Find the accelerations of all three blocks and the tension in the rope

AI Thread Summary
The discussion focuses on solving for the accelerations of three blocks and the tension in a rope using Newton's laws. The initial approach involves setting up three equations based on Newton's second law and a fourth constraint equation. There is a concern about potential sign errors in the equations, specifically regarding the correct formulation of the constraint equation. It is suggested that to maintain consistency, the constraint should be adjusted to reflect the positive accelerations or the signs of the first two equations should be reversed. The system is confirmed to have four equations and four unknowns, making it solvable.
rashida564
Messages
220
Reaction score
7
Homework Statement
In the system shown above m1 > m2 . The pulleys are massless and frictionless, and the
rope joining the blocks has no mass. The coefficient of static friction between the blocks
and the tables is greater than the coefficient of kinetic friction: µs > µk . The downward
acceleration of gravity is g .
Now consider the case where, when released from rest, all three blocks
begin to move. Find the accelerations of all three blocks and the tension in
the rope.
Relevant Equations
F=ma
I am thinking about solving it this way.
The first three equation is from Newton law.
Then the forth one is the constrain equation, after that I simplified the first three equations, and I am thinking about plugging them into the forth equation and solve for T after that I get the value of each acceleration, but I have 10% scared that I am having sign wrong in one of them.
Sorry about writing the equation as picture I want to improve my writing, but going to learn LaTex after exams.
 

Attachments

  • Solving.PNG
    Solving.PNG
    18.9 KB · Views: 221
  • Capture.PNG
    Capture.PNG
    7.8 KB · Views: 230
Physics news on Phys.org
I believe in order to be consistent with your first 3 equations, the fourth equation should read ##a_{1} + a_{2} = 2a_{3}##, since all of the accelerations are taken to be positive. Either that or reverse the signs of the first two Newton II relations, to be consistent with the directions of ##x_{1}## and ##x_{2}## on your diagrams. It doesn't matter which you choose.

Then, the system has 4 equations and 4 unknowns so it should be possible to solve.
 
Thanks mate, now I see my mistake.
 
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