# Newton's law problem

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1. Aug 16, 2015

### Titan97

1. The problem statement, all variables and given/known data
Two Blocks A and B, each of mass M are attached by a thin inextensible string through a frictionless pulley. Block B is released from rest. Block A hits the pullty in time tA and block B hits the vertical surface at time tB. Find the relation between tA and tB.

2. Relevant equations
$F=ma$

3. The attempt at a solution
Length of the string from pulley to block B increases as it moves down. If the string does not become slack, and since its inextensible, I think tA=tB since acceleration of both ends of string should have equal acceleration. But I think its wrong. (I have attached the solution given by the author, but I want to solve it myself)
Iet block B descend by a distance $y$ and cover an angle $\theta$. Let the length of string be $R$ which can be expressed as a function of $y$ and $\theta$.
$R=ycosec\theta$.
For block A, $T=Ma$.
For block B, along y-direction, $Mg-Tsin\theta=Ma_y$ and $Tcos\theta=Ma_x$
From this, substituting value of T with Ma, $a_x=acos\theta$ and $a_y=g-asin\theta$
Then, net acceleration of Block B is $a_B=\sqrt{a^2+g^2-2agsin\theta}$
$|2agsin\theta|\le 2ag$.
But now, acceleration of both ends are different.
Here is the solution given in the textbook.

2. Aug 16, 2015

### Qwertywerty

What is the relation between the acceleration of the block B along the string with that of block A's acceleration ?

3. Aug 16, 2015

### Titan97

They must be equal...

4. Aug 16, 2015

### Qwertywerty

And block B has acceleration along string , and perpendicular to string , so - ?

5. Aug 16, 2015

### Titan97

It wont move along a circle.

6. Aug 16, 2015

### Qwertywerty

And acceleration of block B with respect to block A would be - ?

7. Aug 16, 2015

### Titan97

Why should i do that? I already found acceleration of block B

8. Aug 16, 2015

### Qwertywerty

I have been trying to establish that acceleration of block A is lesser than that of block B , from -

9. Aug 16, 2015

### Staff: Mentor

As your analysis stands now, your kinematic relationships between the accelerations of blocks A and block B are incorrect.

Have you been learning about using cylindrical coordinates yet? If so, this problem is much easier to solve using cylindrical (r-θ) coordinates, with the origin of the coordinate system for block B located at the pulley. Do you know the equations for the radial acceleration and the angular acceleration if both r and theta are both changing with time?

Let r be the radial distance of block B from the pulley and let theta be the same theta you have been using.

Chet

10. Aug 16, 2015

### PeroK

That solution makes no sense. Instead, you could go back to your initial approach. Hint: You can solve this without any calculation.

Perhaps think about the cases where block B is much heavier or much lighter than block A. What happens in those cases?

Then you might see what happens when the two blocks are of equal mass.

Last edited: Aug 16, 2015
11. Aug 16, 2015

### Titan97

Firstly @Chestermiller , I only know the definition of cylindrical coordinates. But I will try to use it.
Secondly, @PeroK , If block B is much heavier than block A, the time it takes to fall down wont change right? A peice of feather and a block of iron take the same time to fall in vacuum. If block A is heavier, since there is no friction, I still don't think there will be any change.

12. Aug 16, 2015

### PeroK

If block B is much heavier then it will fall straight down and pull block A into the pulley.

If block B is much lighter, it will swing into the wall like a pendulum and A will hardly move.

The question is, if they are the same mass, which one gets pulled more, or do they get pulled equally?

13. Aug 17, 2015

### Titan97

@PeroK , here is how I tried to solve it. The force acting along horizontal for block B is $Tsin\theta$ and the force acting along horizontal on block A is $T$.
$Tsin\theta\le T$. Hence, the acceleration along horizontal direction for block A is more and hence it hits the pulley first or takes lesser time.

14. Aug 17, 2015

### Staff: Mentor

I'm embarrassed to admit that I tried a much more complicated approach to this problem and was not able to reach a conclusion. That's because I violated my own first three most important rules of modeling: 1. keep it simple, 2. keep it simple, 3. keep it simple.

Nice job Titan97.

Chet