Solving Physics Homework: Find Acceleration & Tension

AI Thread Summary
A 5.00-kg object on a frictionless table is connected via a pulley to a 9.00-kg hanging object, prompting a calculation for acceleration and tension. The initial approach used a net force equation but raised questions about the correct application of mass in this scenario. It was suggested to draw free body diagrams for both objects and apply Newton's laws separately to derive two equations for the unknowns. The calculated acceleration of 4.55 m/s² is confirmed to be the same for both objects. The discussion emphasizes the importance of using proper methods to solve for tension and acceleration in such systems.
Husker70
Messages
89
Reaction score
0

Homework Statement


A 5.00-kg object placed on a frictionless, horizontal table is connected to a cable that
passes over a pulley and then is fastened to a hangng 9.00-kg object. Find the acceleration of the two objects and the tension in the string.


Homework Equations


I am using Net Force = (m2)-1/2(m1)(g)
a = force/total mass

The Attempt at a Solution


I am using Net Force = (m2)-1/2(m1)(g) = (9.00)-1/2(5.00)(9.8) = 63.7N
I used 1/2 of m1 because if the were hanging like on an atwood machine it
would be -m1. Since this is horizontal surface would this be correct?
I then divided the force(63.7N) by Total Mass 14.0-kg to get 4.55m/s2
They both should be travling at the same acceleration, I believe
I'm working on the tension.
Am I starting this right?
Thanks,
Kevin
 
Physics news on Phys.org
Hi Husker70,

Husker70 said:

Homework Statement


A 5.00-kg object placed on a frictionless, horizontal table is connected to a cable that
passes over a pulley and then is fastened to a hangng 9.00-kg object. Find the acceleration of the two objects and the tension in the string.


Homework Equations


I am using Net Force = (m2)-1/2(m1)(g)
a = force/total mass

The Attempt at a Solution


I am using Net Force = (m2)-1/2(m1)(g) = (9.00)-1/2(5.00)(9.8) = 63.7N
I used 1/2 of m1 because if the were hanging like on an atwood machine it
would be -m1. Since this is horizontal surface would this be correct?

I do not believe this is correct. Rather than try doing this in one step, I would suggest you draw a free body diagram for each of the objects. Then you can write down Newton's law in the vertical direction for the hanging object and in the horizontal direction for the object on the table; with two equations and two unknowns here you can solve for both unknowns. What do you get?


I then divided the force(63.7N) by Total Mass 14.0-kg to get 4.55m/s2
They both should be travling at the same acceleration, I believe

Yes, they will have the same acceleration here.
 
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