Question about a simple free body diagram

In summary: This problem is ideal for Lagrangian mechanics. Let ##x## be the dispacement of the small top block to the right towards the pullley (which is also the vertical displacement of the side block). And, let ##X## be the displacement of the large block to the left (##X## will be negative). Then, the displacement of the top block is ##x + X## and the displacement of the side block is ##(X, -x)##. We have:$$T = \frac 1 2 M \dot X^2 + \frac 1 2 m(\dot x + \dot X)^2 + \frac 1...$$
  • #71
jbriggs444 said:
The interaction with the pulley is more interesting. The cord exerts a uniform radial force along a 90 degree arc of the ideal pulley's surface. The net of this is a force whose magnitude is ##\sqrt{2}## times tension at an angle 45 degrees below the horizontal.
And, we can test this claim! This says that we have a horizontal force from the pulley on the large block of ##T##. (And, likewise, a vertical component of ##T## downwards.) This would give the horizontal acceleration of the large block and side block of:
$$\ddot X = -\frac{T}{M+m} =-\frac{mg}{2M + 3m}$$As expected.
 
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  • #72
jbriggs444 said:
The cord between pulley and bottom block is vertical. It exerts a vertical force at its attachment point on the side block.
That's true initially. But when motion occurs, the cord between pulley and side block is not vertical. If it were, there would be no horizontal component of tension to give the side block its horizontal acceleration.

jbriggs444 said:
The interaction with the pulley is more interesting. The cord exerts a uniform radial force along a 90 degree arc of the ideal pulley's surface. The net of this is a force whose magnitude is ##\sqrt{2}## times tension at an angle 45 degrees below the horizontal.
.
We have a leftward force of magnitude T plus a downward force of magnitude T plus an up-and-to-the-rightward force of magnitude ##\sqrt{2}T##. They sum to zero. Life is still good and the laws of physics still hold.
I think that, when acceleration occurs, there are different tensions in the horizontal and near-vertical sections of the cord. Also the block may not have a square cross-section. So finding the force (magnitude and direction) on the pulley is quite tricky.

Edited.
 
  • #73
Steve4Physics said:
That's true initially. But when motion occurs, the cord between pulley and side block is not vertical. If it were, there would be no horizontal component of tension to give the side block its horizontal acceleration.
Yes indeed.

Steve4Physics said:
I think that, when acceleration occurs, there are different tensions in the horizontal and near-vertical sections of the cord. Also the block may not have a square cross-section. So finding the force (magnitude and direction) on the pulley is quite tricky.
For an ideal pulley, the magnitudes of the two tensions will always remain identical. That is what ideal pulleys do. They equalize the magnitudes of two tensions while allowing their directions to change.

Of course the angle between the directions will shift from 90 degrees through a range of other values as the side block swings.

Edit: I may have misread your claim as saying that the tensions in the two cord segments will differ from each other. I claim that they will always be equal to each other. However, I agree that the shared tension value will vary over time.
 
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  • #74
If we look for the equilibrium angle ##\theta##, where the side block and the large block both have horizontal acceleration ##\ddot X## and the side block has vertical acceleration ##\ddot y##, with ##y = x \cos \theta##, then I get the following equations:
$$m(\ddot x + \ddot X) = T$$$$M\ddot X = -T\cos \theta$$$$m\ddot X = -T\sin \theta$$$$m\ddot y = mg - T\cos \theta \ \Rightarrow \ m\ddot x \cos \theta = mg -T\cos \theta$$This gives the solution:
$$\tan \theta = \frac m M$$$$T = \frac g 2(\frac m M)\sqrt{M^2 + m^2}$$$$\ddot x = \frac g 2(\frac 1 M)\sqrt{M^2 + m^2}$$$$\ddot X = -\frac g 2(\frac 1 M)\frac{\sqrt{M^2 + m^2}}{1 + \sqrt{M^2 + m^2}}$$
 
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  • #75
jbriggs444 said:
For an ideal pulley, the magnitudes of the two tensions will always remain identical. That is what ideal pulleys do. They equalize the magnitudes of two tensions while allowing their directions to change.
.
Edit: I may have misread your claim as saying that the tensions in the two cord segments will differ from each other. I claim that they will always be equal to each other. However, I agree that the shared tension value will vary over time.
Yes. You’re right. Apologies. For some reason my brain added friction at the pulley.
 
  • #76
PeroK said:
$$m\ddot X = -T\sin \theta$$
I think that equation is wrong. In order for ##\theta## to be constant, the side block must be moving away from the large block at an acceleration of ##\ddot x \sin \theta##. That force equation ought to be:
$$m(\ddot X + \ddot x \sin \theta) = -T\sin \theta$$Which makes the solution more complicated.
 
  • #77
It just gets a lot messier with the revised equations. E.g. I get:
$$\frac m M = \frac{2\tan \theta}{1 - \sin \theta}$$$$\ddot X = -\frac{2 \sin \theta}{1 + \sin \theta}\ddot x$$$$\big ( \frac M m + \frac{1 + \sin \theta}{2\tan \theta} \big )\ddot X = -g$$I don't propose to spend any more time on this!

PS although the equilibrium solution is quite simple:$$\ddot X = -(\tan \theta)g$$$$\ddot x = \frac{1 + \sin \theta}{2\cos \theta}g$$Where ##\theta## can be obtained numerically from ##\frac M m## using the above equation.
 
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