Find the force so that all blocks are at rest with respect to each other

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SUMMARY

The discussion centers on calculating the force required to keep three blocks at rest with respect to each other, involving a cart, two blocks, and a pulley. Key equations derived include the friction force for Block 1 as 6 N and the acceleration calculated as 4.12 m/s². Two different approaches to find the external force F yielded results of 61.8 N and 43.4 N, highlighting discrepancies due to omitted forces acting on the cart. The conversation emphasizes the importance of considering all contact forces and the mass of the pulley in the calculations.

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songoku
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Homework Statement
(diagram given below)

Block 1 is put on top of a cart having mass 10 kg and block 2 is connected to block 1 through ideal pulley as shown in diagram (there is contact between cart and block 2). If coefficient of friction between block 1 and cart and block 2 and cart is 0.2, find the force F and acceleration of the system so block 1 and 2 are at rest with respect to the cart (mass of block 1 = 2 kg, mass of block 2 = 3 kg, the floor is frictionless and take acceleration due to gravity = 10 N/kg)
Relevant Equations
ΣF = m.a
friction = μ.N
Untitled 1.png

Free body diagram one each object:
Block 1:
Normal force upwards, weight downwards, tension to the right, friction to the left

Block 2:
Normal force to the right, tension upwards, weight downwards, friction upwards

Cart:
Contact force with block 1 downwards, contact force with block 2 to the left, external force to the right, weight downwards, friction with block 1 to the rightEquation:
Block 1:
friction = μ.N1 = 0.2 x 30 = 6 N

ΣF = m.a
T - friction = 3a
T = 3a + 6Block 2:
ΣF = m.a
N2 = 2a

T + friction2 = W
3a + 6 + 0.2 x 2a = 20
a = 4.12 ms-2Cart:
I did 2 different working to find F but ended up with different results.
(1) Taking all the 3 objects as one system, so the horizontal force acting on the system is F only
ΣF = mtotal.a
F = 15a = 61.8 N

(2) Only considering forces acting on the cart:
ΣF = m.a
F + friction with cart 1 - N2 = m.a
F = 10a + 2a - 6 = 43.4 NWhere is my mistake? Thanks
 
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songoku said:
(2) Only considering forces acting on the cart:
ΣF = m.a
F + friction with cart 1 - N2 = m.a
I believe you left out one of the horizontal forces acting on the cart. Anything that touches the cart can exert a contact force on the cart.
 
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TSny said:
I believe you left out one of the horizontal forces acting on the cart. Anything that touches the cart can exert a contact force on the cart.
You mean the contact force between cart and pulley?
 
songoku said:
You mean the contact force between cart and pulley?
Yes, assuming that you are not considering the pulley as part of the cart.

Or, you may consider the pulley as part of the cart. Then you would consider any forces that the string exerts on the pulley as forces acting on the cart.

Either approach is fine.
 
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TSny said:
Yes, assuming that you are not considering the pulley as part of the cart.

Or, you may consider the pulley as part of the cart. Then you would consider any forces that the string exerts on the pulley as forces acting on the cart.

Either approach is fine.
If I consider pulley and cart separately, should I put mass of pulley when I use my 1st approach?

(1) Taking all the 3 objects as one system, so the horizontal force acting on the system is F only
ΣF = mtotal.a

So total mass is the sum of masses of 4 objects?
 
songoku said:
friction to the left
How did you decide that? Are you looking for the minimum, maximum acceleration or range?
 
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haruspex said:
How did you decide that? Are you looking for the minimum, maximum acceleration or range?
I am assuming the force given is minimum so acceleration is minimum
 
songoku said:
If I consider pulley and cart separately, should I put mass of pulley when I use my 1st approach?
The pulley can be considered to be massless.
 
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TSny said:
The pulley can be considered to be massless.
It can be considered massless but it still can exert contact force on the cart in this question?
 
  • #10
songoku said:
It can be considered massless but it still can exert contact force on the cart in this question?
Yes.
 
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  • #11
haruspex said:
Yes.
Why? Because I think if an object is considered massless it won't exert contact force

Thanks
 
  • #12
songoku said:
Why? Because I think if an object is considered massless it won't exert contact force

Thanks
You can't conceive of a strut that is strong but so light that its mass is irrelevant?

But in the present case, for the purposes of linear acceleration its mass can be included in the mass of the block. The only reason for wanting to say it is massless is because of its rotational acceleration. So we could instead take it to be very small, so that its rotational inertia is negligible.
 
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  • #13
The problem would be the same if you replaced the pulley with a protrusion of the cart such that the rope slides without friction over the protrusion.

1579276733792.png
 
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  • #14
TSny said:
The problem would be the same if you replaced the pulley with a protrusion of the cart such that the rope slides without friction over the protrusion.

View attachment 255724
Except for the moment of inertia of the pulley leading to a difference in tensions.
 
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  • #15
haruspex said:
Except for the moment of inertia of the pulley leading to a difference in tensions.
But the pulley was stated to be "ideal". I take that to mean no inertia.
 
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  • #16
TSny said:
But the pulley was stated to be "ideal". I take that to mean no inertia.
If it's ideal it has no mass, right? I took your post to be explaining how it doesn't matter if it has mass, so not ideal.
 
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  • #17
haruspex said:
If it's ideal it has no mass, right? I took your post to be explaining how it doesn't matter if it has mass, so not ideal.
OK. I didn't mean to imply that you could replace the pulley by a slippery surface if the pulley has mass.
 
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  • #18
TSny said:
OK. I didn't mean to imply that you could replace the pulley by a slippery surface if the pulley has mass.
Ah, ok.
 
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  • #19
TSny said:
OK. I didn't mean to imply that you could replace the pulley by a slippery surface if the pulley has mass.
Well, if the string is not accelerating tangentially across the pulley (and it isn't in this case) then the moment of inertia of the pulley will not matter.
 
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  • #20
jbriggs444 said:
Well, if the string is not accelerating tangentially across the pulley (and it isn't in this case) then the moment of inertia of the pulley will not matter.
Yes, I forgot it is not accelerating.
 
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  • #21
I get it. Thank you very much for all the help TSny, haruspex, jbriggs444
 

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