How does friction affect data in a modified Atwood machine?

Click For Summary
SUMMARY

The discussion focuses on the effects of friction and mass changes in a modified Atwood machine setup. The system consists of a vertical mass hanger with a mass of 2.0 kg and a horizontal cart with four 1.0 kg masses. The analysis reveals that doubling the mass on the vertical hanger does not double the acceleration, as the correct formula for acceleration is a = (m / (M + m)) * g. Additionally, if one of the cart's wheels rubs, it skews the position-time data, leading to an inaccurate representation of acceleration due to unaccounted friction forces.

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Familiarity with free-body diagrams
  • Basic knowledge of kinematics and acceleration
  • Concept of friction in mechanical systems
NEXT STEPS
  • Explore the implications of friction in mechanical systems
  • Learn about the dynamics of Atwood machines
  • Investigate the effects of mass distribution on acceleration
  • Study experimental methods for measuring acceleration and forces
USEFUL FOR

Physics students, educators, and anyone interested in understanding the dynamics of mechanical systems and the impact of friction on motion.

dbakg00
Messages
20
Reaction score
1

Homework Statement



I have a modified atwood machine which is made up of a vertical mass hanger (hanging over the side of a table) connected to a horizontal cart (on the table) via a rope (un-stretchable) and a pulley (massless, frictionless, etc). The cart wheels have negligible friction. There are four equal masses stacked on the cart, each weighing 1.0kg. The mass hanger has a mass of 2.0 kg. The questions are:

(a) if I were to double the mass on the vertical hanger, without changing the mass of the entire system, would the acceleration double?

(b) how would the data be affected if the frame of the cart were bent so that one of the wheels were rubbing



Homework Equations



F_{net}=ma



The Attempt at a Solution



(a) After drawing free-body diagrams for both the cart and the mass hanger, I used F=ma to derive the following equation for the acceleration:

a=\frac{m}{M+m}*g

where "M" = the mass of the cart/weights on top and "m" is the mass of the vertical hanger and any weights it may contain

After plugging a few values in this formula, it appears that the acceleration will double if I double the mass on the hanger without changing the mass of the system.

Did I get the formula right?

(b)

If one of the wheels were rubbing on the cart, that would skew the position-time data for the experiment. It would look like there was a greater amount of mass on the cart that what was there in reality. This would make your acceleration appear slower than it actually was. Also, if you didn't realize that the wheel was rubbing, you wouldn't include the friction force on the free-body diagram; therefore, your skewed data could induce the erroneous conclusion that F_{net}\neq ma.

Is my reasoning correct? Did I miss anything?

Thanks
 
Physics news on Phys.org
You wrote,

"(a) if I were to double the mass on the vertical hanger, without changing the mass of the entire system, would the acceleration double?"

Check your math, I got different results.
 
I see where I made the mistake now, I was making the denominator the entire mass of the system instead of adding the vertical mass to the entire system mass. The acceleration will increase, but not double.

Did part b look ok?

Thanks
 
Part b looked right to me.
 
Thanks for your help.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

Replies
2
Views
2K
Replies
1
Views
4K
  • · Replies 19 ·
Replies
19
Views
7K
  • · Replies 13 ·
Replies
13
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 5 ·
Replies
5
Views
5K
Replies
17
Views
9K
  • · Replies 5 ·
Replies
5
Views
4K
Replies
3
Views
4K
  • · Replies 29 ·
Replies
29
Views
5K