Mass on incline will it slip or stick?

Click For Summary
A block of mass m on a 20-degree slope is connected to a 2.0 kg hanging mass via a pulley, with coefficients of static and kinetic friction of 0.80 and 0.50, respectively. The discussion centers on calculating the minimum mass m that will not slip down the incline and the acceleration if it is nudged. The initial attempt to solve for the minimum mass incorrectly omitted the gravitational force acting on the block on the slope. For part B, it is emphasized that the kinetic friction coefficient should be used instead of the static one. Accurate calculations require considering all forces acting on the system.
charan1
Messages
40
Reaction score
0

Homework Statement


Figure shows a block of mass m resting on a 20* slope. The block has coefficients of friction uS=0.80 and uK=0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.

Part A:
What is the minimum mass m that will stick and not slip?

Part B:
If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

Homework Equations


Fnet=ma
us=F/N

The Attempt at a Solution



I really have no idea how to do Part B, but I tried part A and was real confident with my answer but it turned out wrong. This is what I did.

(2kg)(9.8)=19.6N for the hanging mass

assuming that the Hanging mass = to the tension force on the block with mass m I then did this...

19.6N=m(9.8)cos20*(.8)
m=2.66kg and this is wrong

Were did I go wrong?

Thanks
 

Attachments

  • knight_Figure_08_36.jpg
    knight_Figure_08_36.jpg
    6.1 KB · Views: 487
Physics news on Phys.org
Welcome to PF!

charan1 said:
Figure shows a block of mass m resting on a 20* slope. The block has coefficients of friction uS=0.80 and uK=0.50 with the surface. It is connected via a massless string over a massless, frictionless pulley to a hanging block of mass 2.0 kg.

Part A:
What is the minimum mass m that will stick and not slip?

Part B:
If this minimum mass is nudged ever so slightly, it will start being pulled up the incline. What acceleration will it have?

(2kg)(9.8)=19.6N for the hanging mass

19.6N=m(9.8)cos20*(.8)
m=2.66kg and this is wrong

Hi charan1! Welcome to PF! :smile:

(Useful tip: don't bother to multiply everything by 9.8 … all the terms have 9.8, so just call it g, and cancel it at the end. :wink:)

(oh … and have a mu: µ :smile:)

I take it you were applying Newton's second law in the direction of the slope.

That means that you must include all the forces … you left out the force due to gravity. :smile:

(For part B, use µK instead of µS)
 
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 1 ·
Replies
1
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
Replies
6
Views
3K
  • · Replies 11 ·
Replies
11
Views
5K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
21K
Replies
8
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K