go and sit on the M1 block, I can mark a fictitious force on M2 to the left with magnitude (M2.F)/M1.

I believe this is incorrect here. According to this expression, the overall horizontal acceleration would be given by F=(m1) a, which is not true.

Instead, begin by considering the entire setup as one mass to find the horizontal acceleration, and then continue from there as you already have done.
 
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