Why Is My Calculation for the Time to Drop an Object Incorrect?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
7 replies · 2K views
aa_o
Messages
24
Reaction score
4

Homework Statement


upload_2017-12-11_16-19-22.png


Homework Equations


F = ma

The Attempt at a Solution


The gravity pulls on m with a force Fg:
Fg = m*g
This force is directly translated into pulling the total mass m+M with an acceleration a:
(m+M)*a = Fg <=> a = Fg / (m+M) = m*g / (m+M)
with this acceleration, we can calculate the time it takes to drop the distance, d:
d = 1/2 * a * t^2 <=> t = sqrt(2*d / a)
This yields a time of t = 1.73 s, but the answers says 1.0 s.

What am doing wrong!?
 

Attachments

  • upload_2017-12-11_16-19-22.png
    upload_2017-12-11_16-19-22.png
    11.8 KB · Views: 863
Physics news on Phys.org
aa_o said:
What am doing wrong!?
Your assumption that mg is the force that pulls the total mass to the right is not correct.

Draw a free body diagram for each mass and set up Newton's second law for each mass.
 
  • Like
Likes   Reactions: aa_o
Ahh, the the tension is of course decreasing as the mass, m, is accelerating. But this means that the force on the total mass is actually less than assumed above T = (g-a)*m, giving me an even longer time?
More specifically:
T = Fx = m*(g-a) = (m+M)*a <=> a = m*g / (2*m + M)
with the same solution for t as above
 
Last edited:
How many horizontal tension forces act on the total mass?

Does the vertical acceleration of m equal the horizontal acceleration of the system?
 
  • Like
Likes   Reactions: aa_o
Okay, i see that we acutally have the tension force acting in the horizontal in 4 places, giving us:
a = 4*m*g / (5*m + M).
Is this correct? (It gives the right answer, but it could be a coincidence!).

It seems that i need to practice my FBD skills, especially when it comes to tension forces.
 
There are not 4 horizontal tensions acting on the (m+M) system. Also, have you taken into account that the vertical acceleration of m is different from the horizontal acceleration of the system?
 
  • Like
Likes   Reactions: aa_o
TSny said:
There are not 4 horizontal tensions acting on the (m+M) system. Also, have you taken into account that the vertical acceleration of m is different from the horizontal acceleration of the system?
Eureka! Of course, both the upper and lower horizontal parts of the string contributes to the acceleration, giving a vertical acceleration that's 2 x that of the system. And then we have 2 horizontal tension forces pulling the system. This gives the same result as above, but with the right physical explanation. Thanks a lot!