• Support PF! Buy your school textbooks, materials and every day products Here!

Acceleration of a Block

  • Thread starter Abid Rizvi
  • Start date
  • #1
20
0

Homework Statement


What horizontal force must be applied to a large block of mass M shown in the figure below so that the tan blocks remain stationary relative to M? Assume all surfaces and the pulley are frictionless. Notice that the force exerted by the string accelerates m2. (Use the following as necessary: m1, m2, M, and g.)

5-p-067.gif


Homework Equations


F = ma

The Attempt at a Solution


So I found the right answer to be (M+m1+m2)* ## \frac{m1g}{m2}##, but I don't understand why. My way was:
T = ##m_1##a
##m_2##g-T = ##m_2##a
##m_2##g = ##m_2##a + ##m_1##a
so therefore a = ##\frac{m_2 g}{m_1+m_2}##
now we know that F = ma, and I know that the entire system must have the same acceleration as the acceleration of the two blocks so that they don't move relative to the big block. Therefore (M+m1+m2)* ##\frac{m_2 g}{m_1+m_2}## was my answer. So what is going wrong here is my acceleration. But why? Also I found two pdf's from two different universities containing the solution to the problem. Both have different solutions...

http://www.physics.oregonstate.edu/~mcintyre/COURSES/ph211_S04/PHpdfs/hw5_solns.pdf
http://opencourses.emu.edu.tr/pluginfile.php/6813/mod_resource/content/1/tutorial_The_Laws_of_Motion.pdf [Broken]

Interestingly enough my solution was more similar to the first pdf (which was wrong) but both pdf's say ##m_2##g-T = 0 where I had ##m_2##g-T = ##m_2##a. The only difference between the two are that the fraction at the end of the answer is flipped
 
Last edited by a moderator:

Answers and Replies

  • #2
ehild
Homework Helper
15,492
1,874

Homework Statement


What horizontal force must be applied to a large block of mass M shown in the figure below so that the tan blocks remain stationary relative to M? Assume all surfaces and the pulley are frictionless. Notice that the force exerted by the string accelerates m2. (Use the following as necessary: m1, m2, M, and g.)

5-p-067.gif


Homework Equations


F = ma

The Attempt at a Solution


So I found the right answer to be (M+m1+m2)* ## \frac{m1g}{m2}##, but I don't understand why. My way was:
T = ##m_1##a
##m_2##g-T = ##m_2##a
##m_2##g = ##m_2##a + ##m_1##a
so therefore a = ##\frac{m_2 g}{m_1+m_2}##
now we know that F = ma, and I know that the entire system must have the same acceleration as the acceleration of the two blocks so that they don't move relative to the big block. Therefore (M+m1+m2)* ##\frac{m_2 g}{m_1+m_2}## was my answer. So what is going wrong here is my acceleration.
You are right that the whole system has the same acceleration, a. The acceleration is in the horizontal direction. When writing Newtons's second law, you have to take those force into account which act directly on the body.
What are the forces acting on block m1? It is the tension (upward), gravity (downward), and the normal force from the big block (horizontal, forward). You have two equations, one for the vertical acceleration, and the other for the horizontal acceleration. But the block accelerates only horizontally. How are the tension and the force of gravity related?
In case of the block m2, it is pulled by the string, and the only horizontal force applied is the tension. Gravity cancels with the normal force from the big block.
 
  • #3
20
0
Oh I defined the forces for the wrong blocks, I accidentally switched them. It should be T = m2a and m1g-T = m1a. However, I'm not sure I understand why it only accelerates horizontally. If block m2 starts accelerating to the right, then wouldn't block m1 start accelerating vertically since it starts to fall downward, or are we supposed to define the forces as if block m1 is stationary therefore resulting in that zero acceleration? Also, which way is block m2 accelerating? If the entire system is to remain stationary, then block m2 needs to have an opposite force to cancel out tension right? So relative to the big cube M, it is accelerating to the left when the big cube is accelerating to the right? Thanks in advance.

Disregard the PDF thing, I accidentally mixed the two blocks
 
  • #4
haruspex
Science Advisor
Homework Helper
Insights Author
Gold Member
33,454
5,406
If block m2 starts accelerating to the right, then wouldn't block m1 start accelerating vertically?
The condition is not that m2 remains stationary, but that it remains stationary with respect to M. In particular, it remains stationary with respect to the pulley mounted on M. What does that tell you about the distance from m2 to the pulley? What does that tell you about the distance from m1 to the pulley?
 
  • #5
ehild
Homework Helper
15,492
1,874
Oh I defined the forces for the wrong blocks, I accidentally switched them. It should be T = m2a and m1g-T = m1a. However, I'm not sure I understand why it only accelerates horizontally. If block m2 starts accelerating to the right, then wouldn't block m1 start accelerating vertically since it starts to fall downward, or are we supposed to define the forces as if block m1 is stationary therefore resulting in that zero acceleration? Also, which way is block m2 accelerating? If the entire system is to remain stationary, then block m2 needs to have an opposite force to cancel out tension right? So relative to the big cube M, it is accelerating to the left when the big cube is accelerating to the right?
The problem asks:
What horizontal force must be applied to a large block of mass M shown in the figure below so that the the blocks remain stationary relative to M?
Stationary relative to M means that the blocks move together with the big block, as if they were glued together... If you sit in a car you are stationary with respect to the car, but you travel with the velocity of the car with respect to the ground.
You know that the acceleration of the centre of mass of a system of bodies is determined by the resultant of the external forces and the the total mass of the system. The resultant force is F and the total mass is M+m1+m2, a=F/(M+m1+m2).

As m1 and m2 do not move with respect to M, they must have the same acceleration as M. This acceleration is horizontal. You also know that the acceleration and the force are parallel.
The only horizontal force applied to the block m2 is the tension in the string. You wrote correctly that T=m2a.
The hanging block accelerates also horizontally. It touches M and the normal force from M pushes it to the right.
Does it fall downward if it does not move with respect to the big block?
Gravity acts on you if you sit in your car. Do you fall downward while driving the car?
If there is no vertical acceleration what is the sum of the vertical forces? You wrote that m1g-T = m1a, but it is wrong as m1g and T are vertical forces, while the acceleration, a, is horizontal. So what is the correct equation?
 
  • #6
20
0
The condition is not that m2 remains stationary, but that it remains stationary with respect to M. In particular, it remains stationary with respect to the pulley mounted on M. What does that tell you about the distance from m2 to the pulley? What does that tell you about the distance from m1 to the pulley?
So this tells me that the distance remains constant

The problem asks:


Stationary relative to M means that the blocks move together with the big block, as if they were glued together... If you sit in a car you are stationary with respect to the car, but you travel with the velocity of the car with respect to the ground.
You know that the acceleration of the centre of mass of a system of bodies is determined by the resultant of the external forces and the the total mass of the system. The resultant force is F and the total mass is M+m1+m2, a=F/(M+m1+m2).

As m1 and m2 do not move with respect to M, they must have the same acceleration as M. This acceleration is horizontal. You also know that the acceleration and the force are parallel.
The only horizontal force applied to the block m2 is the tension in the string. You wrote correctly that T=m2a.
The hanging block accelerates also horizontally. It touches M and the normal force from M pushes it to the right.
Does it fall downward if it does not move with respect to the big block?
Gravity acts on you if you sit in your car. Do you fall downward while driving the car?
If there is no vertical acceleration what is the sum of the vertical forces? You wrote that m1g-T = m1a, but it is wrong as m1g and T are vertical forces, while the acceleration, a, is horizontal. So what is the correct equation?
So it doesn't fall downward, and I don't fall down while driving. (But...I fall down if i push my seat down :D) So we get m1g-T = 0. Okay, I think I understand now how this works. Also I asked if block m2 is accelerating to the left relative to block M, but to try and answer my own question: this is synonymous to a person falling on a roller coaster and a penny levitating in front of you. Both accelerate in the same direction so it is not that block m2 is pulling on the string to the left but that it's not sliding fast enough to move relative to M. Is this correct? Thank you!
 
  • #7
ehild
Homework Helper
15,492
1,874
So this tells me that the distance remains constant



So it doesn't fall downward, and I don't fall down while driving. (But...I fall down if i push my seat down :D) So we get m1g-T = 0. Okay, I think I understand now how this works. Also I asked if block m2 is accelerating to the left relative to block M, but to try and answer my own question: this is synonymous to a person falling on a roller coaster and a penny levitating in front of you. Both accelerate in the same direction so it is not that block m2 is pulling on the string to the left but that it's not sliding fast enough to move relative to M. Is this correct? Thank you!
I think you got it. :) But m2 pulls the string just as the string pulls m2, but that pull is not enough to give more acceleration as that of the big mass M.
 

Related Threads on Acceleration of a Block

  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
945
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
3
Views
2K
  • Last Post
Replies
6
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
13
Views
6K
  • Last Post
Replies
1
Views
524
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
2
Views
3K
Top