What is the Relationship Between Acceleration of a Block and a Pulley?

In summary: It is easy to see that the pulley is moving. Let s_1 be the distance between pulley and A, s_2 be the distance between pulley and B. During the pull, s_2 stays constant. For s_1, it is decreasing right? But that section goes to the upper side of the pulley, so we can set 2s_1 equals also a constant. the reason of 2s_1 comes from the initial situation, we ignore the upper portion that is left of the originial position of A.Hence 2s_1+s_2=constant, differentiate twice yields your desired result.Oh. IDK why I thought the pulley was in equilibrium.
  • #1
imatreyu
82
0

Homework Statement



A horizontal force F is applied to a frictionless pulley of mass m2. The horizontal surface is smooth. Show that the acceleration of the block of mass m1 is twice the acceleration of the pulley.

LOOKS LIKE THIS: http://cnx.org/content/m14060/latest/npq1.gif
But WITHOUT block B and its string.

Homework Equations


F=ma


The Attempt at a Solution


I drew separate force diagrams for m1 (the block) and m2 (the pulley. In the x direction, the block is only being acted on by T1 going in the pos. x direction. In the x direction, the pulley is being acted on by 2T1 and F. 2T1 is going in the neg. x direction. F, the opposite.

I have to show that a1= 2a2

So:

The pulley is in equilibrium:
F-2T1 = 0
m2a2 - 2(m1a1)=0

. . .and I don't know where to go from here. . . .I can't eliminate mass. . .


Thank you in advance!
 
Physics news on Phys.org
  • #2
It is easy to see that the pulley is moving. Let s_1 be the distance between pulley and A, s_2 be the distance between pulley and B. During the pull, s_2 stays constant. For s_1, it is decreasing right? But that section goes to the upper side of the pulley, so we can set 2s_1 equals also a constant. the reason of 2s_1 comes from the initial situation, we ignore the upper portion that is left of the originial position of A.

Hence 2s_1+s_2=constant, differentiate twice yields your desired result.
 
  • #3
Oh. IDK why I thought the pulley was in equilibrium.

Thank you so much!
 
  • #4
imatreyu said:
Oh. IDK why I thought the pulley was in equilibrium.

Thank you so much!

I made the same mistake as you when I was having a introductory mechanics class.
The method I present here is sometimes referred to no-stretch assumption. I don't know why the method is always not mentioned in the textbooks. Is it too obvious for the authors?
 
  • #5
I suppose they think so! The textbook (College Physics, 3rd ed. Serway & Faughn) says nothing about using distances and time derivatives . . . xD

Thank you!
 
Last edited:

What is the acceleration of a pulley?

The acceleration of a pulley is a measure of how quickly its velocity is changing over time. It is typically expressed in meters per second squared (m/s²).

How do you calculate the acceleration of a pulley?

The acceleration of a pulley can be calculated using the formula a = (2T - mg)/m, where T is the tension in the rope, m is the mass of the pulley, and g is the gravitational acceleration (9.8 m/s²).

What factors affect the acceleration of a pulley?

The acceleration of a pulley is affected by the mass of the pulley, the tension in the rope, and the gravitational acceleration. Friction, air resistance, and the shape and size of the pulley can also affect its acceleration.

What is the relationship between acceleration and mass in a pulley system?

In a pulley system, the acceleration is inversely proportional to the mass of the pulley. This means that as the mass of the pulley increases, the acceleration decreases, and vice versa.

How does the direction of the rope affect the acceleration of a pulley?

The direction of the rope does not affect the acceleration of a pulley, as long as the tension in the rope remains constant. This is because the acceleration is determined by the difference in tension on each side of the pulley, not the direction of the rope.

Similar threads

  • Introductory Physics Homework Help
Replies
2
Views
922
  • Introductory Physics Homework Help
Replies
17
Views
1K
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
986
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
33
Views
2K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
2
Replies
66
Views
5K
Back
Top