Why is this wrong?

  • Thread starter suspenc3
  • Start date
  • #1
402
0
I just recently got my midterm back and I felt this was graded wrong...

I dont have a picture..but just picture a ramp inclined to [tex] \phi=30[/tex] One mass is sitting on the incline...And the other is dangling from a rope which runs over a pulley

Two masses [tex]m_1[/tex] & [tex]m_2[/tex] both have a mass of 2Kg. If [tex]m_1[/tex] slide up the ramp at a constant velocity, what is the kinetic frictional force on [tex]m_1[/tex]? Assume that the masses of the rope and pulley are negligible, and that the pulley is frictionless.
How I did it...
[tex]m_1 x/ T-f_k = m_1a[/tex] (Eq1)
[tex] y/ N-mgsin \phi=0[/tex]
[tex]m_2 x/ T-mgsin \phi=m_2a[/tex]
[tex] y/ T=mgsin \phi[/tex] (Eq2)
-------------------------------------------------
isolate T in Eq1
[tex]T-f_k=m_1a[/tex] Since a=0
[tex]T=f_k[/tex]
Sub Eq2 into Eq1
[tex]mgsin \phi=f_k[/tex]
[tex]2Kg(9.8m/s)sin \phi=f_k[/tex]
[tex]f_k=9.8N[/tex]
Although i got the right answer he says that it needs to be expressed like this
[tex]T-f_k-mgsin \phi=0[/tex]
[tex]f_k=T-mgsin \phi[/tex] T=mg
[tex]f_k=mg-mgsin \phi = 9.8N[/tex]
Why is my way wrong?
 
Last edited:

Answers and Replies

  • #2
mezarashi
Homework Helper
650
0
Could you explain what your variables x, y, and T mean?

You should analyze it through the use of force diagrams. The hanging mass is there only to tell you that the tension in the string, T is equal to mg. From there you free body diagram should focus entirely on the mass on the incline.
 
  • #3
3,763
9
suspenc3 said:
[tex]m_1 x/ T-f_k = m_1a[/tex] (Eq1)
What is this x ? Is that x=gsin(30°) ?

Besides, in the x-direction (ie along the incline) you have 3 forces
1) T [tex]m_2g[/tex]
2) friction [tex] \mu_kN[/tex]
3) the x-component of gravity : gsin(30°)
I did not put in the right signs of these forces...
I only see two forces in your equation

What is x/T ?

The acceleration of object 1 is indeed 0

You should have [tex]m_1a_1=0=m_2g - \mu_kN -mgsin(30°)[/tex]

[tex] y/ N-mgsin \phi=0[/tex]
[tex]m_2 x/ T-mgsin \phi=m_2a[/tex]
What is m_2 x/T ???
[tex] y/ T=mgsin \phi[/tex] (Eq2)
This is incorrect.

In the y direction (perpendicular to the incline) you have
[tex]m_1a_1 = 0 =-mgcos(30°) + N[/tex]

Where N is the normal force.


But you only need the equation in the x direction to get your answer

marlon
 
Last edited:

Related Threads on Why is this wrong?

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
7
Views
8K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
2
Views
1K
Replies
3
Views
1K
  • Last Post
Replies
5
Views
1K
Replies
1
Views
2K
Replies
2
Views
2K
Replies
4
Views
1K
Replies
17
Views
1K
Top