How does friction affect data in a modified Atwood machine?

AI Thread Summary
In a modified Atwood machine setup, doubling the mass on the vertical hanger does not double the acceleration, as the correct formula accounts for the total mass of the system. The acceleration is derived from the equation a = (m / (M + m)) * g, where M is the mass of the cart and weights, and m is the mass of the hanger. If one of the cart's wheels rubs, it skews the position-time data, making the system appear to have more mass and resulting in a slower acceleration than actual. This frictional force must be included in the free-body diagram to avoid incorrect conclusions about net force and acceleration. Overall, the discussions highlight the importance of accurate calculations and the impact of friction on experimental data.
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.
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top