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Q: What should be the mass of block A so that the system stays at rest ?

enter image description here Conditions: i) No normal force is present at point of 2kg mass. It doesn’t mean N=0. ii) No friction , mass of string & pulley.

What I have done till now:

  1. FBD of blocks.

enter image description here

1.5) Since system is at rest , acc=0. Therefore , no tension force present too.

  1. We know: (mA)(g) = (2gsin30)+(2g)+….

I am having difficulty in thinking how or where to write 2gcos30.

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  • $\begingroup$ Could you please edit your question with a clearer sketch of your situation? I honestly can't understand what's going on. Are those blocks on a flat surface? Are they connected to each other? What are the boundary conditions? Why is the weight vector for the middle block broken tilted by 30 degrees? The question is unfortunately unanswerable as it stands, so I'm closing it for now. Once you've edited it to clarify what's going on, it'll be reopened. $\endgroup$
    – Wasabi
    Mar 30, 2022 at 16:24
  • $\begingroup$ @Wasabi I have posted the diagram. Extremely Sorry , I just forgot about the diagram. I’m sure it will be clearer now. $\endgroup$
    – S.M.T
    Mar 30, 2022 at 16:42

3 Answers 3

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i) No normal force is present at point of 2kg mass.

This is a bit weird. That would mean the fact the ramp is there is irrelevant and the pulleys could just be on posts and nothing needs to actually be underneath the center block with it just suspended in the air. Are you sure about this? Is this a condition you made up or did the problem give it to you?

Because if the center block is really supposed to be experiencing no normal force, then why do you have normal force at all in your free-body diagrams? Without a normal force the ramp cannot guide the center block down the ramp which means there is no component parallel to the ramp surface due to the weight. The center block would only experience the downward pull of gravity and the ropes which leads to...

Therefore , no tension force present too.

I think you would agree that something hanging in the air from a rope exerts tension on the rope so why do you think there is no tension here (in all cases, with or without the normal force, and with or without the ramp)?

Another problem is you have multiple 2kg masses so you can't just say refer to a 2kg mass and expect us to know which one you are talking about.

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  • $\begingroup$ I drew normal force in the diagram , correct but I wrote on there specifically , that we will be considering it to not be present. $\endgroup$
    – S.M.T
    Mar 31, 2022 at 4:49
  • $\begingroup$ Normal is not equal to 0 but it is not present. $\endgroup$
    – S.M.T
    Mar 31, 2022 at 6:34
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Revised (after @DKNguyen's comment)

The force diagram below contains both cases, which should result in the same conclusion for the mass of block A. (For both cases, the only effect left from the block on the slope is the inertia force, which will cause additional tension, $\Delta T$, on the upper rope.)

However, for case i), without knowing the friction coefficient, the solution may not be correct, because the shear friction caused by sliding may overcome the inertia force. In such case, A = 2 kg.

enter image description here

ADD: Complete Force Diagram of the System

enter image description here

The gravity force ($mg$) of a block on the slope has two potentials - 1) sliding down the slope if $S < F_x$, and 2) rotation about the upper roller support if the bearing capacity (essentially the normal force $N$) is weaker than $F_y$. Note that the contact surface is usually considered infinite rigid, thus rotational instability will not occur.

Important notes:

  • The tension in the rope on the sides of the roller support are equal in magnitude, but in opposite direction.

  • The shear friction ($S = \mu N$), as indicated by the equation, is dependent on the normal force ($N$). The stipulated Condition i) of the problem - "normal force is not present", effectively eliminates both $N$ and $S$. The stated Condition 2) - no friction (I do not understand the terms thereafter), clearly eliminates $S$. As discussed in the paragraph above, $N$ was shown no effect on moving the block on the slope, so for both conditions, it is a matter of figuring out the forces required to maintain the equilibrium of the block (specifically $T_C$, take into account of the force due to the potential to slide, $F_x$).

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  • $\begingroup$ It is a bit unintuitive to think that if there was no ramp then the scenario would be the same as if the pulleys were level with each other. And also that the sag of the rope has no bearing on the actual question asked. $\endgroup$
    – DKNguyen
    Mar 30, 2022 at 21:26
  • $\begingroup$ Response to edit: It was stated that there is no friction in the problem. $\endgroup$
    – DKNguyen
    Mar 30, 2022 at 22:55
  • $\begingroup$ @DKNguyen Friction occurs on the surface (a contact force), sliding/rolling is caused by the inertial force (a body force) due to the block on the inclined surface. On a frictionless surface, the block will slide downhill freely, which causes tension on the upper rope to counter the sliding force. $\endgroup$
    – r13
    Mar 30, 2022 at 23:17
  • $\begingroup$ @r13 If Normal force is present , then answer is A=2+2gsin30 where the normal force gets cancelled out with 2gcos30. But in case if Normal force is not present , can we say that 2gcos30 will cause any effect ? Like in some way , 2gcos30 has to be added to A. $\endgroup$
    – S.M.T
    Mar 31, 2022 at 5:01
  • $\begingroup$ No. "mgcos30" is acting normal to the surface, it must be zero per the condition given. $\endgroup$
    – r13
    Mar 31, 2022 at 5:10
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Note that for convenience, units are ignored in this answer !
Free body diagram of the inclined mass: enter image description here

Assuming zero acceleration and Coulomb friction $F=\mu N$ enter image description here
Friction opposes motion so we can write $$\begin{align}\sum F_x&=T_2-T_1-2g\sin(30^\circ)\pm F=0\\\sum F_y&=N-2g\cos(30^\circ)=0\end{align}$$ where $T_1=2g$ and $T_2=Ag$. This gives $$A=2(1+\sin(30^\circ)\pm \mu\cos(30^\circ))$$ Hence $A$ can take any value in the interval $I=\langle 3-2\mu\cos(30^\circ), 3+2\mu\cos(30^\circ)\rangle$ whereas the system stays at rest. When $\mu=0$ then $A=3$.

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