How to find the net acceleration?

  • Thread starter Thread starter navneet9431
  • Start date Start date
  • Tags Tags
    Acceleration Net
Click For Summary
To find the net acceleration in this problem, the equations for tension and forces acting on both the monkey and the block are crucial. The tension in the rope must be greater than the weight of the block to lift it, while also considering the monkey's force on the rope. The initial equations presented were incorrect; the force applied by the monkey should be represented as -T, aligning with Newton's third law. By combining the inequalities for tension, the final solution for net acceleration can be derived effectively. Understanding these relationships is essential for solving the problem accurately.
navneet9431
Gold Member
Messages
107
Reaction score
9

Homework Statement


IMG_20180710_093219.jpg

This is the question(16) I am going to talk about.

Homework Equations



Acceleration=Force/Mass

The Attempt at a Solution


IMG_20180710_092322.jpg

I drew a Free Body Diagram for this question.Here,The red mark shows the point on the rope at which the monkey applies the force(10a).
So for the block to move up T<10a+10g and T>15g.
So,At the red mark
10(a+g)-T=10*a (Net Acceleration)...(i)
And for the block,
T-15g=15*a (Net Acceleration)....(ii)
So I have successfully find two equation.
What should I do next to get the final answer?

I will be thankful for help!
 

Attachments

  • IMG_20180710_093219.jpg
    IMG_20180710_093219.jpg
    33.4 KB · Views: 779
  • IMG_20180710_092322.jpg
    IMG_20180710_092322.jpg
    16.6 KB · Views: 832
Physics news on Phys.org
It is usual in problems like this to ignore the acceleration of the block that is needed to raise it a couple of millimetres off the floor. As long as the tension in the rope is greater than the weight of the block, by however small a margin, the block will raise off the floor, if we ignore friction. So we can effectively regard the block as being stationary, and so the rope is also stationary, while the monkey scrambles up the rope.

So the problem becomes:

(1) what tension must the rope have in it to support the stationary block off the ground but not accelerate it upwards other than negligibly? and
(2) how fast will the monkey accelerate upwards, climbing the rope, if the rope is pulling up on him by that tension (and he is pulling back down on the rope by the same tension)?
 
How could you say that the monkey is pulling back on the rope by the same tension?
andrewkirk said:
It is usual in problems like this to ignore the acceleration of the block that is needed to raise it a couple of millimetres off the floor. As long as the tension in the rope is greater than the weight of the block, by however small a margin, the block will raise off the floor, if we ignore friction. So we can effectively regard the block as being stationary, and so the rope is also stationary, while the monkey scrambles up the rope.

So the problem becomes:

(1) what tension must the rope have in it to support the stationary block off the ground but not accelerate it upwards other than negligibly? and
(2) how fast will the monkey accelerate upwards, climbing the rope, if the rope is pulling up on him by that tension (and he is pulling back down on the rope by the same tension)?
 
Equation (ii) seems correct. Inequality T>15g is correct.

But equation (i) and inequality T<10a+10g are incorrect. The force that the monkey applies to the rope is not 10a it is -10(a+g)=-T. To prove this make a FBD for the monkey and use Newton's second law. Also use Newton's 3rd law to prove that the force that the monkey applies to the rope equals -T.
 
  • Like
Likes navneet9431
navneet9431 said:
How could you say that the monkey is pulling back on the rope by the same tension?

This is Newton's 3rd Law: The rope applies to the monkey force T, hence the monkey applies to the rope force -T.
 
As @Delta² notes, this
navneet9431 said:
T<10a+10g
should really be an equality. But what you had here:
navneet9431 said:
T<10a+10g and T>15g.
was enough to solve the problem. Just combine them to eliminate T.
 
  • Like
Likes navneet9431
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 12 ·
Replies
12
Views
2K
Replies
31
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 18 ·
Replies
18
Views
1K
  • · Replies 14 ·
Replies
14
Views
1K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
3
Views
484
  • · Replies 13 ·
Replies
13
Views
1K
Replies
8
Views
1K
  • · Replies 13 ·
Replies
13
Views
2K