A not-so-easy dynamics problem
In the system below, the coefficient of friction between all surfaces is the same and equal to 1. When you apply force F on the green body, the three bodies move with different accelerations with respect to the ground.
What is the acceleration of the blue body?
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3 solutions
Thank you for posting this problem. I wanted to revise on this concept :)
That`s pretty interesting.Thank you for posting!( ͡° ͜ʖ ͡°)✧
Thank you!
Just to clarify the relation between the accelerations:
Let A, B and C be points that move along with each block and x A x A and x C the x positions of each point. Then let's call the length of upper portion of the rope in the beginning of the movement L 0 and the length of the bottom portion l 0 .
First, we can write the x coordinates for each point in time by setting the origin in the point A:
x A ( t ) = a 2 t 2
x B ( t ) = l 0 + a r 2 t 2 (Assuming that a r < 0 )
x C ( t ) = L 0 + a g 2 t 2
The length of the upper and bottom parts in time can be defined as the distance between the x coordinates of the points:
L ( t ) = x C ( t ) − x A ( t ) = L 0 + a g 2 t 2 − a 2 t 2
l ( t ) = x B ( t ) − x A ( t ) = l 0 + a r 2 t 2 − a 2 t 2
We know that the rope doesn't get any longer or shorter, then the sum L ( t ) + l ( t ) should not vary over time:
L ( t ) + l ( t ) = L 0 + l 0 ⇒ L 0 + a g 2 t 2 − a 2 t 2 + l 0 + a r 2 t 2 − a 2 t 2 = L 0 + l 0
( a g + a r − 2 a ) 2 t 2 = 0 ⇒ a = 2 a g + a r
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Nice contribution, thanks!
That is how I thought about it as well. But you don't need to make any assumptions about constant acceleration motion. The two key equations are L = x c − x a and l = x b − x a , which come straight from your diagram. Since the length of string is constant, we get ( x c − x a ) + ( x b − x a ) = L 0 + l 0 = c o n s t a n t . Take one derivative to get a velocity constraint or two derivatives to get an acceleration constraint: a c + a b − 2 a a = 0 .
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@Eric Roberts – You're right! I had all my a's and c's backwards. Fixed.
I was avoiding using calculus since this problem could be solved by beginners with no calculus knowledge. But you're right, you could prove it without assuming constant acceleration (which is a totally fair assumption in this problem by the way) and then take the derivatives in the end.
sir i think you haven’t considered string constraint in this problem
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Yes he did, the string constraint is embedded in the relation between the accelerations and more specifically proven bellow in my comment.
What about the gearing ratio of the pulleys? The attachment point of the blue box (in the centre) has a radius far smaller than the attachment point of the green box. So won't it's acceleration be that same fraction less than that of the green box?
Yes, you are right. But I think the color of the left box is purple, not blue!
To find mutual relationship for individual accelerations you have to realize they are constant vectors (no change with time). Thus, if blue block moves with acceleration a, the red block has to move with acceleration a + x while the green block is moving with acceleration a - x due to string constrain. This is equivalent to 2*a(blue)=a(red) + a(green). Plugging these into x-direction equations gives a(blue)=g/2 and x=3/2g, thus a(red) is 2g and A(green) is -1g.
Can you explain how you related the acceleration of blocks
Why T is 6Mg? (9Mg-Mg=8Mg) Is it wrong?
I got it sorry.
Zero. There is no blue body. That one is lavender.
The equation for red box is incorrect you assumed that its going to left but equation suggests its going to the right
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Notice I use the same reference frame for all forces and accelerations (positive to the right), so when I solve the system I get the right sign for a r , which is negative. The equation is correct.
I think there's a problem if a（B）=（a（r）+a（g））/2,then follow newton's second law,we'll get T-F（r）-F（g）=2Ma（r） the key is to understand the direction of acceleration of"r" Here is my anwser first according to newton'second law,we can get T-3Mg-Mg=2Ma（r）① F-T-Mg=Ma(g）② 2T-8Mg=8M(b)③ (Here,all the accelerations are positive,so all the accelerations acctually are their size) since the red one and the green one are connected to each other with the rope,their acceleration is the same(I mean their size) a（g）=a（r) so with①②④we can get a（g）=a（r)=4g/3,T=20Mg/3 then with③ we can get a（B）=2g/3 maybe I am false,so I hope you can help me find out if I am false
Why is F r 3 mg. Shouldn't it be 2 mg? Thanks Mudit
If the red block is moving left, the frictional forces on the green block act in opposite directions and cancel each other out. The frictional force on the red block is (2M+M) g = 3 Mg and is the only force opposing the 9Mg force. Therefore the tension in the rope pulling the red block is 9Mg - 3 Mg = 6 Mg. The force acting to the right on the blue block is 12 Mg, as the tension 6 Mg acts doubly. The frictional force acting to the left on the blue block is 8Mg. Therefore the resultant force on the blue block is 12 Mg -8 Mg= 4 Mg to the right, and the acceleration is 4Mg/ 8M= 1/2g
The red one goes left,so the f is right and is 3Mg. (9Mg+3Mg-8Mg)/8M=1/2g English not very good sorry
Would need to show that the red box goes left.
Nice holistic solution! But yes if red doesnt move the answer is 1/8g
However if g=0 then the answer can also be simplified to g thereby giving a possible right answer as well depending on whose g value we use.
I treated the pulley system on its own at first...
The pulley system will accelerate clockwise, with respect to the the pulley. There is a "driving force" of 9Mg and an external "retarding" force between the red block and the floor of 3Mg (because coefficient of friction is 1). So the pulley system's, of total mass 3M (green + red), net force equation is:
Fnet = 9Mg - 3MG = 3Ma
Note: When you solve "regular" pulley systems this is a standard method to avoid dealing with the internal forces (i.e. Tension). Here, we have additional internal forces such as the friction between the red and green blocks...
So anyway, we get the acceleration of the pulley system with RESPECT TO THE PULLEY (this is an important distinction) is a = 2g
Then we need to find the value of the Tension acting on the green and red blocks. So I used the green block. Fnet = driving force - tension - friction = 9Mg - T - 3Mg = Ma
But acceleration is +2g; therefore 9Mg - T - 3Mg = 2Mg
This leads to T = 6Mg
Net force on the pulley is zero; hopefully, you realize that the Tension force on the blue block has to be twice the Tension we just mentioned above.
So the Net Force question on the blue block is Fnet = 2(tension on green block) - friction = 2(6Mg) - 8Mg = 8Ma
So a = g/2
All vertical accelerations are 0 and μ = 1 :
N b − 8 M g = 0 N g − M g = 0 N r − N g − 2 M g = 0 ⟹ ⟹ ⟹ N b = 8 M g ; N g = M g ; N r = 3 M g ; F b = μ N b = 8 M g F g = μ N g = M g F r = μ N r = 3 M g
The x-axis equations are:
2 T − F b = 8 M a b F − F g − T = M a g F r + F g − T = 2 M a r ⟹ ⟹ ⟹ + M g 2 T − 8 M g = 8 M a b 9 M g − M g − T = M a g 3 M g + M g − T = 2 M a r
The pulley causes the three accelerations a b , a g and a r to be related in the following way:
a b = 2 a g + a r
(Convince yourself of this by considering, for example, what would happen if the red body did not slide.)
Solving the resulting linear system for T , a g , a r and a b we get:
T = 6 M g ( T > F g + F r , the tension is large enough to make the red body slide to the left. The initial supposition was correct!)
a g = 2 g , a r = − g and a b = 2 g